Solutions to Tao’s An Introduction to Measure Theory

The non-vanishing set is \(\bigcup_{n \ge 1} A_n\) with \(A_n := \{\alpha : x_\alpha > 1/n\}\), and each \(A_n\) is finite.

Write \(S := \sum_{\alpha \in A} x_\alpha < \infty\). By the definition (0.1) of the sum as \(\sup\{\sum_{\alpha \in F} x_\alpha : F \subset A \text{ finite}\}\),

\begin{equation*} \sum_{\alpha \in F} x_\alpha \le S \qquad \text{for every finite } F \subset A. \tag{\(\star\)} \end{equation*}

If \(A_n\) had \(k\) distinct elements, take \(F \subset A_n\) with \(|F| = k\); monotonicity of finite sums and \((\star)\) give \(k/n \le \sum_{\alpha \in F} x_\alpha \le S\), so \(k \le nS\). Hence \(|A_n| \le nS < \infty\).

If \(x_\alpha \ne 0\) then \(x_\alpha > 1/n\) for some \(n\) (any \(n > 1/x_\alpha\), or \(n = 1\) when \(x_\alpha = +\infty\)), so \(\alpha \in A_n\); conversely each \(A_n\) consists of indices with \(x_\alpha > 0\). Thus \(\{\alpha : x_\alpha \ne 0\}\) is a countable union of finite sets, hence at most countable. No cardinality assumption on \(A\) entered. \(\square\)

Write \(S := \sum_{(n,m) \in A \times B} x_{n,m}\), \(y_n := \sum_{m \in B} x_{n,m}\) and \(T := \sum_{n \in A} y_n\); we prove \(S = T\), and the third sum follows by symmetry. The hint is not needed.

All sums are the supremum (0.1) over finite subsets of the index set. Three consequences, used silently below: on a finite index set (0.1) is attained at \(F = A\) and so agrees with the ordinary finite sum (which is therefore freely regroupable and monotone in each summand); \(A^{\prime} \subset A\) forces \(\sum_{A^{\prime}} \le \sum_{A}\), as the competing finite subsets are fewer; and for non-empty \(P_1, \dots, P_k \subset [0,+\infty]\),

\begin{equation*} \sup\{p_1 + \cdots + p_k : p_j \in P_j\} = \sup P_1 + \cdots + \sup P_k \tag{\(\dagger\)} \end{equation*}

(reduce to \(k=2\); \(\le\) is monotonicity of addition, and for \(\ge\) either some \(\sup P_j = +\infty\), making the left side \(+\infty\), or pick \(p_j > \sup P_j - \varepsilon/2\)).

(i) \(S \le T\). For finite \(F \subset A \times B\), put \(G := \{n : (n,m) \in F\}\) and \(F_n := \{m : (n,m) \in F\}\), so \(F\) is the disjoint union of the \(\{n\} \times F_n\), \(n \in G\). Regrouping and then enlarging \(F_n\) to \(B\) and \(G\) to \(A\),

\begin{equation*} \sum_{(n,m) \in F} x_{n,m} = \sum_{n \in G} \sum_{m \in F_n} x_{n,m} \le \sum_{n \in G} y_n \le T . \end{equation*}

Taking the supremum over \(F\) gives \(S \le T\).

(ii) \(T \le S\). It suffices that \(\sum_{n \in G} y_n \le S\) for every finite \(G = \{n_1, \dots, n_k\} \subset A\). Let \(H_1, \dots, H_k \subset B\) be finite and \(H := H_1 \cup \cdots \cup H_k\); since the \(n_j\) are distinct, \(G \times H\) is a finite subset of \(A \times B\) whose sum regroups into rows, so

\begin{equation*} \sum_{j=1}^{k} \sum_{m \in H_j} x_{n_j,m} \le \sum_{j=1}^{k} \sum_{m \in H} x_{n_j,m} = \sum_{(n,m) \in G \times H} x_{n,m} \le S . \end{equation*}

The \(H_j\) vary independently, so \((\dagger)\) applied to \(P_j := \{\sum_{m \in H_j} x_{n_j,m} : H_j \subset B \text{ finite}\}\), non-empty and with \(\sup P_j = y_{n_j}\) by (0.1), gives \(\sum_{j} y_{n_j} \le S\).

For the third expression, set \(\tilde{x}_{m,n} := x_{n,m}\). The bijection \(\phi(m,n) := (n,m)\) and the relabelling formula (0.2) give \(S = \sum_{(m,n) \in B \times A} \tilde{x}_{m,n}\), which by (i)–(ii) with the roles of \(A, B\) swapped equals \(\sum_{m \in B} \sum_{n \in A} x_{n,m}\). \(\square\)

Everything reduces to two facts about boxes: \(B \cap B^{\prime}\) is a box and \(B \backslash B^{\prime}\) is elementary.

(i) Intervals. An interval in the sense of Definition 1.1.1 is exactly a bounded convex subset of \(\mathbb{R}\) (empty set included, as the degenerate \((a,a)\)), and convexity is preserved by intersection, so \(I \cap J\) is an interval. If \(J\) has endpoints \(c \le d\) then \(\mathbb{R} \backslash J = L \cup R\), where \(L\) is \(\{x < c\}\) or \(\{x \le c\}\) according as \(c \in J\) or not and \(R\) is \(\{x > d\}\) or \(\{x \ge d\}\) likewise (points of \((c,d)\) lie in \(J\) since \(J\) is an interval with those endpoints, and the endpoints are assigned by inspection). Both are convex, so

\begin{equation*} I \backslash J = (I \cap L) \cup (I \cap R) \end{equation*}

is a union of at most two intervals.

(ii) Boxes. For \(B = I_1 \times \dots \times I_d\) and \(B^{\prime} = J_1 \times \dots \times J_d\), coordinatewise membership gives \(B \cap B^{\prime} = (I_1 \cap J_1) \times \dots \times (I_d \cap J_d)\), a box by (i). Splitting on the least index \(k\) with \(x_k \notin J_k\),

\begin{equation*} B \backslash B^{\prime} = \bigcup_{k=1}^{d} (I_1 \cap J_1) \times \dots \times (I_{k-1} \cap J_{k-1}) \times (I_k \backslash J_k) \times I_{k+1} \times \dots \times I_d , \end{equation*}

and each \(I_k \backslash J_k\) splits into at most two intervals by (i); so \(B \backslash B^{\prime}\) is a union of at most \(2d\) boxes, hence elementary.

(iii) Boolean operations. Write \(E = B_1 \cup \dots \cup B_n\), \(F = B_1^{\prime} \cup \dots \cup B_m^{\prime}\). Then \(E \cup F\) is again a finite union of boxes. Distributing,

\begin{equation*} E \cap F = \bigcup_{i=1}^{n} \bigcup_{j=1}^{m} (B_i \cap B_j^{\prime}), \qquad E \backslash F = \bigcup_{i=1}^{n} \bigcap_{j=1}^{m} (B_i \backslash B_j^{\prime}), \end{equation*}

the second because \(x \in E \backslash F\) iff \(x\) lies in some \(B_i\) and in no \(B_j^{\prime}\). Each \(B_i \cap B_j^{\prime}\) is a box and each \(B_i \backslash B_j^{\prime}\) is elementary by (ii); the intersection display shows finite intersections of elementary sets are elementary, so each inner intersection is elementary, and the outer union of finitely many finite unions of boxes is elementary. Finally \(E \Delta F = (E \backslash F) \cup (F \backslash E)\) is a union of two elementary sets.

(iv) Translation. \(I + t\) is an interval of the same type, so \(B + x = (I_1 + x_1) \times \dots \times (I_d + x_d)\) is a box, and translation commutes with unions: \(E + x = \bigcup_i (B_i + x)\). \(\square\)

Given two partitions \(E = B_1 \cup \dots \cup B_k = B_1^{\prime} \cup \dots \cup B_{k^{\prime}}^{\prime}\) into disjoint boxes, both are refined by a single grid of product boxes, and each partition’s volumes sum to the grid total.

(i) One dimension. Let \(I_1, \dots, I_n \subset \mathbb{R}\) be intervals, the empty ones discarded, and let \(a_0 < \dots < a_p\) list their distinct endpoints. The \(2p+1\) atoms

\begin{equation*} \{a_0\}, \dots, \{a_p\}, \quad (a_0,a_1), \dots, (a_{p-1},a_p) \end{equation*}

are pairwise disjoint intervals (singletons are the degenerate \([a_s,a_s]\)), and each \(I_r\), say with endpoints \(a_u \le a_v\), is the union of the atoms it contains: a point \(x \in I_r\) is either some \(a_s\), lying in the atom \(\{a_s\} \subset I_r\), or satisfies \(a_s < x < a_{s+1}\) for a unique \(s\) with \(u \le s \le v-1\), and then \((a_s,a_{s+1}) \subset (a_u,a_v) \subset I_r\). The open atoms inside \(I_r\) are exactly those \((a_s,a_{s+1})\), \(u \le s \le v-1\), and the singletons have length \(0\), so the lengths telescope:

\begin{equation*} \sum_{J \text{ atom},\, J \subset I_r} |J| = \sum_{s=u}^{v-1} (a_{s+1} - a_s) = a_v - a_u = |I_r| . \end{equation*}

(ii) The grid. Write \(B_i = I_{i,1} \times \dots \times I_{i,d}\) and \(B_j^{\prime} = I_{j,1}^{\prime} \times \dots \times I_{j,d}^{\prime}\), and for each coordinate \(\ell\) let \(\mathcal{J}_\ell\) be the atoms of the family of all \(k + k^{\prime}\) \(\ell\)-th factors. The grid boxes are the products \(Q = J_1 \times \dots \times J_d\) with \(J_\ell \in \mathcal{J}_\ell\); finitely many, and pairwise disjoint since two distinct ones have disjoint factors in some coordinate. Discard the empty ones. Distributing each \(I_\ell = \bigcup\{J \in \mathcal{J}_\ell : J \subset I_\ell\}\) over the product exhibits any one of our boxes \(B = I_1 \times \dots \times I_d\) as the disjoint union of exactly the grid boxes contained in it, and by (i) with the distributive law,

\begin{equation*} \sum_{Q \subset B} |Q| = \sum_{J_\ell \subset I_\ell} \prod_{\ell=1}^{d} |J_\ell| = \prod_{\ell=1}^{d} \Big( \sum_{J \in \mathcal{J}_\ell,\, J \subset I_\ell} |J| \Big) = \prod_{\ell=1}^{d} |I_\ell| = |B| . \tag{\(\ast\)} \end{equation*}

(iii) Conclusion. Let \(\mathcal{Q}_E\) be the (non-empty) grid boxes contained in \(E\). Each such \(Q\) lies in exactly one \(B_i\): it meets some \(B_i\), which is a union of grid boxes, and distinct grid boxes are disjoint, so \(Q \subset B_i\), with \(i\) unique as the \(B_i\) are disjoint. Conversely every grid box inside some \(B_i\) lies in \(\mathcal{Q}_E\). Hence by \((\ast)\),

\begin{equation*} \sum_{i=1}^{k} |B_i| = \sum_{i=1}^{k} \sum_{Q \subset B_i} |Q| = \sum_{Q \in \mathcal{Q}_E} |Q| , \end{equation*}

and \(\mathcal{Q}_E\) depends only on \(E\) and the grid, so the same computation for the second partition gives \(\sum_j |B_j^{\prime}| = \sum_{Q \in \mathcal{Q}_E} |Q|\). This is Lemma 1.1.2(ii). \(\square\)

The constant is \(c := m^{\prime}([0,1)^d) \ge 0\).

Two consequences of the axioms. Finite additivity at \(E = F = \emptyset\) gives \(m^{\prime}(\emptyset) = 0\) and, by induction, additivity over any finite disjoint union; and for elementary \(E \subset F\) the set \(F \backslash E\) is elementary (Exercise 1.1.1), so \(m^{\prime}(F) = m^{\prime}(E) + m^{\prime}(F \backslash E) \ge m^{\prime}(E)\) by non-negativity, i.e. \(m^{\prime}\) is monotone.

(i) Cubes. Put \(Q_n := [0,\tfrac1n)^d\). The translates \(Q_n + \tfrac{k}{n}\), \(k \in \mathbb{Z}^d\), tile \(\mathbb{R}^d\) disjointly (\(x\) lies in the one with \(k_j = \lfloor n x_j \rfloor\)), and the \(n^d\) of them with \(k \in \{0,\dots,n-1\}^d\) tile \([0,1)^d\). Additivity and translation invariance give \(c = n^d m^{\prime}(Q_n)\), so \(m^{\prime}(Q_n) = c n^{-d} = c\,m(Q_n)\), and hence for any disjoint union \(S\) of \(N\) translates of \(Q_n\),

\begin{equation*} m^{\prime}(S) = Nc/n^d = c\,m(S). \tag{1} \end{equation*}

(ii) Boxes. Let \(B = I_1 \times \dots \times I_d\) with \(I_j\) of endpoints \(a_j \le b_j\) and \(L_j := b_j - a_j\). Let \(A_n\) (resp. \(C_n\)) be the union of the grid cubes contained in (resp. meeting) \(B\); as \(B\) is bounded these are finite disjoint unions of translates of \(Q_n\), and \(A_n \subset B \subset C_n\). The admissible \(k\) range independently over coordinates, so it suffices to count in one coordinate \(j\):

• \([\tfrac{k}{n},\tfrac{k+1}{n}) \subset I_j\) forces \(n a_j \le k \le n b_j - 1\), while \(n a_j < k < n b_j - 1\) suffices; so the count \(\alpha_{j,n}\) obeys \(\max(0, nL_j - 2) \le \alpha_{j,n} \le nL_j\).
• \([\tfrac{k}{n},\tfrac{k+1}{n}) \cap I_j \ne \emptyset\) forces \(n a_j - 1 < k \le n b_j\), so the count \(\gamma_{j,n} \le nL_j + 1\).

Hence

\begin{equation*} \prod_{j=1}^d \max\Big(0, L_j - \tfrac{2}{n}\Big) \le m(A_n) = \frac{1}{n^d}\prod_j \alpha_{j,n}, \qquad m(C_n) = \frac{1}{n^d}\prod_j \gamma_{j,n} \le \prod_{j=1}^d \Big( L_j + \tfrac{1}{n}\Big), \end{equation*}

and both outer bounds tend to \(|B|\). Monotonicity of \(m^{\prime}\) with (1) squeezes

\begin{equation*} c\,m(A_n) = m^{\prime}(A_n) \le m^{\prime}(B) \le m^{\prime}(C_n) = c\,m(C_n), \end{equation*}

so \(m^{\prime}(B) = c|B|\).

(iii) General \(E\). Partition \(E = B_1 \cup \dots \cup B_k\) into disjoint boxes (Lemma 1.1.2(i)); then

\begin{equation*} m^{\prime}(E) = \sum_{i=1}^{k} m^{\prime}(B_i) = c \sum_{i=1}^{k} |B_i| = c\,m(E), \end{equation*}

the last step by Lemma 1.1.2(ii). Normalising \(m^{\prime}([0,1)^d) = 1\) gives \(c = 1\), i.e. \(m^{\prime} \equiv m\). \(\square\)

Partition \(E_1 = B_1 \cup \dots \cup B_n\) and \(E_2 = B_1^{\prime} \cup \dots \cup B_m^{\prime}\) into disjoint boxes (Lemma 1.1.2(i)); the rectangles \(B_i \times B_j^{\prime}\) then partition \(E_1 \times E_2\).

A product of a box in \(\mathbb{R}^{d_1}\) and a box in \(\mathbb{R}^{d_2}\) is a product of \(d_1 + d_2\) intervals, hence a box, with \(|B \times B^{\prime}| = |B|\,|B^{\prime}|\) directly from Definition 1.1.1. Since \((x,y) \in E_1 \times E_2\) iff \(x \in B_i\) and \(y \in B_j^{\prime}\) for some \(i, j\),

\begin{equation*} E_1 \times E_2 = \bigcup_{i=1}^{n}\bigcup_{j=1}^{m} (B_i \times B_j^{\prime}), \end{equation*}

a finite union of boxes, so \(E_1 \times E_2\) is elementary. These boxes are disjoint: for \((i,j) \ne (i^{\prime},j^{\prime})\) the intersection is \((B_i \cap B_{i^{\prime}}) \times (B_j^{\prime} \cap B_{j^{\prime}}^{\prime})\), one of whose factors is empty. Hence by Lemma 1.1.2(ii), applied first to this partition and then to the two given ones,

\begin{equation*} m_{d_1+d_2}(E_1 \times E_2) = \sum_{i,j} |B_i|\,|B_j^{\prime}| = \Big(\sum_{i} |B_i|\Big)\Big(\sum_{j} |B_j^{\prime}|\Big) = m_{d_1}(E_1)\, m_{d_2}(E_2). \qquad \square \end{equation*}

We prove (1) \(\Leftrightarrow\) (2) and (2) \(\Leftrightarrow\) (3). Two standing facts: \(m_{*,(J)}(E) \le m^{*,(J)}(E)\), since \(A \subset E \subset B\) forces \(m(A) \le m(B)\); and \(m^{*,(J)}\) is monotone with \(m^{*,(J)}( C) \le m( C)\) for elementary \(C\) (take \(B = C\) in the infimum). All Boolean combinations below are elementary by Exercise 1.1.1.

(1) \(\Rightarrow\) (2). Pick elementary \(A \subset E\) with \(m(A) \ge m_{*,(J)}(E) - \varepsilon/2\) and elementary \(B \supset E\) with \(m(B) \le m^{*,(J)}(E) + \varepsilon/2\). Then \(A \subset B\), so finite additivity on \(B = A \cup (B \backslash A)\) and the hypothesis \(m_{*,(J)}(E) = m^{*,(J)}(E)\) give

\begin{equation*} m(B \backslash A) = m(B) - m(A) \le \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon . \end{equation*}

(2) \(\Rightarrow\) (1). With \(A \subset E \subset B\) and \(m(B \backslash A) \le \varepsilon\), additivity gives \(m(B) \le m(A) + \varepsilon\), and \(A, B\) compete in the two extrema:

\begin{equation*} m^{*,(J)}(E) \le m(B) \le m(A) + \varepsilon \le m_{*,(J)}(E) + \varepsilon . \end{equation*}

Let \(\varepsilon \to 0\) and combine with \(m_{*,(J)} \le m^{*,(J)}\).

(2) \(\Rightarrow\) (3). With \(A \subset E \subset B\) as in (2), \(A \backslash E = \emptyset\), so \(A \Delta E = E \backslash A \subset B \backslash A\) and monotonicity gives

\begin{equation*} m^{*,(J)}(A \Delta E) \le m^{*,(J)}(B \backslash A) \le m(B \backslash A) \le \varepsilon . \end{equation*}

(3) \(\Rightarrow\) (2). Apply (3) with \(\varepsilon/3\) to get elementary \(A\) with \(m^{*,(J)}(A \Delta E) \le \varepsilon/3\), then elementary \(C \supset A \Delta E\) with \(m( C) \le \varepsilon/3 + \varepsilon/3 \le \varepsilon\). Take

\begin{equation*} A^{\prime} := A \backslash C, \qquad B^{\prime} := A \cup C . \end{equation*}

Then \(A^{\prime} \subset E\): \(x \in A \backslash C\) cannot lie in \(A \backslash E \subset C\). And \(E \subset B^{\prime}\): \(x \in E\) either lies in \(A\) or in \(E \backslash A \subset C\). Finally \(B^{\prime} \backslash A^{\prime} \subset C\), since a point of \(A \cup C\) outside \(A \backslash C\) lies in \(C\) either way; so \(m(B^{\prime} \backslash A^{\prime}) \le m( C) \le \varepsilon\). \(\square\)

Everything runs off the criterion of Exercise 1.1.5: a bounded \(S\) is Jordan measurable iff for each \(\varepsilon > 0\) there are elementary \(A \subset S \subset B\) with \(m(B \backslash A) \le \varepsilon\).

(1) Fix \(\varepsilon > 0\) and take elementary \(A \subset E \subset B\), \(A^{\prime} \subset F \subset B^{\prime}\) with \(m(B \backslash A), m(B^{\prime} \backslash A^{\prime}) \le \varepsilon/2\) (Exercise 1.1.5). All sets below are elementary by Exercise 1.1.1, and in each case the gap sits inside \(D := (B \backslash A) \cup (B^{\prime} \backslash A^{\prime})\), which has \(m(D) \le \varepsilon\) by subadditivity:

• Union: \(A \cup A^{\prime} \subset E \cup F \subset B \cup B^{\prime}\), and a point of \(B \cup B^{\prime}\) outside both \(A, A^{\prime}\) lies in \(B \backslash A\) or \(B^{\prime} \backslash A^{\prime}\).
• Intersection: \(A \cap A^{\prime} \subset E \cap F \subset B \cap B^{\prime}\), and a point of \(B \cap B^{\prime}\) missing \(A\) or \(A^{\prime}\) lies in \(D\) correspondingly.
• Difference: \(A \backslash B^{\prime} \subset E \backslash F \subset B \backslash A^{\prime}\) (using \(F \subset B^{\prime}\) and \(A^{\prime} \subset F\)); a point of \(B \backslash A^{\prime}\) outside \(A \backslash B^{\prime}\) either misses \(A\), so lies in \(B \backslash A\), or lies in \(B^{\prime} \backslash A^{\prime}\).
• Symmetric difference: \(E \Delta F = (E \backslash F) \cup (F \backslash E)\), by the difference and union cases.

(2) \(\emptyset\) competes in the inner supremum, so \(m(E) = m_{*,(J)}(E) \ge 0\).

(3) Let \(E \cap F = \emptyset\), so \(E \cup F\) is Jordan measurable by (1). For elementary \(A \subset E\), \(A^{\prime} \subset F\) the union \(A \cup A^{\prime} \subset E \cup F\) is disjoint, and for elementary \(B \supset E\), \(B^{\prime} \supset F\) we have \(B \cup B^{\prime} \supset E \cup F\); so by additivity and subadditivity of elementary measure,

\begin{equation*} m(A) + m(A^{\prime}) = m(A \cup A^{\prime}) \le m_{*,(J)}(E \cup F), \qquad m^{*,(J)}(E \cup F) \le m(B) + m(B^{\prime}). \end{equation*}

Taking the supremum on the left and the infimum on the right over the independent families pins \(m(E \cup F)\) between \(m(E) + m(F)\) and itself.

(4) Every elementary \(A \subset E\) also lies in \(F\), so the inner supremum for \(F\) is over a larger family.

(5) \(E \cup F = E \cup (F \backslash E)\) is disjoint, so (3) and then (4) on \(F \backslash E \subset F\) give \(m(E \cup F) = m(E) + m(F \backslash E) \le m(E) + m(F)\).

(6) \(A \mapsto A + x\) is a measure-preserving bijection of the elementary sets (Exercise 1.1.1 and translation invariance of \(m\)) respecting inclusions, and \((B + x) \backslash (A + x) = (B \backslash A) + x\). So \(A + x \subset E + x \subset B + x\) with \(m((B+x) \backslash (A+x)) \le \varepsilon\) makes \(E + x\) Jordan measurable, and the inner suprema for \(E\) and \(E + x\) range over the same numbers, giving \(m(E + x) = m(E)\). \(\square\)

Both parts use one construction: a partition of \(B\) into disjoint boxes of small diameter, over each of which \(f\) varies by at most \(\varepsilon\).

Write \(B = [a_1,b_1] \times \dots \times [a_d,b_d]\), compact, so \(|f| \le M\) for some \(M\) and \(f\) is uniformly continuous. Given \(\varepsilon > 0\), pick \(\delta > 0\) with \(|f(x) - f(y)| \le \varepsilon\) whenever \(|x-y| \le \delta\). Splitting each \([a_j,b_j]\) into \(n\) intervals of length \((b_j-a_j)/n\), half-open on the right save the last, and taking products gives disjoint non-empty boxes \(B_1, \dots, B_N\) with union \(B\) and diameters \(\operatorname{diam}(B)/n \le \delta\) for \(n\) large. Choose \(x_i \in B_i\), set \(c_i := f(x_i)\); then

\begin{equation*} c_i - \varepsilon \le f(x) \le c_i + \varepsilon \quad (x \in B_i), \qquad \sum_{i=1}^{N} |B_i| = |B| , \tag{\(\ast\)} \end{equation*}

the second by Lemma 1.1.2(ii).

(1) \(G := \{(x,f(x))\} \subset B \times [-M,M]\) is bounded, and by \((\ast)\) the part of \(G\) over \(B_i\) sits in the box \(B_i \times [c_i - \varepsilon, c_i + \varepsilon]\) of volume \(2\varepsilon|B_i|\). Hence the elementary set \(S := \bigcup_i B_i \times [c_i-\varepsilon, c_i+\varepsilon]\) contains \(G\) and, by subadditivity and \((\ast)\),

\begin{equation*} m^{*,(J)}(G) \le m(S) \le 2\varepsilon \sum_{i=1}^{N} |B_i| = 2\varepsilon |B| \xrightarrow[\varepsilon \to 0]{} 0 . \end{equation*}

Since \(0 \le m_{*,(J)}(G) \le m^{*,(J)}(G)\), the set \(G\) is Jordan measurable with \(m(G) = 0\).

(2) Let \(E := \{(x,t) : x \in B,\ 0 \le t \le f(x)\} \subset B \times [0,M]\). With the same partition put \(I := \{i : c_i \ge \varepsilon\} \subset I^{\prime} := \{i : c_i \ge -\varepsilon\}\) and

\begin{equation*} L := \bigcup_{i \in I} B_i \times [0, c_i - \varepsilon] \ \subset \ U := \bigcup_{i \in I^{\prime}} B_i \times [0, c_i + \varepsilon] , \end{equation*}

elementary since the index restrictions make each right endpoint \(\ge 0\). Then \(L \subset E\), as \(t \le c_i - \varepsilon \le f(x)\) by \((\ast)\); and \(E \subset U\), since the unique \(i\) with \(x \in B_i\) has \(0 \le t \le f(x) \le c_i + \varepsilon\), forcing \(i \in I^{\prime}\). The boxes in each union are disjoint (the \(B_i\) are), so Lemma 1.1.2(ii) and additivity give

\begin{equation*} m(U \backslash L) = m(U) - m(L) = \sum_{i \in I} 2\varepsilon |B_i| + \sum_{i \in I^{\prime} \backslash I} |B_i|(c_i + \varepsilon) \le 2\varepsilon \sum_{i=1}^{N} |B_i| = 2\varepsilon|B| , \end{equation*}

using \(0 \le c_i + \varepsilon < 2\varepsilon\) for \(i \in I^{\prime} \backslash I\). Given \(\eta > 0\), running this with \(\varepsilon := \eta/(2|B|+1)\) gives \(m(U \backslash L) \le \eta\), so \(E\) is Jordan measurable by Exercise 1.1.5. \(\square\)

The triangle \(T = T(A,B,C)\), the convex hull of the three points, is Jordan measurable with \(m(T) = \frac12|\Delta|\), where \(\Delta(A,B,C) := (B-A)\wedge(C-A)\). Cutting at the middle vertex reduces everything to the horizontal-edge case.

Two symmetries, used to normalise. Expanding, \(\Delta = (b_1c_2-b_2c_1) - (a_1c_2-a_2c_1) + (a_1b_2-a_2b_1)\) is a \(3\times3\) determinant in the rows \((1,a_1,a_2)\) etc., hence alternating: \(|\Delta|\) is permutation-invariant, and it depends only on \(B-A, C-A\), hence is translation-invariant (as is \(m(T)\), by Exercise 1.1.6(6)). The reflection \(\rho(x,y) := (x,-y)\) carries boxes to boxes of equal volume, hence preserves Jordan inner and outer measure, and \(\rho(T(A,B,C)) = T(\rho A,\rho B,\rho C)\) with \(|\Delta|\) unchanged.

(i) Collinear \(A,B,C\), i.e. \(\Delta = 0\). Then \(T\) is a closed segment: if vertical or a point it lies in a box \(\{x_0\}\times[c,d]\) of volume \(0\); otherwise it is the graph of an affine function on a closed interval, null by Exercise 1.1.7(1). Either way \(m(T) = 0 = \frac12|\Delta|\).

(ii) A horizontal edge. Normalising as above, take vertices \(P = (p,0)\), \(Q = (q,0)\), \(R = (r,s)\) with \(p < q\) and \(s > 0\). A point \(\alpha P + \beta Q + \gamma R\) has height \(\gamma s =: y\), and its abscissa sweeps exactly \([\ell(y), u(y)]\) as \(\alpha + \beta = 1 - \gamma\) splits, where

\begin{equation*} \ell(y) := (1-\tfrac ys)p + \tfrac ys r, \qquad u(y) := (1-\tfrac ys)q + \tfrac ys r , \end{equation*}

so \(T = \{(x,y) : 0 \le y \le s,\ \ell(y) \le x \le u(y)\}\) and \(u(y)-\ell(y) = (1-\tfrac ys)(q-p)\). Both \(\ell, u\) are affine, of slopes \((r-p)/s\), \((r-q)/s\); so over a \(y\)-interval of length \(s/N\) their variations sum to exactly \(K/N\), \(K := |r-p| + |r-q|\).

Fix \(N\), put \(y_j := js/N\), and let \(M_j, m_j\) (resp. \(M_j^{\prime}, m_j^{\prime}\)) be the max and min of \(u\) (resp. \(\ell\)) on \([y_{j-1},y_j]\), attained at the endpoints. The elementary sets

\begin{equation*} A_N := \bigcup_{j} [M^{\prime}_j, m_j] \times [y_{j-1},y_j], \qquad C_N := \bigcup_{j} [m^{\prime}_j, M_j] \times [y_{j-1},y_j] \end{equation*}

(the first box read as \(\emptyset\) when \(m_j < M^{\prime}_j\)) satisfy \(A_N \subset T \subset C_N\), since \(\ell(y) \le M^{\prime}_j \le x \le m_j \le u(y)\) on the \(j\)-th slab, and conversely \(m^{\prime}_j \le \ell(y) \le x \le u(y) \le M_j\). Consecutive slabs share only a horizontal segment, a box of volume \(0\), so

\begin{equation*} m(A_N) = \frac sN \sum_{j=1}^{N} (m_j - M^{\prime}_j)_+, \qquad m(C_N) = \frac sN \sum_{j=1}^{N} (M_j - m^{\prime}_j). \end{equation*}

Slab by slab the two terms differ by at most \((M_j - m_j) + (M^{\prime}_j - m^{\prime}_j) = K/N\), for \(m_j \ge M^{\prime}_j\) this is the identity, and otherwise \(M_j - m^{\prime}_j < K/N\) directly. Hence

\begin{equation*} 0 \le m^{*,(J)}(T) - m_{*,(J)}(T) \le m(C_N) - m(A_N) \le \frac{sK}{N} \xrightarrow[N\to\infty]{} 0, \end{equation*}

so \(T\) is Jordan measurable, with \(m(A_N) \le m(T) \le m(C_N)\). Since \(M_j - m^{\prime}_j\) lies between \(u(y_j) - \ell(y_j)\) and that plus \(K/N\), and \(u(y_j)-\ell(y_j) = (1-\tfrac jN)(q-p)\),

\begin{equation*} m(C_N) = \frac{s(q-p)}{N}\sum_{j=1}^{N}\Big(1 - \frac jN\Big) + O\Big(\frac{sK}{N}\Big) = \frac{s(q-p)}{N}\cdot\frac{N-1}{2} + O\Big(\frac{sK}{N}\Big) \to \frac{s(q-p)}{2}. \end{equation*}

Therefore \(m(T) = \frac12 s(q-p) = \frac12|\Delta(P,Q,R)|\), as \(\Delta(P,Q,R) = (q-p)s\).

(iii) General non-collinear \(A,B,C\). Relabel as \(P_1,P_2,P_3\) with heights \(y_1 \le y_2 \le y_3\); \(y_1 < y_3\), and if \(y_1 = y_2\) or \(y_2 = y_3\) case (ii) applies, so assume \(y_1 < y_2 < y_3\). Put \(\lambda := \frac{y_2-y_1}{y_3-y_1} \in (0,1)\) and \(Q := (1-\lambda)P_1 + \lambda P_3\), at height \(y_2\) and distinct from \(P_2\), so \(P_2Q\) is horizontal of length \(h > 0\). Then \(T_1 := T(P_1,P_2,Q)\) and \(T_2 := T(P_2,P_3,Q)\) fall under (ii), with \(m(T_1) = \frac12 h(y_2-y_1)\) and \(m(T_2) = \frac12 h(y_3-y_2)\).

Now \(T_1 = T \cap \{y \le y_2\}\): convexity of \(T\) and of the half-plane gives \(\subset\), and for \(z = \alpha P_1 + \beta P_2 + \gamma P_3\) with \(y(z) \le y_2\), substituting \(P_3 = \lambda^{-1}(Q - (1-\lambda)P_1)\) gives

\begin{equation*} z = \Big(\alpha - \tfrac{\gamma(1-\lambda)}{\lambda}\Big)P_1 + \beta P_2 + \tfrac{\gamma}{\lambda} Q , \end{equation*}

whose coefficients sum to \(1\) and are non-negative: \(y(z) \le y_2\) rearranges to \(\gamma(y_3-y_2) \le \alpha(y_2-y_1)\), which is exactly \(\gamma(1-\lambda)/\lambda \le \alpha\). Symmetrically \(T_2 = T \cap \{y \ge y_2\}\). So \(T = T_1 \cup T_2\) is Jordan measurable by Exercise 1.1.6(1), and \(T_1 \cap T_2 \subset \{y = y_2\} \cap T\) is a segment, null by (i); hence by Exercise 1.1.6(3),(4),

\begin{equation*} m(T) = m(T_1) + m(T_2 \backslash T_1) = m(T_1) + m(T_2) = \tfrac12 h(y_3 - y_1). \end{equation*}

Finally \(Q - P_1 = \lambda(P_3-P_1)\) and bilinearity give \(\Delta(P_1,P_2,Q) = \lambda \Delta(P_1,P_2,P_3)\), while (ii) gives \(|\Delta(P_1,P_2,Q)| = 2m(T_1) = h(y_2-y_1)\). Hence \(|\Delta(P_1,P_2,P_3)| = \lambda^{-1}h(y_2-y_1) = h(y_3-y_1) = 2m(T)\). \(\square\)

We prove the stronger statement that every compact convex \(C \subset \mathbb{R}^d\) is Jordan measurable; a compact convex polytope is such a set. Two facts, used again in Exercises 1.1.10 and 1.1.11.

Dilation. For \(\theta > 0\) a box dilates to the box \(\theta I_1 \times \dots \times \theta I_d\) of volume \(\theta^d|B|\), and a disjoint box partition (Lemma 1.1.2(i)) dilates to one; so \(\theta E\) is elementary with \(m(\theta E) = \theta^d m(E)\).

Hyperplanes are null. Let \(S\) be a bounded subset of \(H = \{x \cdot v = c\}\), \(v \ne 0\). Permuting coordinates (which preserves boxes, volumes, and hence Jordan measure), assume \(v_d \ne 0\); then \(H\) is the graph of the affine \(f(x) := (c - \sum_{i<d} x_i v_i)/v_d\), so \(S\) sits inside the graph of \(f\) over a large closed box, a Jordan null set by Exercise 1.1.7(1). Monotonicity gives \(m^{*,(J)}(S) = 0\). (For \(d = 1\), \(H\) is a point.)

Now let \(C \ne \emptyset\) be compact and convex.

(i) \(\operatorname{int}( C) = \emptyset\). Then \(C\) has no \(d+1\) affinely independent points: convexity would put the simplex they span inside \(C\), and the linear isomorphism carrying the standard simplex there is a homeomorphism, so \(C\) would have interior. Hence \(C\) lies in an affine hyperplane, and is null by the above.

(ii) \(\operatorname{int}( C) \ne \emptyset\). Translate (Exercise 1.1.6(6)) so that \(B(0,r) \subset C\). For \(\theta \in (0,1)\), \(\theta C \subset \operatorname{int}( C)\): if \(x \in C\) and \(|w - \theta x| < (1-\theta)r\) then \(w = \theta x + (1-\theta)z\) with \(|z| < r\), so \(w \in C\) by convexity. Now \(\theta C\) is compact and covered by the open cubes \(\operatorname{int}(Q_y)\), \(y \in \theta C\), where \(Q_y\) is a small closed cube about \(y\) inside \(\operatorname{int}( C)\); finitely many suffice, and their union \(A\) is elementary with

\begin{equation*} \theta C \subset A \subset C, \qquad \text{hence} \qquad C \subset \theta^{-1}A . \end{equation*}

So \(A\) competes in the inner supremum and \(\theta^{-1}A\) in the outer infimum, giving

\begin{equation*} m^{*,(J)}( C) \le m(\theta^{-1}A) = \theta^{-d}m(A) \le \theta^{-d} m_{*,(J)}( C) . \end{equation*}

As \(C\) is bounded these quantities are finite, so letting \(\theta \to 1^{-}\) gives \(m^{*,(J)}( C) \le m_{*,(J)}( C)\); the reverse always holds. \(\square\)

Take \(c_d := m(\overline{B(0,1)})\), which exists because \(\overline{B(0,1)}\) is compact and convex, hence Jordan measurable by Exercise 1.1.9. Assume \(r > 0\) (for \(r = 0\) both balls are a point or empty, of measure \(0\)).

(1) By the dilation fact from Exercise 1.1.9, \(E \mapsto \theta E\) is an inclusion-preserving bijection of the elementary sets scaling \(m\) by \(\theta^d\), so for bounded \(E\),

\begin{equation*} m_{*,(J)}(\theta E) = \theta^d m_{*,(J)}(E), \qquad m^{*,(J)}(\theta E) = \theta^d m^{*,(J)}(E). \end{equation*}

Applying this to \(\overline{B(0,r)} = r\overline{B(0,1)}\) and then translating (Exercise 1.1.6(6)) gives that \(\overline{B(x,r)}\) is Jordan measurable with measure \(c_d r^d\).

For the open ball, \(\overline{B(0,\theta r)} \subset B(0,r) \subset \overline{B(0,r)}\) for \(\theta \in (0,1)\), so by monotonicity of the inner and outer measures

\begin{equation*} c_d(\theta r)^d \le m_{*,(J)}(B(0,r)) \le m^{*,(J)}(B(0,r)) \le c_d r^d . \end{equation*}

Let \(\theta \to 1^{-}\): all four agree, so \(B(x,r)\) is Jordan measurable with measure \(c_d r^d\) as well.

(2) \([-1/\sqrt d, 1/\sqrt d]^d \subset \overline{B(0,1)} \subset [-1,1]^d\), since \(|y_i| \le 1/\sqrt d\) for all \(i\) forces \(|y|^2 \le 1\), and \(|y| \le 1\) forces \(|y_i| \le 1\). The two cubes are elementary, of measures \((2/\sqrt d)^d\) and \(2^d\), so they compete in the inner supremum and outer infimum respectively:

\begin{equation*} \Big(\frac{2}{\sqrt d}\Big)^d \le c_d \le 2^d . \qquad \square \end{equation*}

The constant is \(D = D(L) := m(L([0,1)^d))\), and it equals \(|\det L|\). Throughout we use the two facts from Exercise 1.1.9: compact convex sets are Jordan measurable, and bounded subsets of hyperplanes are null.

(1) If \(L\) is singular then \(L(\mathbb{R}^d)\) lies in a hyperplane, so \(m(L(E)) = 0\) for all elementary \(E\) and \(D := 0\) works. Let \(L\) be invertible. For a box \(B\), the image \(L(\overline{B})\) is compact and convex, hence Jordan measurable; and

\begin{equation*} L(\overline{B}) \backslash L(B) \subset L(\overline{B} \backslash B^{\circ}), \end{equation*}

which is a bounded subset of the union of the images of the \(2d\) face hyperplanes of \(\overline{B}\), hence null by subadditivity. So \(L(B) = L(\overline{B}) \backslash N\) with \(N\) null is Jordan measurable (Exercise 1.1.6(1)), and \(L(E)\) is a finite union of such sets for elementary \(E\). Now \(m^{\prime}(E) := m(L(E))\) satisfies the three hypotheses of Exercise 1.1.3: non-negativity is Exercise 1.1.6(2); injectivity of \(L\) makes \(L(E), L(F)\) disjoint when \(E, F\) are, giving finite additivity via Exercise 1.1.6(3); and \(L(E+x) = L(E) + L(x)\) gives translation invariance via Exercise 1.1.6(6). Exercise 1.1.3 supplies \(D \ge 0\) with \(m(L(E)) = D\,m(E)\).

(2) For singular \(L\) both sides vanish, as \(L(E)\) is again a bounded subset of a hyperplane. For invertible \(L\) and Jordan measurable \(E\), take elementary \(A \subset E \subset B\) with \(m(B \backslash A) \le \varepsilon\) (Exercise 1.1.5). Bijectivity gives \(L(A) \subset L(E) \subset L(B)\) and \(L(B) \backslash L(A) = L(B \backslash A)\), so by part (1) and Exercise 1.1.6(1),(3),

\begin{equation*} m(L(B)) - m(L(A)) = m(L(B \backslash A)) = D\,m(B \backslash A) \le D\varepsilon . \end{equation*}

Monotonicity then squeezes \(m^{*,(J)}(L(E)) - m_{*,(J)}(L(E)) \le D\varepsilon\), so \(L(E)\) is Jordan measurable; and since \(D\,m(A) \le m(L(E)) \le D\,m(B)\) while \(m(A) \le m(E) \le m(B)\) with \(m(B) - m(A) \le \varepsilon\), we get \(|m(L(E)) - D\,m(E)| \le D\varepsilon\) for all \(\varepsilon\).

(3) \(D\) is multiplicative: by (2), \(D(L_1L_2) = m(L_1(L_2([0,1)^d))) = D(L_1)D(L_2)\), and \(D(\mathrm{id}) = 1\). Singular \(L\) has \(D = 0 = |\det L|\); otherwise \(L\) is a product of elementary matrices (Gaussian elimination), and since \(|\det|\) is multiplicative too it suffices to check the three types.

• Scaling \(S\) (\(i\)-th coordinate times \(c \ne 0\)): \(S([0,1)^d)\) is a box with \(i\)-th side \([0,c)\) or \((c,0]\), so \(D(S) = |c| = |\det S|\).
• Transposition \(P\): \(P([0,1)^d) = [0,1)^d\), so \(D(P) = 1 = |\det P|\).
• Shear \(T = T_{ij}( c) : x \mapsto x + cx_j e_i\) (\(i \ne j\)), with \(\det T = 1\). Here \(T_{ij}( c)T_{ij}(c^{\prime}) = T_{ij}(c+c^{\prime})\), so \(T^n = T_{ij}(nc)\); and \(T_{ij}(c^{\prime})([0,1)^d)\) sits in a box whose \(i\)-th side has length \(1 + |c^{\prime}|\) and the rest \(1\), so monotonicity gives \(D(T)^n = D(T_{ij}(nc)) \le 1 + n|c|\). Geometric growth would outrun this, so \(D(T) \le 1\); the same for \(T^{-1} = T_{ij}(-c)\), and \(D(T)D(T^{-1}) = 1\) forces \(D(T) = 1\). \(\square\)

Monotonicity of Jordan outer measure does it. Let \(S \subset N\) with \(N\) Jordan null; \(N\) is bounded, hence so is \(S\). Every elementary set containing \(N\) contains \(S\), so

\begin{equation*} 0 \le m_{*,(J)}(S) \le m^{*,(J)}(S) \le m^{*,(J)}(N) = 0, \end{equation*}

the first inequality because \(\emptyset \subset S\) competes in the inner supremum, the second because \(A \subset S \subset B\) forces \(m(A) \le m(B)\). Hence \(S\) is Jordan measurable with \(m(S) = 0\). \(\square\)

Squeeze the count between two elementary sets, on each of which (1.1) is the discretisation argument of Lemma 1.1.2(ii).

For an interval \(I\) with endpoints \(a \le b\), the points of \(\frac1N \mathbb{Z}\) in \(I\) correspond to the integers in an interval of length \((b-a)N\), of which there are between \((b-a)N - 1\) and \((b-a)N + 1\); dividing by \(N\) gives \(|I| = \lim_N \frac1N \#(I \cap \frac1N\mathbb{Z})\). A grid point lies in a box iff each coordinate lies in the corresponding factor, so taking products and then summing over a disjoint box partition (Lemma 1.1.2(i)) gives (1.1) for every elementary \(F\).

Now let \(E\) be Jordan measurable (hence bounded, so every count is finite) and \(\varepsilon > 0\). By Exercise 1.1.5 take elementary \(A \subset E \subset B\) with \(m(B \backslash A) \le \varepsilon\); then \(m(A) \le m(E) \le m(B) \le m(A) + \varepsilon\), so \(m(A) \ge m(E) - \varepsilon\) and \(m(B) \le m(E) + \varepsilon\). Counting grid points through \(A \subset E \subset B\) and letting \(N \to \infty\),

\begin{equation*} m(E) - \varepsilon \le m(A) \le \liminf_{N} \frac{\#(E \cap \frac1N\mathbb{Z}^d)}{N^d} \le \limsup_{N} \frac{\#(E \cap \frac1N\mathbb{Z}^d)}{N^d} \le m(B) \le m(E) + \varepsilon . \end{equation*}

Let \(\varepsilon \to 0\). \(\square\)

The two counts are elementary measures: for bounded \(S\) write \(A_n(S)\) (resp. \(B_n(S)\)) for the union of the dyadic cubes of sidelength \(2^{-n}\) contained in (resp. meeting) \(S\). These cubes are disjoint of volume \(2^{-dn}\) and tile \(\mathbb{R}^d\), and only finitely many meet the bounded \(S\), so \(A_n(S), B_n(S)\) are elementary with

\begin{equation*} m(A_n(S)) = 2^{-dn}\mathcal{E}_*(S,2^{-n}), \qquad m(B_n(S)) = 2^{-dn}\mathcal{E}^*(S,2^{-n}), \end{equation*}

and \(A_n(S) \subset S \subset B_n(S)\). Write \(A_n := A_n(E)\), \(B_n := B_n(E)\), so that

\begin{equation*} m(A_n) \le m_{*,(J)}(E) \le m^{*,(J)}(E) \le m(B_n) . \tag{\(\ast\)} \end{equation*}

If. Suppose \(m(B_n) - m(A_n) \to 0\). Then \((\ast)\) forces \(m^{*,(J)}(E) = m_{*,(J)}(E)\), so \(E\) is Jordan measurable, and \(m(A_n) \le m(E) \le m(B_n)\) makes both \(2^{-dn}\mathcal{E}_*\) and \(2^{-dn}\mathcal{E}^*\) converge to \(m(E)\).

Only if. First, for a box \(Q = I_1 \times \dots \times I_d\) with \(I_j\) of endpoints \(a_j \le b_j\):

\begin{equation*} \prod_{j=1}^d \max\big(0, (b_j-a_j) - 2^{1-n}\big) \le m(A_n(Q)) \le |Q| \le m(B_n(Q)) \le \prod_{j=1}^d \big((b_j-a_j) + 2^{1-n}\big). \end{equation*}

Indeed \([k2^{-n},(k+1)2^{-n}) \subset I_j\) whenever \(a_j2^n < k\) and \(k+1 < b_j2^n\), and the open interval \((a_j2^n, b_j2^n-1)\) holds at least \((b_j-a_j)2^n - 2\) integers; conversely if that dyadic interval meets \(I_j\) at \(x\) then \(k2^{-n} \le x \le b_j\) and \((k+1)2^{-n} > x \ge a_j\), putting \(k\) in \((a_j2^n-1,\ b_j2^n]\), which holds at most \((b_j-a_j)2^n + 2\) integers. Products of these one-dimensional unions bracket \(A_n(Q)\) and \(B_n(Q)\), so Exercise 1.1.4 and monotonicity give the display; both outer products tend to \(|Q|\), so \(m(A_n(Q)), m(B_n(Q)) \to |Q|\).

For elementary \(F = Q_1 \cup \dots \cup Q_k\) (disjoint boxes, Lemma 1.1.2(i)), the \(A_n(Q_i) \subset A_n(F)\) are disjoint and every dyadic cube meeting \(F\) meets some \(Q_i\), so additivity and subadditivity give

\begin{equation*} \sum_{i} m(A_n(Q_i)) \le m(A_n(F)) \le m(F) \le m(B_n(F)) \le \sum_{i} m(B_n(Q_i)), \end{equation*}

whose outer sums both tend to \(m(F)\); hence \(m(A_n(F)), m(B_n(F)) \to m(F)\) for every elementary \(F\).

Now let \(E\) be Jordan measurable, \(\varepsilon > 0\), and take elementary \(A \subset E \subset B\) with \(m(B \backslash A) \le \varepsilon\) (Exercise 1.1.5), so \(m(A) \ge m(E) - \varepsilon\) and \(m(B) \le m(E) + \varepsilon\). Monotonicity of the constructions gives \(m(A_n) \ge m(A_n(A))\) and \(m(B_n) \le m(B_n(B))\), so for \(n\) large,

\begin{equation*} m(E) - 2\varepsilon \le m(A_n) \le m(E) \le m(B_n) \le m(E) + 2\varepsilon , \end{equation*}

whence \(0 \le m(B_n) - m(A_n) \le 4\varepsilon\). As \(\varepsilon\) was arbitrary, the entropy difference tends to \(0\) and both counts converge to \(m(E)\). \(\square\)

Restrict \(m^{\prime}\) to the elementary sets and apply Exercise 1.1.3: since \(\mathcal{E}(\mathbb{R}^d) \subset \mathcal{J}(\mathbb{R}^d)\) with the two measures agreeing there, and \(\mathcal{E}(\mathbb{R}^d)\) is closed under the Boolean operations and translations (Exercise 1.1.1), the restriction inherits the three axioms, so

\begin{equation*} m^{\prime}(A) = c\,m(A) \qquad (A \text{ elementary}), \qquad c := m^{\prime}([0,1)^d). \end{equation*}

Also \(m^{\prime}\) is monotone on \(\mathcal{J}(\mathbb{R}^d)\): for \(A \subset B\) Jordan measurable, \(B \backslash A\) is Jordan measurable (Exercise 1.1.6(1)) and \(m^{\prime}(B) = m^{\prime}(A) + m^{\prime}(B \backslash A) \ge m^{\prime}(A)\).

Now let \(E\) be Jordan measurable and \(\varepsilon > 0\), and take elementary \(A \subset E \subset B\) with \(m(B) - m(A) = m(B \backslash A) \le \varepsilon\) (Exercise 1.1.5). Monotonicity of \(m^{\prime}\) and of \(m\) (Exercise 1.1.6(4)) place both \(m^{\prime}(E)\) and \(c\,m(E)\) in

\begin{equation*} [\,c\,m(A),\ c\,m(B)\,], \qquad \text{an interval of length } c(m(B)-m(A)) \le c\varepsilon , \end{equation*}

so \(|m^{\prime}(E) - c\,m(E)| \le c\varepsilon\) for all \(\varepsilon > 0\). Normalising \(m^{\prime}([0,1)^d) = 1\) gives \(c = 1\). \(\square\)

Sandwich the product by products of the approximants from Exercise 1.1.5.

Let \(0 < \varepsilon \le 1\) and take elementary \(A_i \subset E_i \subset B_i\) with \(m(B_i) - m(A_i) = m(B_i \backslash A_i) \le \varepsilon\) (\(i = 1,2\)); monotonicity gives \(m(A_i) \le m_{d_i}(E_i) \le m(B_i) \le m_{d_i}(E_i) + 1\). By Exercise 1.1.4 the sets \(A_1 \times A_2 \subset E_1 \times E_2 \subset B_1 \times B_2\) are elementary with measures \(m(A_1)m(A_2)\) and \(m(B_1)m(B_2)\), so

\begin{equation*} \begin{aligned} m\big((B_1 \times B_2) \backslash (A_1 \times A_2)\big) &= m(B_1)\big(m(B_2) - m(A_2)\big) + \big(m(B_1) - m(A_1)\big)m(A_2) \\ &\le \big(m_{d_1}(E_1) + m_{d_2}(E_2) + 1\big)\varepsilon =: C\varepsilon . \end{aligned} \end{equation*}

As \(C \ge 1\) is independent of \(\varepsilon\), running this with \(\varepsilon := \min(1, \eta/C)\) meets criterion (2) of Exercise 1.1.5, so \(E_1 \times E_2\) is Jordan measurable.

Both \(m_{d_1+d_2}(E_1 \times E_2)\) and \(m_{d_1}(E_1)m_{d_2}(E_2)\) lie in \([m(A_1)m(A_2),\, m(B_1)m(B_2)]\), an interval of length at most \(C\varepsilon\) by the display; letting \(\varepsilon \to 0\) equates them. \(\square\)

Both dissections are measure-additive, and each rigid motion preserves measure, so \(m(P) = \sum_i m(P_i) = \sum_i m(Q_i) = m(Q)\). We read the hypothesis in the classical dissection sense, \(P = P_1 \cup \dots \cup P_n\) with \(P_i \cap P_j \subset \partial P_i\) and \(Q = Q_1 \cup \dots \cup Q_n\) likewise with \(Q_i := \rho_i(P_i) + w_i\); a literal disjoint partition of a connected \(P\) into closed pieces would force \(n = 1\), and “cover” must mean exact cover, since two halves of \([0,1]^2\) left in place cover \([0,\frac12]^2\) with no change of pieces. Polytopes are Jordan measurable by Exercise 1.1.9, and each \(Q_i\) is again one, being the image of an intersection of half-spaces under an invertible affine map.

(i) \(\partial R\) is null for every polytope \(R = \bigcap_{k \le K}\{x \cdot v_k \le c_k\}\). Discard the \(v_k = 0\) constraints (each is trivial or makes \(R\) empty). If \(x \cdot v_k < c_k\) for all \(k\) then the strict inequalities persist nearby, so \(x \in R^{\circ}\); since \(R\) is closed this gives

\begin{equation*} \partial R \subset \bigcup_{k=1}^{K} \big(R \cap \{x \cdot v_k = c_k\}\big), \end{equation*}

a finite union of bounded subsets of hyperplanes, each null by Exercise 1.1.9; so \(\partial R\) is null by Exercises 1.1.6 and 1.1.12.

(ii) Additivity across a dissection. Let \(R = R_1 \cup \dots \cup R_n\) with \(R_i \cap R_j \subset \partial R_i\) for \(i \ne j\), and put \(R_i^{\prime} := R_i \backslash (R_1 \cup \dots \cup R_{i-1})\). These are Jordan measurable (Exercise 1.1.6(1)), pairwise disjoint with union \(R\), so \(m( R) = \sum_i m(R_i^{\prime})\) by Exercise 1.1.6(3). And

\begin{equation*} R_i \backslash R_i^{\prime} = \bigcup_{j<i}(R_i \cap R_j) \subset \partial R_i \end{equation*}

is null by (i) and Exercise 1.1.12, so \(m(R_i^{\prime}) = m(R_i)\) and \(m( R) = \sum_i m(R_i)\).

(iii) Rigid motions. By Exercise 1.1.6(6) and Exercise 1.1.11(2),(3), \(m(Q_i) = m(\rho_i(P_i)) = |\det \rho_i|\,m(P_i) = m(P_i)\), a rotation being orthogonal. Applying (ii) to \(P\) and to \(Q\) and summing gives \(m(P) = m(Q)\). \(\square\)

Everything follows from one observation: an elementary set may be closed up or opened up without changing its measure. Writing \(B = B_1 \cup \dots \cup B_k\) as disjoint boxes (Lemma 1.1.2(i)), the closures \(\overline{B_j}\) and interiors \(B_j^{\circ}\) are boxes with \(|\overline{B_j}| = |B_j| = |B_j^{\circ}|\), so \(\overline{B}\) and \(B^{\flat} := \bigcup_j B_j^{\circ}\) are elementary (the latter open, \(\subset B\)) with

\begin{equation*} m(\overline{B}) = m(B) = m(B^{\flat}) . \end{equation*}

We also use freely that \(m^{*,(J)}\) is monotone and finitely subadditive (cover \(S\), \(T\) by near-optimal elementary sets and unite), that \(m_{*,(J)}\) is monotone, and that \(m_{*,(J)} \le m^{*,(J)}\).

(1) \(m^{*,(J)}(E) \le m^{*,(J)}(\overline{E})\) by monotonicity. Conversely, an elementary \(B \supset E\) has \(\overline{B} \supset \overline{E}\) elementary with \(m(\overline{B}) = m(B)\), so \(m^{*,(J)}(\overline{E}) \le m(B)\); take the infimum over \(B\).

(2) \(m_{*,(J)}(E^{\circ}) \le m_{*,(J)}(E)\) by monotonicity. Conversely, for elementary \(A \subset E\) the set \(A^{\flat}\) is open and inside \(E\), hence inside \(E^{\circ}\), so \(m_{*,(J)}(E^{\circ}) \ge m(A^{\flat}) = m(A)\); take the supremum over \(A\).

(3) ( \(\Rightarrow\) ) Take elementary \(A \subset E \subset B\) with \(m(B \backslash A) \le \varepsilon\) (Exercise 1.1.5). Then \(A^{\flat} \subset E^{\circ}\) and \(\overline{E} \subset \overline{B}\), so

\begin{equation*} \partial E = \overline{E} \backslash E^{\circ} \subset \overline{B} \backslash A^{\flat}, \end{equation*}

elementary with \(m(\overline{B} \backslash A^{\flat}) = m(B) - m(A) \le \varepsilon\). Hence \(m^{*,(J)}(\partial E) = 0\).

( \(\Leftarrow\) ) Let \(\varepsilon > 0\) and take elementary \(C \supset \partial E\) with \(m( C) \le \varepsilon\). Fattening each box of \(C\) by \(\delta\) in every direction (its volume \(\prod_i(b_i - a_i + 2\delta) \to |B_j|\) as \(\delta \to 0\)) gives an open elementary \(U \supset C\) with \(m(U) \le 2\varepsilon\). Now \(K := \overline{E} \backslash U\) is compact and, since \(\overline{E} = E^{\circ} \cup \partial E\) with \(\partial E \subset U\), lies in the open set \(E^{\circ}\); covering \(K\) by finitely many open boxes inside \(E^{\circ}\) yields an elementary \(A\) with \(K \subset A \subset E\). Hence, with \(\overline{E} \subset K \cup U\), subadditivity and (1) give

\begin{equation*} m^{*,(J)}(E) = m^{*,(J)}(\overline{E}) \le m(A) + m(U) \le m_{*,(J)}(E) + 2\varepsilon . \end{equation*}

Let \(\varepsilon \to 0\).

(4) Let \(S := [0,1]^2 \cap \mathbb{Q}^2\) and \(T := [0,1]^2 \backslash \mathbb{Q}^2\). Density of \(\mathbb{Q}\) gives \(\overline{S} = [0,1]^2\); density of the irrationals gives \(\overline{T} = [0,1]^2\) (perturb the first coordinate to an irrational). No non-empty open set lies in either, since every one contains both a rational point and a point with an irrational coordinate, so \(S^{\circ} = T^{\circ} = \emptyset\). Since \([0,1]^2\) is a box of volume \(1\), parts (1) and (2) give

\begin{equation*} m^{*,(J)}(S) = m^{*,(J)}(T) = 1, \qquad m_{*,(J)}(S) = m_{*,(J)}(T) = 0, \end{equation*}

so neither set is Jordan measurable. \(\square\)

\(\le\) is subadditivity; \(\ge\) is where elementarity of \(F\) enters, via the split of an arbitrary elementary \(B \supset E\).

\(\le\): given \(\varepsilon > 0\), take elementary \(B_1 \supset E \cap F\) and \(B_2 \supset E \backslash F\) within \(\varepsilon\) of the respective outer measures. Then \(B_1 \cup B_2\) is elementary (Exercise 1.1.1) and contains \(E\), so

\begin{equation*} m^{*,(J)}(E) \le m(B_1) + m(B_2) \le m^{*,(J)}(E \cap F) + m^{*,(J)}(E \backslash F) + 2\varepsilon . \end{equation*}

\(\ge\): for elementary \(B \supset E\), the sets \(B \cap F\) and \(B \backslash F\) are elementary (Exercise 1.1.1), disjoint with union \(B\), and contain \(E \cap F\) and \(E \backslash F\) respectively. So they compete in the two infima, and additivity of elementary measure gives

\begin{equation*} m^{*,(J)}(E \cap F) + m^{*,(J)}(E \backslash F) \le m(B \cap F) + m(B \backslash F) = m(B). \end{equation*}

Take the infimum over \(B\). \(\square\)

Pass to the common refinement \(I_i \cap I_j^{\prime}\) of two such partitions \((I_i, c_i)\) and \((I_j^{\prime}, c_j^{\prime})\).

Each \(I_i \cap I_j^{\prime}\) is an interval, since intervals are exactly the bounded convex subsets of \(\mathbb{R}\) and convexity survives intersection. For fixed \(i\), these intervals are disjoint with union \(I_i\), so Lemma 1.1.2(ii) gives \(|I_i| = \sum_j |I_i \cap I_j^{\prime}|\), and symmetrically \(|I_j^{\prime}| = \sum_i |I_i \cap I_j^{\prime}|\). Hence

\begin{equation*} \sum_{i} c_i |I_i| = \sum_{i,j} c_i |I_i \cap I_j^{\prime}|, \qquad \sum_{j} c_j^{\prime} |I_j^{\prime}| = \sum_{i,j} c_j^{\prime}|I_i \cap I_j^{\prime}| , \end{equation*}

and the two right-hand sides agree term by term: if \(I_i \cap I_j^{\prime} = \emptyset\) both terms vanish, and otherwise any \(x\) in it gives \(c_i = f(x) = c_j^{\prime}\). \(\square\)

Work on a common partition. As in Exercise 1.1.20, the intersections of a partition adapted to \(f\) with one adapted to \(g\) form a single partition \(J_1, \dots, J_m\) of \([a,b]\) into intervals, with \(f \equiv \alpha_k\) and \(g \equiv \beta_k\) on \(J_k\); by Exercise 1.1.20 the integrals may be computed from it.

(1) \(cf \equiv c\alpha_k\) and \(f + g \equiv \alpha_k + \beta_k\) on \(J_k\), so both are piecewise constant and

\begin{equation*} \mathrm{p.c.}\!\int_a^b cf = \sum_k c\alpha_k|J_k| = c\,\mathrm{p.c.}\!\int_a^b f, \qquad \mathrm{p.c.}\!\int_a^b (f+g) = \sum_k (\alpha_k + \beta_k)|J_k|, \end{equation*}

the latter splitting into the two integrals.

(2) For each non-empty \(J_k\), any \(x \in J_k\) gives \(\alpha_k = f(x) \le g(x) = \beta_k\), so \(\alpha_k|J_k| \le \beta_k|J_k|\) (and both vanish if \(J_k = \emptyset\)); sum over \(k\).

(3) Partition \(E = J_1 \cup \dots \cup J_r\) and \([a,b] \backslash E = J_{r+1} \cup \dots \cup J_s\) into disjoint intervals (Lemma 1.1.2(i), the complement being elementary by Exercise 1.1.1). Together these partition \([a,b]\), with \(1_E \equiv 1\) on the first \(r\) and \(\equiv 0\) on the rest, so \(1_E\) is piecewise constant and, by Exercise 1.1.20 and Lemma 1.1.2(ii),

\begin{equation*} \mathrm{p.c.}\!\int_a^b 1_E = \sum_{k=1}^{r}|J_k| = m(E). \qquad \square \end{equation*}

The bridge is that Riemann sums of a piecewise constant function converge to its piecewise constant integral. (For \(a = b\) both integrals are \(0\) by convention; assume \(a < b\).)

Lemma. If \(g\) is piecewise constant with \(|g| \le M\) and \(g \equiv c_j\) on the intervals \(I_1,\dots,I_k\) partitioning \([a,b]\), then every tagged partition \(\mathcal{P}\) of mesh \(\delta := \Delta(\mathcal{P})\) satisfies

\begin{equation*} \Big|\mathcal{R}(g,\mathcal{P}) - \mathrm{p.c.}\!\int_a^b g\Big| \le 2Mk\delta . \end{equation*}

Each tag lies in exactly one \(I_j\), so grouping gives \(\mathcal{R}(g,\mathcal{P}) = \sum_j c_j S_j\) with \(S_j := \sum_{x_i^{*} \in I_j} \delta x_i\), against \(\mathrm{p.c.}\int_a^b g = \sum_j c_j|I_j|\) (Exercise 1.1.20); so it suffices that \(|S_j - |I_j|| \le 2\delta\). Upper: a subinterval with its tag in \(I_j\) meets \(I_j\) and has length \(\le \delta\), so it lies in the \(\delta\)-enlargement of \(I_j\), of length \(|I_j| + 2\delta\); the subintervals have disjoint interiors. Lower: with \(I_j^{-\delta} := \{x \in I_j : \operatorname{dist}(x, \mathbb{R}\backslash I_j) > \delta\}\), of length \(\ge |I_j| - 2\delta\), any subinterval meeting it lies in \(I_j\), hence has its tag in \(I_j\) and contributes to \(S_j\); such subintervals cover \(I_j^{-\delta}\). \(\square\)

Darboux \(\Rightarrow\) Riemann. Let \(I\) be the common Darboux value and \(\varepsilon > 0\). Choose piecewise constant \(g \le f \le h\) with \(\mathrm{p.c.}\int g > I - \varepsilon\) and \(\mathrm{p.c.}\int h < I + \varepsilon\); both are bounded, so the Lemma gives \(\delta > 0\) with both Riemann sums within \(\varepsilon\) of their integrals once \(\Delta(\mathcal{P}) \le \delta\). Since \(\mathcal{R}(g,\mathcal{P}) \le \mathcal{R}(f,\mathcal{P}) \le \mathcal{R}(h,\mathcal{P})\),

\begin{equation*} I - 2\varepsilon < \mathcal{R}(f,\mathcal{P}) < I + 2\varepsilon \qquad \text{whenever } \Delta(\mathcal{P}) \le \delta . \end{equation*}

Riemann \(\Rightarrow\) Darboux. Let \(I := \int_a^b f\), \(\varepsilon > 0\), and \(\delta\) with \(|\mathcal{R}(f,\mathcal{P}) - I| \le \varepsilon\) for \(\Delta(\mathcal{P}) \le \delta\). Take the uniform partition \(x_i := a + i(b-a)/n\) with \((b-a)/n \le \delta\), and \(M_i, m_i\) the sup and inf of \(f\) on \([x_{i-1},x_i]\). Tagging with \(f(x_i^{*}) \ge M_i - \varepsilon/(b-a)\),

\begin{equation*} \sum_{i} M_i\,\delta x_i \le \mathcal{R}(f,\mathcal{P}) + \varepsilon \le I + 2\varepsilon . \end{equation*}

The function equal to \(M_i\) on \((x_{i-1},x_i]\) and \(M_1\) at \(a\) is piecewise constant, dominates \(f\), and has integral \(\sum_i M_i \delta x_i\), so \(\overline{\int_a^b} f \le I + 2\varepsilon\). Symmetrically \(\underline{\int_a^b} f \ge I - 2\varepsilon\). Let \(\varepsilon \to 0\) and use \(\underline{\int} \le \overline{\int}\). \(\square\)

By Exercise 1.1.22 it suffices to make the upper and lower Darboux integrals differ by at most a constant multiple of \(\varepsilon\). The tool throughout: for a partition of an interval into \(J_1,\dots,J_n\), the piecewise constant functions \(\sum_i (\sup_{J_i} f)1_{J_i}\) and \(\sum_i (\inf_{J_i} f)1_{J_i}\) sandwich \(f\) with integrals differing by \(\sum_i (\sup_{J_i} f - \inf_{J_i} f)|J_i|\), by Exercises 1.1.20 and 1.1.21.

Continuous \(f\) (assume \(a < b\)). Compactness gives boundedness and uniform continuity: pick \(\eta > 0\) with \(|f(x)-f(y)| \le \varepsilon\) for \(|x-y| \le \eta\), and partition \([a,b]\) into \(\{a\}, (x_0,x_1], \dots, (x_{n-1},x_n]\) with \(x_i := a + i(b-a)/n\) and \((b-a)/n \le \eta\). Each piece has oscillation at most \(\varepsilon\), so

\begin{equation*} \overline{\int_a^b} f - \underline{\int_a^b} f \le \varepsilon \sum_i |J_i| = \varepsilon(b-a) . \end{equation*}

Piecewise continuous \(f\), \(|f| \le M\), continuous on each of the intervals \(I_1,\dots,I_k\) partitioning \([a,b]\). Let \(I_j\) have endpoints \(\alpha_j \le \beta_j\) and set \(K_j := [\alpha_j+\eta, \beta_j-\eta] \cap I_j\), a compact subinterval of \(I_j\) (as \([\alpha_j+\eta,\beta_j-\eta] \subset (\alpha_j,\beta_j) \subset I_j\)) with \(|I_j| - |K_j| \le 2\eta\). The continuous case on \(K_j\), run with a small enough parameter, gives piecewise constant \(g_j \le f \le h_j\) there whose integrals differ by at most \(\varepsilon/k\). The leftover \(R := \bigcup_j (I_j \backslash K_j)\) is a finite union of intervals of total length at most \(2k\eta\). Put \(g := g_j\), \(h := h_j\) on \(K_j\) and \(g := -M\), \(h := +M\) on \(R\); refining the finitely many intervals involved shows both are piecewise constant, and \(g \le f \le h\). Additivity of the piecewise constant integral over the pieces gives

\begin{equation*} \mathrm{p.c.}\!\int_a^b h - \mathrm{p.c.}\!\int_a^b g = \sum_{j=1}^{k}\Big(\mathrm{p.c.}\!\int_{K_j} h_j - \mathrm{p.c.}\!\int_{K_j} g_j\Big) + 2M|R| \le \varepsilon + 4Mk\eta , \end{equation*}

which is at most \(2\varepsilon\) once \(\eta \le \varepsilon/(4Mk+1)\). \(\square\)

Work throughout in the Darboux formulation (legitimate by Exercise 1.1.22), using Exercise 1.1.21 for linearity, monotonicity and the indicator property of \(\mathrm{p.c.}\int\); sums, scalar multiples and maxima of piecewise constant functions are piecewise constant (common refinement).

(1) For \(c > 0\), \(g \le f\) iff \(cg \le cf\), so the piecewise constant minorants of \(f\) and of \(cf\) correspond with integrals scaled by \(c\): \(\underline{\int} cf = c\underline{\int} f\), and likewise for the upper integrals. For \(c = 0\) both sides are \(0\). For \(c = -1\), since \(h \ge -f\) iff \(-h \le f\) and \(\mathrm{p.c.}\int h = -\mathrm{p.c.}\int(-h)\),

\begin{equation*} \overline{\int_a^b} (-f) = \inf_{-h \le f} \Big( -\mathrm{p.c.}\!\int (-h) \Big) = -\sup_{g \le f} \mathrm{p.c.}\!\int g = -\underline{\int_a^b} f, \end{equation*}

and symmetrically \(\underline{\int}(-f) = -\overline{\int} f\); a general \(c < 0\) is \(|c| \cdot (-f)\). For additivity, piecewise constant \(g_1 \le f\), \(g_2 \le g\) give \(g_1 + g_2 \le f+g\) with additive p.c. integral, and dually with majorants \(h_1 \ge f\), \(h_2 \ge g\), so

\begin{equation*} \int_a^b f + \int_a^b g \le \underline{\int_a^b}(f+g) \le \overline{\int_a^b}(f+g) \le \int_a^b f + \int_a^b g ; \end{equation*}

all four coincide, and \(f+g\) is bounded, hence integrable with the stated value.

(2) Every piecewise constant \(h \ge g\) also satisfies \(h \ge f\), so the infimum defining \(\overline{\int} f\) runs over a larger family: \(\int_a^b f = \overline{\int} f \le \overline{\int} g = \int_a^b g\).

(3) Given \(\varepsilon > 0\), Exercise 1.1.5 supplies elementary \(A \subset E \subset B\) with \(m(B \setminus A) \le \varepsilon\), and replacing \(B\) by \(B \cap [a,b]\) we may take \(A \subset B \subset [a,b]\). Then \(1_A, 1_B\) are piecewise constant with p.c. integrals \(m(A), m(B)\) (Exercise 1.1.21(3)), and \(1_A \le 1_E \le 1_B\), so

\begin{equation*} m(A) \le \underline{\int_a^b} 1_E \le \overline{\int_a^b} 1_E \le m(B), \qquad m(A) \le m(E) \le m(B), \end{equation*}

the second chain by monotonicity of Jordan measure (Exercise 1.1.6). Finite additivity gives \(m(B) - m(A) = m(B \setminus A) \le \varepsilon\), so both Darboux integrals of \(1_E\) lie within \(\varepsilon\) of \(m(E)\); let \(\varepsilon \to 0\).

Uniqueness. Let \(L\) obey (1), (2), (3). A piecewise constant \(g \equiv c_j\) on intervals \(I_1,\dots,I_n\) partitioning \([a,b]\) is Riemann integrable (bounded piecewise continuous, Exercise 1.1.23) and equals \(\sum_j c_j 1_{I_j}\), with each \(I_j\) Jordan measurable of measure \(|I_j|\), so (1) and (3) give

\begin{equation*} L(g) = \sum_{j=1}^n c_j |I_j| = \mathrm{p.c.}\!\int_a^b g(x)\,dx . \end{equation*}

For Riemann integrable \(f\) and \(\varepsilon > 0\), pick piecewise constant \(g \le f \le h\) with \(\mathrm{p.c.}\int g \ge \int_a^b f - \varepsilon\) and \(\mathrm{p.c.}\int h \le \int_a^b f + \varepsilon\); property (2) gives \(L(g) \le L(f) \le L(h)\), so \(|L(f) - \int_a^b f| \le \varepsilon\). Let \(\varepsilon \to 0\).

Assume \(a < b\) (for \(a = b\) both sets sit in a degenerate box and the integral is \(0\)) and fix \(M\) with \(|f| \le M\).

Slice lemma. Let \(A = \bigcup_{r=1}^N P_r \times Q_r \subset [a,b] \times \mathbb{R}\) be elementary, all \(P_r, Q_r\) non-empty (so \(P_r \subset [a,b]\)). The endpoints of the \(P_r\), together with \(a,b\), cut \([a,b]\) into intervals \(I_1,\dots,I_n\) each of which is contained in or disjoint from every \(P_r\); hence \(A \cap (I_j \times \mathbb{R}) = I_j \times C_j\) with \(C_j := \bigcup_{r : I_j \subset P_r} Q_r\) elementary and equal to the slice \(A_x\) for every \(x \in I_j\), so by Exercise 1.1.4 and finite additivity

\begin{equation*} m^2(A) = \sum_{j=1}^n |I_j| \, m(C_j). \end{equation*}

(i) \(f \ge 0\). Here \(E_{-} \subset [a,b] \times \{0\}\), a degenerate box, so \(E_{-}\) is Jordan measurable with \(m^2(E_{-}) = 0\); we claim \(m_{*,(J)}(E_{+}) = \underline{\int_a^b} f\) and \(m^{*,(J)}(E_{+}) = \overline{\int_a^b} f\). Given piecewise constant \(g \le f\), replace \(g\) by \(\max(g,0)\) (still \(\le f\), p.c. integral not decreased) and write \(g \equiv d_j \ge 0\) on \(I_1,\dots,I_n\); then \(A := \bigcup_j I_j \times [0,d_j] \subset E_{+}\) is elementary and the union is disjoint, so \(m^2(A) = \mathrm{p.c.}\int_a^b g\), giving \(m_{*,(J)}(E_{+}) \ge \underline{\int} f\). Conversely for elementary \(A \subset E_{+}\), the lemma gives \(C_j = A_x \subset [0,f(x)]\) for \(x \in I_j\), so \(m(C_j) \le \inf_{I_j} f\) and \(g := \sum_j m(C_j) 1_{I_j} \le f\) is piecewise constant with \(\mathrm{p.c.}\int g = m^2(A)\); hence \(m_{*,(J)}(E_{+}) \le \underline{\int} f\). Dually, piecewise constant \(h \equiv c_j \ge f \ge 0\) yields elementary \(B := \bigcup_j I_j \times [0,c_j] \supset E_{+}\) with \(m^2(B) = \mathrm{p.c.}\int h\); and any elementary \(B \supset E_{+}\), intersected with \([a,b] \times [0,M]\) (which only shrinks \(m^2(B)\)), has \(f(x) \le m(C_j)\) for \(x \in I_j\), so \(h := \sum_j m(C_j) 1_{I_j} \ge f\) has \(\mathrm{p.c.}\int h = m^2(B)\). Thus \(E_{+}\) is Jordan measurable iff the two Darboux integrals agree, i.e. (Exercise 1.1.22) iff \(f\) is Riemann integrable, and then \(m^2(E_{+}) = \int_a^b f\).

(ii) General \(f\). Put \(f_{\pm} := \max(\pm f, 0)\) and \(Z := [a,b] \times \{0\}\); \(m^2(Z) = 0\), so every subset of \(Z\) is Jordan measurable of measure \(0\).

• Comparing slices, \(E_{+}(f) \subset E_{+}(f_{+})\) with the difference inside \(Z\), so by Boolean closure and additivity of Jordan measure (Exercise 1.1.6) one is Jordan measurable iff the other is, with the same measure.
• The reflection \(\sigma(x,t) := (x,-t)\) maps boxes to boxes of equal volume, hence preserves Jordan inner and outer measure, and \(\sigma(E_{-}(f))\) differs from \(E_{+}(f_{-})\) only inside \(Z\); so \(E_{-}(f)\) is Jordan measurable iff \(E_{+}(f_{-})\) is, with the same measure.
• \(f\) is Riemann integrable iff \(f_{+}, f_{-}\) are. Given \(\varepsilon > 0\) and piecewise constant \(g \le f \le h\) with \(\mathrm{p.c.}\int(h-g) \le \varepsilon\) (Exercise 1.1.22), the piecewise constant \(g^{+} := \max(g,0) \le f_{+} \le h^{+} := \max(h,0)\) satisfy \(0 \le h^{+} - g^{+} \le h-g\) (as \(t \mapsto \max(t,0)\) is non-decreasing and \(1\)-Lipschitz), so \(\overline{\int} f_{+} - \underline{\int} f_{+} \le \varepsilon\) by Exercise 1.1.21; apply this to \(-f\) and use \(f_{-} = (-f)_{+}\). The converse is linearity (Exercise 1.1.24(1)).

Chaining these with (i):

\begin{equation*} \begin{aligned} f \text{ Riemann integrable} &\iff f_{+}, f_{-} \text{ Riemann integrable} \\ &\iff E_{+}(f_{+}), E_{+}(f_{-}) \text{ Jordan measurable} \\ &\iff E_{+}, E_{-} \text{ Jordan measurable}, \end{aligned} \end{equation*}

and in that case, by (i) and then linearity (Exercise 1.1.24(1)),

\begin{equation*} \begin{aligned} m^2(E_{+}) - m^2(E_{-}) &= m^2(E_{+}(f_{+})) - m^2(E_{+}(f_{-})) \\ &= \int_a^b f_{+}(x)\,dx - \int_a^b f_{-}(x)\,dx = \int_a^b f(x)\,dx . \end{aligned} \end{equation*}

Fix \(d \ge 1\) and a box \(B = \prod_{l=1}^d I_l \subset \mathbb{R}^d\), \(|B| = \prod_l |I_l|\).

Definitions. A partition of \(B\) is a finite family of pairwise disjoint boxes with union \(B\); a tagged partition \(\mathcal{P}\) adds tags \(x_i^{*} \in \overline{B_i} \cap B\), and carries \(\Delta(\mathcal{P}) := \max_i \mathrm{diam}(B_i)\) and \(\mathcal{R}(f,\mathcal{P}) := \sum_i f(x_i^{*})|B_i|\). Call \(f\) Riemann integrable with integral \(\int_B f\) if for each \(\varepsilon > 0\) there is \(\delta > 0\) with \(|\mathcal{R}(f,\mathcal{P}) - \int_B f| \le \varepsilon\) whenever \(\Delta(\mathcal{P}) \le \delta\) (and declare every \(f\) integrable with integral \(0\) when \(|B| = 0\)). Call \(g\) piecewise constant if \(g \equiv c_i\) on the boxes of some partition, and set \(\mathrm{p.c.}\int_B g := \sum_i c_i|B_i|\); this is well defined, since for two such partitions the common refinement \(\{B_i \cap B_j^{\prime}\}\) is again a box partition and finite additivity gives \(\sum_i c_i|B_i| = \sum_{i,j} c_i|B_i \cap B_j^{\prime}| = \sum_j c_j^{\prime}|B_j^{\prime}|\) (analogue of Exercise 1.1.20). The Darboux integrals are

\begin{equation*} \underline{\int_B} f := \sup_{g \le f \text{ p.c.}} \mathrm{p.c.}\!\int_B g, \qquad \overline{\int_B} f := \inf_{h \ge f \text{ p.c.}} \mathrm{p.c.}\!\int_B h , \end{equation*}

and \(f\) is Darboux integrable when they agree. For Jordan measurable \(\Omega \subset \mathbb{R}^d\) and bounded \(f : \Omega \to \mathbb{R}\), put \(\int_\Omega f := \int_B \tilde f\) for any box \(B \supset \Omega\) and \(\tilde f\) the extension by \(0\); this is independent of \(B\), as minorants and majorants may be taken to vanish off \(\Omega\). For \(d = 1\) this agrees with Definition 1.1.5, whose partition \(a = x_0 < \dots < x_n = b\) corresponds to the disjoint \(\{a\},(x_0,x_1],\dots,(x_{n-1},x_n]\) with the same tags; both conventions are shown Darboux-equivalent below.

Exercise 1.1.21. Linearity and monotonicity of \(\mathrm{p.c.}\int_B\) pass to the common refinement as above. For elementary \(E = \bigcup_{r \le N} R_r \subset B\), the endpoints of the \(l\)-th sides of \(B, R_1,\dots,R_N\) generate in each coordinate a grid whose product partitions \(B\) into boxes each contained in or disjoint from every \(R_r\), hence from \(E\); so \(1_E\) is piecewise constant with \(\mathrm{p.c.}\int_B 1_E = m^d(E)\) by finite additivity.

Exercise 1.1.22 (Riemann equals Darboux). The key lemma is that \(\mathcal{R}(g,\mathcal{P}) \to \mathrm{p.c.}\int_B g\) for piecewise constant \(g \equiv c_j\) on \(B_1,\dots,B_k\). Grouping by which \(B_j\) holds the tag, \(\mathcal{R}(g,\mathcal{P}) = \sum_j c_j S_j\) with \(S_j := \sum_{i : x_i^{*} \in B_j}|P_i|\) over the boxes \(P_i\) of \(\mathcal{P}\), so it suffices that \(S_j \to |B_j|\) uniformly as \(\delta := \Delta(\mathcal{P}) \to 0\). If \(x_i^{*} \in B_j\) then \(\mathrm{diam}(P_i) \le \delta\) forces \(P_i\) into the \(\delta\)-enlargement of \(B_j\), and the \(P_i\) are disjoint, so \(S_j \le \prod_l(s_{j,l}+2\delta)\) with \(s_{j,l}\) the side lengths of \(B_j\); conversely any \(P_i\) meeting \(B_j^{-2\delta} := \{x \in B_j : \mathrm{dist}(x,\mathbb{R}^d \setminus B_j) > 2\delta\}\) has \(\overline{P_i} \subset B_j\), hence \(x_i^{*} \in B_j\), and such \(P_i\) cover \(B_j^{-2\delta}\), so \(S_j \ge \prod_l \max(s_{j,l}-4\delta,0)\). Both bounds tend to \(|B_j|\). Given the lemma the two implications run as in Exercise 1.1.22: if \(f\) is Darboux integrable with value \(I\), sandwiching \(f\) between piecewise constant \(g \le f \le h\) with integrals within \(\varepsilon\) of \(I\) gives \(\mathcal{R}(g,\mathcal{P}) \le \mathcal{R}(f,\mathcal{P}) \le \mathcal{R}(h,\mathcal{P})\), pinning \(\mathcal{R}(f,\mathcal{P})\) within \(2\varepsilon\) of \(I\); conversely, on the uniform grid of diameter \(\le \delta\), with \(M_i := \sup_{\overline{P_i} \cap B} f\) and tags chosen so that \(f(x_i^{*}) \ge M_i - \varepsilon/(|B|+1)\), the function \(h := \sum_i M_i 1_{P_i} \ge f\) has \(\mathrm{p.c.}\int_B h \le I + 2\varepsilon\), so \(\overline{\int_B} f \le I + 2\varepsilon\), and symmetrically.

Exercise 1.1.23. A continuous \(f\) on the compact box \(\overline{B}\) is uniformly continuous, so the uniform grid of diameter below the modulus of continuity for \(\varepsilon\) gives \(g := \sum_i (\inf_{P_i} f)1_{P_i} \le f \le h := \sum_i (\sup_{P_i} f)1_{P_i}\) with \(\mathrm{p.c.}\int_B (h-g) \le \varepsilon|B|\). For \(f\) piecewise continuous (continuous on each box of some partition) and bounded by \(M\), shrink each \(B_j\) to a compact \(K_j \subset B_j\) with \(|B_j| - |K_j|\) small, apply the continuous case on each \(K_j\), and use \(\pm M\) on the leftover elementary set \(B \setminus \bigcup_j K_j\) of small measure.

Exercise 1.1.24. Linearity, monotonicity and \(\int_B 1_E = m^d(E)\) for Jordan measurable \(E \subset B\) have the same Darboux proofs; for the indicator, Exercise 1.1.5 supplies elementary \(A \subset E \subset B^{\prime}\) with \(m^d(B^{\prime} \setminus A) \le \varepsilon\) (replace \(B^{\prime}\) by \(B^{\prime} \cap B\)) and one applies the indicator property above. Uniqueness likewise: piecewise constant functions are finite linear combinations of indicators of boxes, which are Jordan measurable, so any functional with the three properties agrees with \(\mathrm{p.c.}\int_B\) there and hence with \(\int_B\) by squeezing. All of this transfers to Jordan measurable \(\Omega\) via \(\int_\Omega f = \int_B \tilde f\).

Exercise 1.1.25. With \(E_{+} := \{(x,t) : x \in B;\ 0 \le t \le f(x)\}\) and \(E_{-} := \{(x,t) : x \in B;\ f(x) \le t \le 0\}\) in \(\mathbb{R}^{d+1}\), a bounded \(f\) is Riemann integrable iff \(E_{+}, E_{-}\) are Jordan measurable, and then \(\int_B f = m^{d+1}(E_{+}) - m^{d+1}(E_{-})\). The proof is Exercise 1.1.25 with interval partitions replaced by box partitions: the slice lemma becomes that any elementary \(A \subset B \times \mathbb{R}\), refined by the grid generated in the first \(d\) coordinates by \(B\) and the boxes constituting \(A\), is \(\bigsqcup_j P_j \times C_j\) with \(\{P_j\}\) a box partition of \(B\) and \(C_j\) elementary equal to the slice \(A_x\) for \(x \in P_j\), so \(m^{d+1}(A) = \sum_j |P_j| m^1(C_j)\) by Exercise 1.1.4; the four inequalities for \(f \ge 0\) and the reduction of general \(f\) via \(f_{\pm}\), the Jordan null set \(B \times \{0\}\) and the reflection \((x,t) \mapsto (x,-t)\) are unchanged. Exercises 1.1.4, 1.1.5 and 1.1.6 are already stated in arbitrary dimension.

Enumerate \(\mathbb{Q} \cap [0,1] = \{q_1, q_2, \ldots\}\) and take

\begin{equation*} E_n := \{q_n\}, \qquad \text{respectively} \qquad E_n := [0,1] \setminus \{q_1,\ldots,q_n\} . \end{equation*}

Each is elementary, hence Jordan measurable (Exercise 1.1.5), of measure \(0\) respectively \(1\), while

\begin{equation*} \bigcup_{n=1}^{\infty} E_n = \mathbb{Q} \cap [0,1], \qquad \bigcap_{n=1}^{\infty} E_n = [0,1] \setminus \mathbb{Q}, \end{equation*}

both bounded. Neither limit set is Jordan measurable. Indeed, an elementary subset \(A\) of either one is a finite union of intervals none of which is non-degenerate (a non-degenerate interval contains both a rational and an irrational), so \(A\) is a finite union of degenerate intervals and \(m(A) = 0\) by Lemma 1.1.2(ii); taking suprema, both sets have Jordan inner measure \(0\). Both are also dense in \([0,1]\), hence have closure \([0,1]\) and, by Exercise 1.1.18(1),

\begin{equation*} m^{*,(J)}(\mathbb{Q} \cap [0,1]) = m^{*,(J)}([0,1] \setminus \mathbb{Q}) = m^{*,(J)}([0,1]) = 1 \ne 0 . \end{equation*}

Take \(f_n := 1_{\{q_1,\ldots,q_n\}}\) on \([0,1]\), where \(\mathbb{Q} \cap [0,1] = \{q_1, q_2, \ldots\}\). Then \(0 \le f_n \le 1\), and \(\{q_1,\ldots,q_n\}\) is elementary of measure \(0\), so \(f_n\) is piecewise constant with \(\mathrm{p.c.}\int_0^1 f_n = 0\) (Exercise 1.1.21(3)); taking \(g = h = f_n\) makes both Darboux integrals \(0\), so \(f_n\) is Riemann integrable with \(\int_0^1 f_n = 0\) (Exercise 1.1.22). Pointwise \(f_n \to f := 1_{\mathbb{Q} \cap [0,1]}\), since \(f_n(x) = 0 = f(x)\) for irrational \(x\) and \(f_n(q_N) = 1 = f(q_N)\) for \(n \ge N\); and \(0 \le f \le 1\).

But \(f\) is not Riemann integrable, its Darboux integrals differing. If piecewise constant \(g \equiv c_i \le f\) on intervals \(I_1,\ldots,I_k\) partitioning \([0,1]\), each non-degenerate \(I_i\) contains an irrational, forcing \(c_i \le 0\), while degenerate \(I_i\) have \(|I_i| = 0\); so \(\mathrm{p.c.}\int_0^1 g \le 0\), attained at \(g \equiv 0\). Dually each non-degenerate \(I_i\) contains a rational, so \(h \ge f\) forces \(c_i \ge 1\) there and

\begin{equation*} \mathrm{p.c.}\int_0^1 h \ge \sum_{I_i \text{ non-degenerate}} |I_i| = 1, \end{equation*}

attained at \(h \equiv 1\). Hence \(\underline{\int_0^1} f = 0 \ne 1 = \overline{\int_0^1} f\).

Under uniform convergence no counterexample exists: \(f\) is Riemann integrable with \(\int_0^1 f = \lim_n \int_0^1 f_n\). Given \(\varepsilon > 0\) pick \(N\) with \(|f_n - f| \le \varepsilon\) on \([0,1]\) for all \(n \ge N\), and fix such an \(n\); \(f\) is bounded because \(f_n\) is. For piecewise constant \(g \le f_n\), the function \(g - \varepsilon \le f\) is piecewise constant, so \(\underline{\int_0^1} f \ge \mathrm{p.c.}\int_0^1 g - \varepsilon\) (Exercise 1.1.21(1)); taking the supremum over \(g\) and using Darboux integrability of \(f_n\),

\begin{equation*} \int_0^1 f_n - \varepsilon \ \le\ \underline{\int_0^1} f \ \le\ \overline{\int_0^1} f \ \le\ \int_0^1 f_n + \varepsilon , \end{equation*}

the upper bound symmetrically from \(h \ge f_n\) piecewise constant and \(h + \varepsilon \ge f\), and the middle inequality by Exercise 1.1.21(2). The outer gap is \(2\varepsilon\) for every \(\varepsilon > 0\), so \(f\) is Darboux and hence Riemann integrable (Exercise 1.1.22), and the same chain gives \(|\int_0^1 f - \int_0^1 f_n| \le \varepsilon\) for all \(n \ge N\).

(i) The cover \(B_n := \emptyset\) has total volume \(0\), and every admissible total volume is non-negative, so \(m^*(\emptyset) = 0\).

(ii) Every countable box cover of \(F\) also covers \(E\), so the infimum defining \(m^*(E)\) ranges over a set containing the one defining \(m^*(F)\); hence \(m^*(E) \le m^*(F)\).

(iii) Write \(E := \bigcup_n E_n\) and assume \(\sum_n m^*(E_n) < \infty\) (otherwise there is nothing to prove). Given \(\varepsilon > 0\), countable choice supplies for each \(n\) boxes \(B_{n,1}, B_{n,2}, \ldots\) with

\begin{equation*} E_n \subset \bigcup_{m=1}^{\infty} B_{n,m}, \qquad \sum_{m=1}^{\infty} |B_{n,m}| \le m^*(E_n) + \frac{\varepsilon}{2^n} , \end{equation*}

each infimum being over a non-empty set of reals. The doubly indexed family \(\{B_{n,m}\}\) is countable, re-enumerates as a single sequence of boxes, and covers \(E\); its volumes are non-negative, so by Tonelli for series the total is the iterated sum, independent of the enumeration:

\begin{equation*} \sum_{n,m} |B_{n,m}| = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} |B_{n,m}| \le \sum_{n=1}^{\infty} m^*(E_n) + \varepsilon . \end{equation*}

This is an admissible cover of \(E\), so \(m^*(E) \le \sum_n m^*(E_n) + \varepsilon\); let \(\varepsilon \to 0\).

If \(E\) or \(F\) is empty then \(\mathrm{dist}(E,F) := \inf\{|x-y| : x \in E, y \in F\}\) is an infimum over the empty set, hence \(+\infty\); otherwise assume by symmetry that \(E\) is compact. The map \(x \mapsto \mathrm{dist}(x,F)\) is \(1\)-Lipschitz, hence continuous, and strictly positive on \(E\) (as \(\mathrm{dist}(x,F) = 0\) would put \(x\) in \(\overline{F} = F\), against disjointness), so it attains its infimum on the compact \(E\), at \(x_0\) say:

\begin{equation*} \mathrm{dist}(E,F) = \inf_{x \in E} \mathrm{dist}(x,F) = \mathrm{dist}(x_0,F) > 0 . \end{equation*}

Compactness cannot be dropped: in \(\mathbb{R}\) take

\begin{equation*} E := \{1, 2, 3, \ldots\}, \qquad F := \left\{ n + \tfrac{1}{2n} : n = 1, 2, 3, \ldots \right\} . \end{equation*}

Each meets any bounded set in finitely many points, hence has no limit point and is closed; they are disjoint since \(0 < \frac{1}{2n} < 1\); neither is compact, being unbounded; yet

\begin{equation*} \mathrm{dist}(E,F) \le \inf_{n \ge 1} \tfrac{1}{2n} = 0 . \end{equation*}

For general \(d\), take \(E \times \{0\}^{d-1}\) and \(F \times \{0\}^{d-1}\).

Write \(E = \bigcup_{n=1}^{\infty} B_n\) with the boxes \(B_n\) almost disjoint, so that \(m^*(E) = \sum_n |B_n|\) by Lemma 1.2.9, and take the unbounded Jordan inner measure to be \(m_{*,(J)}(E) := \sup\{m(A) : A \subset E \text{ elementary}\} \in [0,+\infty]\).

For any set \(E\), an elementary \(A \subset E\) satisfies \(m(A) = m^*(A) \le m^*(E)\) by Lemma 1.2.6 and monotonicity (Exercise 1.2.3(ii)); taking the supremum, \(m_{*,(J)}(E) \le m^*(E)\). Conversely each \(A_N := B_1 \cup \ldots \cup B_N \subset E\) is elementary, and the boxes being almost disjoint, (1.3) gives \(m(A_N) = \sum_{n \le N} |B_n|\), so

\begin{equation*} m_{*,(J)}(E) \ \ge \ \sup_N \sum_{n=1}^{N} |B_n| \ = \ \sum_{n=1}^{\infty} |B_n| \ = \ m^*(E) , \end{equation*}

both sides being \(+\infty\) when the series diverges.

Take \(d = 1\) and \(E := [0,1] \setminus \mathbb{Q}\), for which \(m^*(E) = 1\) while the only open \(U \subset E\) is \(\emptyset\).

Indeed \(m^*(E) \le m^*([0,1]) = 1\) by monotonicity (Exercise 1.2.3(ii)) and Lemma 1.2.6, while \(\mathbb{Q} \cap [0,1]\) is countable, hence null (Example 1.2.1), so finite subadditivity (Exercise 1.2.3(i),(iii)) applied to \([0,1] = E \cup (\mathbb{Q} \cap [0,1])\) gives

\begin{equation*} 1 = m^*([0,1]) \le m^*(E) + m^*(\mathbb{Q} \cap [0,1]) = m^*(E) . \end{equation*}

A non-empty open \(U \subset E\) would contain a non-degenerate interval and hence a rational, which is not in \(E\); so the supremum is \(m^*(\emptyset) = 0 \ne 1\).

Statement (ii) is verbatim Definition 1.2.2, so (i) \(\Leftrightarrow\) (ii), and the remaining implications close the two chains (ii) \(\Rightarrow\) (iii) \(\Rightarrow\) (vi) \(\Rightarrow\) (i) and (i) \(\Rightarrow\) (iv) \(\Rightarrow\) (v) \(\Rightarrow\) (vi).

• (ii) \(\Rightarrow\) (iii): \(E \subset U\) gives \(U \Delta E = U \setminus E\).
• (iii) \(\Rightarrow\) (vi): open sets are measurable (Lemma 1.2.13(i)), so take \(E_\varepsilon := U\).
• (i) \(\Rightarrow\) (iv): \(E^c\) is measurable (Lemma 1.2.13(v)), so Definition 1.2.2 supplies open \(U \supset E^c\) with \(m^*(U \setminus E^c) \le \varepsilon\); then \(F := \mathbb{R}^d \setminus U\) is closed with \(F \subset E\) and \(E \setminus F = E \cap U = U \setminus E^c\).
• (iv) \(\Rightarrow\) (v): \(F \subset E\) gives \(F \Delta E = E \setminus F\).
• (v) \(\Rightarrow\) (vi): closed sets are measurable (Lemma 1.2.13(ii)).

(vi) \(\Rightarrow\) (i). First, the approximating set may be taken to contain \(E\). Put \(\delta := \varepsilon/3\) and use (vi) to get measurable \(A\) with \(m^*(A \Delta E) \le \delta\); by monotonicity \(m^*(E \setminus A)\) and \(m^*(A \setminus E)\) are both \(\le \delta\), and since \(m^*(E \setminus A)\) is finite, outer regularity (Lemma 1.2.12) supplies an open \(V \supset E \setminus A\) with \(m^*(V) \le m^*(E \setminus A) + \delta \le 2\delta\). Then \(E^{\prime}_\varepsilon := A \cup V\) is measurable (Lemma 1.2.13(i),(vi)), contains \(E = (E \cap A) \cup (E \setminus A)\), and satisfies \(E^{\prime}_\varepsilon \setminus E \subset (A \setminus E) \cup V\), so by Exercise 1.2.3,

\begin{equation*} m^*(E^{\prime}_\varepsilon \setminus E) \le m^*(A \setminus E) + m^*(V) \le 3\delta = \varepsilon . \end{equation*}

Now set \(E^{\prime\prime} := \bigcap_{n \ge 1} E^{\prime}_{1/n}\), measurable by Lemma 1.2.13(vii) and containing \(E\); as \(E^{\prime\prime} \setminus E \subset E^{\prime}_{1/n} \setminus E\) for every \(n\), monotonicity gives \(m^*(E^{\prime\prime} \setminus E) \le 1/n \to 0\). Hence \(N := E^{\prime\prime} \setminus E\) is null, so measurable (Lemma 1.2.13(iii)), and \(E = E^{\prime\prime} \cap (\mathbb{R}^d \setminus N)\) is measurable by Lemma 1.2.13(v),(vii).

Every elementary set is Lebesgue measurable: a box \(B\) is \(B^\circ \cup (B \setminus B^\circ)\) with \(B^\circ\) open and \(B \setminus B^\circ \subset \partial B\) a finite union of degenerate boxes of elementary measure \(0\), hence null by Lemma 1.2.6 and monotonicity, so \(B\) is measurable by Lemma 1.2.13(i),(iii),(vi), and finite unions of boxes likewise by Lemma 1.2.13(vi).

Now let \(E\) be Jordan measurable and \(\varepsilon > 0\). The definitions of Jordan inner and outer measure supply elementary \(A \subset E \subset B\) with \(m(A) \ge m(E) - \varepsilon\) and \(m(B) \le m(E) + \varepsilon\), so \(B \setminus A\) is elementary with \(m(B \setminus A) = m(B) - m(A) \le 2\varepsilon\) by finite additivity. Here \(A\) is Lebesgue measurable and \(A \Delta E = E \setminus A \subset B \setminus A\), so by monotonicity (Exercise 1.2.3) and Lemma 1.2.6,

\begin{equation*} m^*(A \Delta E) \leq m^*(B \setminus A) = m(B \setminus A) \leq 2\varepsilon . \end{equation*}

By Exercise 1.2.7, (vi) \(\Rightarrow\) (i), the set \(E\) is Lebesgue measurable.

For \(a \in \{0,2\}^n\) put \(p_a := \sum_{i \le n} a_i 3^{-i}\) and \(J_a := [p_a, p_a + 3^{-n}]\), so \(I_n = \bigcup_{a \in \{0,2\}^n} J_a\) is a union of \(2^n\) closed intervals of length \(3^{-n}\).

Compact: each \(I_n\) is a finite union of closed intervals, hence closed, so \(C = \bigcap_n I_n\) is closed and bounded (\(C \subset [0,1]\)); Heine-Borel applies.

Null: for distinct \(a, b \in \{0,2\}^n\) first differing at index \(i\), say \(a_i = 0\), \(b_i = 2\),

\begin{equation*} p_a + 3^{-n} \leq \sum_{j<i} \frac{a_j}{3^j} + \sum_{j>i} \frac{2}{3^j} + 3^{-n} = \sum_{j<i} \frac{a_j}{3^j} + 3^{-i} \leq p_b , \end{equation*}

so \(J_a\) lies to the left of \(J_b\); the \(J_a\) are almost disjoint and \(m(I_n) = 2^n 3^{-n}\). Since \(C \subset I_n\), monotonicity (Exercise 1.2.3) and Lemma 1.2.6 give \(m^*( C) \le m(I_n) = (2/3)^n\) for every \(n\), so \(m^*( C) = 0\).

Uncountable: \(\Phi((a_i)_{i \ge 1}) := \sum_{i \ge 1} a_i 3^{-i}\), absolutely convergent, maps \(\{0,2\}^{\mathbb{N}}\) into \(C\), since for each \(n\), writing \(a^{(n)} := (a_1,\dots,a_n)\),

\begin{equation*} p_{a^{(n)}} \leq \Phi(a) = p_{a^{(n)}} + \sum_{i>n} \frac{a_i}{3^i} \leq p_{a^{(n)}} + 3^{-n}, \end{equation*}

so \(\Phi(a) \in J_{a^{(n)}} \subset I_n\). It is injective: if \(i\) is the least index with \(a_i \ne b_i\), say \(a_i = 0\), \(b_i = 2\), then with \(P := \sum_{j<i} a_j 3^{-j}\),

\begin{equation*} \Phi(a) \leq P + \sum_{j>i} \frac{2}{3^j} = P + 3^{-i} < P + 2 \cdot 3^{-i} \leq \Phi(b) . \end{equation*}

Finally \(\{0,2\}^{\mathbb{N}}\) is uncountable, since given \(a^{(1)}, a^{(2)}, \ldots\) the element \(c_i := 2 - a^{(i)}_i\) differs from every \(a^{(i)}\); so \(C\) contains an injective image of an uncountable set.

We prove the stronger statement: \([0,1)\) is not the union of a countable family of at least two pairwise disjoint non-empty closed subsets of \(\mathbb{R}\). (A one-member family is excluded because \([0,1)\) is not closed, and empty members may be discarded, so this contains the stated claim about closed intervals.)

Suppose \([0,1) = \bigsqcup_{n \in N} F_n\) with \(N\) countable, the \(F_n\) non-empty, closed and pairwise disjoint, and at least two indices present. Picking \(p \in F_1\), \(q \in F_2\) and setting \(c := \min(p,q) < d := \max(p,q)\) gives a non-degenerate compact \([c,d] \subset [0,1)\), and \(C_n := F_n \cap [c,d]\) expresses

\begin{equation*} X := [c,d] = \bigsqcup_{n \in N} C_n \end{equation*}

as a countable disjoint union of closed sets, at least two of them non-empty (those containing \(c\) and \(d\)); discard the empty ones. Interiors below are relative to \(X\).

(i) Finitely many pieces is impossible: if \(X = C_1 \sqcup \dots \sqcup C_k\) with \(k \ge 2\), each \(C_j\) is also relatively open, being the complement in \(X\) of the closed set \(\bigcup_{i \ne j} C_i\), so \(C_1\) and \(C_2 \cup \dots \cup C_k\) disconnect the interval \(X\).

(ii) So assume \(X = \bigsqcup_{n \ge 1} C_n\) and put \(U := \bigcup_n \operatorname{int}_X(C_n)\), \(B := X \setminus U\), which is closed in \(X\), hence compact and a complete metric space.

\(B \ne \emptyset\): otherwise the relatively open \(\operatorname{int}_X(C_n)\) cover the compact \(X\); taking a finite subcover with non-empty members, \(k = 1\) forces every other \(C_n\) to be empty, while \(k \ge 2\) disconnects \(X\) as in (i).

Each \(B \cap C_n\) has empty interior in \(B\): otherwise \(\emptyset \ne B \cap V \subset C_n\) for some open \(V \subset \mathbb{R}\); pick \(x \in B \cap V\) and \(r > 0\) with \((x-r,x+r) \subset V\). As \(x \in C_n\) but \(x \notin \operatorname{int}_X(C_n)\), some \(y \in (x-r/2, x+r/2) \cap X\) lies outside \(C_n\), say \(y \in C_m\) with \(m \ne n\) and \(y > x\) (the case \(y < x\) is symmetric). Then

\begin{equation*} z := \inf \{ s \in [x,y] : s \in C_m \} \end{equation*}

lies in \(C_m \cap [x,y]\), the set being non-empty (it contains \(y\)) and closed; \(z > x\) since \(x \in C_n\) and \(C_n \cap C_m = \emptyset\), and \([x,z) \cap C_m = \emptyset\) by definition of the infimum. Every relative neighbourhood of \(z\) therefore meets \([x,z) \subset X\) outside \(C_m\), so \(z \notin \operatorname{int}_X(C_m)\), while \(z \notin C_k\) for \(k \ne m\) gives \(z \notin \operatorname{int}_X(C_k)\); hence \(z \in B\). But \(z \in [x,y] \subset V\), so \(z \in B \cap V \subset C_n\), contradicting \(z \in C_m\).

Thus \(B = \bigcup_{n \ge 1} (B \cap C_n)\) writes the non-empty complete metric space \(B\) as a countable union of closed nowhere dense subsets, against the Baire category theorem.

(i) Put \(E_0 := \emptyset\) and \(A_n := E_n \setminus E_{n-1}\), measurable by Lemma 1.2.13(v),(vii). They are pairwise disjoint (for \(n < n^{\prime}\), \(A_n \subset E_n \subset E_{n^{\prime}-1}\), which misses \(A_{n^{\prime}}\)) and \(\bigcup_{n \le N} A_n = E_N\) (for \(x \in E_N\) take the least \(n\) with \(x \in E_n\)), hence \(\bigcup_n A_n = \bigcup_n E_n\). Countable additivity (Lemma 1.2.15(ii)) and then finite additivity give

\begin{equation*} m\Big( \bigcup_{n=1}^{\infty} E_n \Big) = \sum_{n=1}^{\infty} m(A_n) = \lim_{N \to \infty} \sum_{n=1}^{N} m(A_n) = \lim_{N \to \infty} m(E_N), \end{equation*}

the middle equality being the definition of a sum of non-negative terms. No finiteness was needed.

(ii) Discarding the first \(n_0 - 1\) sets changes neither side, so assume \(m(E_1) < \infty\). The sets \(F_n := E_1 \setminus E_n\) are measurable and non-decreasing with \(\bigcup_n F_n = E_1 \setminus \bigcap_n E_n\) (de Morgan), so (i) gives \(m(E_1 \setminus \bigcap_n E_n) = \lim_n m(E_1 \setminus E_n)\). For measurable \(A \subset E_1\), finite additivity on \(E_1 = A \sqcup (E_1 \setminus A)\) with \(m(E_1) < \infty\) gives \(m(E_1 \setminus A) = m(E_1) - m(A)\); applying this with \(A = E_n\) and with \(A = \bigcap_n E_n\),

\begin{equation*} m(E_1) - m\Big( \bigcap_{n=1}^{\infty} E_n \Big) = \lim_{n \to \infty} \big( m(E_1) - m(E_n) \big) = m(E_1) - \lim_{n \to \infty} m(E_n), \end{equation*}

the limit existing as \(m(E_n)\) is non-increasing and non-negative. Cancel the finite \(m(E_1)\).

(iii) Take \(E_n := \{x \in \mathbb{R}^d : |x| \ge n\}\), closed hence measurable (Lemma 1.2.13(ii)) and non-increasing. Each contains boxes of arbitrarily large volume, so \(m(E_n) = +\infty\) by monotonicity and Lemma 1.2.6, while \(\bigcap_n E_n = \emptyset\) has measure \(0 \ne +\infty\).

Let \(\mu\) map the Lebesgue measurable sets to \([0,+\infty]\) with \(\mu(\emptyset) = 0\) and \(\mu(\bigcup_n E_n) = \sum_n \mu(E_n)\) for pairwise disjoint measurable \(E_n\); the measurable sets form a \(\sigma\)-algebra (Lemma 1.2.13), so every set below is measurable.

Finite additivity first: padding disjoint \(E_1,\dots,E_k\) with \(E_n := \emptyset\) for \(n > k\) leaves the union unchanged, so

\begin{equation*} \mu(E_1 \cup \dots \cup E_k) = \sum_{n=1}^{\infty} \mu(E_n) = \sum_{n=1}^{k} \mu(E_n). \end{equation*}

Monotonicity: for measurable \(E \subset F\), the set \(F \setminus E = F \cap (\mathbb{R}^d \setminus E)\) is measurable (Lemma 1.2.13(v),(vii)) and \(F = E \sqcup (F \setminus E)\), so \(\mu(F) = \mu(E) + \mu(F \setminus E) \ge \mu(E)\), the sum being taken in \([0,+\infty]\).

Countable subadditivity: disjointify by \(A_n := E_n \setminus \bigcup_{n^{\prime} < n} E_{n^{\prime}}\), measurable by Lemma 1.2.13(v),(vi),(vii) and pairwise disjoint (for \(n < n^{\prime}\), \(A_n \subset E_n\) while \(A_{n^{\prime}} \cap E_n = \emptyset\)), with \(A_n \subset E_n\) and \(\bigcup_n A_n = \bigcup_n E_n\) (given \(x\) in the right side, take the least \(n\) with \(x \in E_n\)). Countable additivity and then monotonicity termwise give

\begin{equation*} \mu\Big( \bigcup_{n=1}^{\infty} E_n \Big) = \sum_{n=1}^{\infty} \mu(A_n) \leq \sum_{n=1}^{\infty} \mu(E_n). \end{equation*}

Put \(A_N := \bigcap_{n \ge N} E_n\) and \(B_N := \bigcup_{n \ge N} E_n\), measurable by Lemma 1.2.13(vi),(vii), with \(A_N\) non-decreasing, \(B_N\) non-increasing and \(A_N \subset E_N \subset B_N\).

(i) Indicators take only the values \(0\) and \(1\), so \(1_{E_n}(x) \to 1_E(x)\) says the sequence is eventually constant with value \(1_E(x)\); thus \(x \in E\) iff \(x \in E_n\) for all large \(n\), and dually \(x \notin E\) iff \(x\) lies in only finitely many \(E_n\), giving

\begin{equation*} E = \bigcup_{N=1}^{\infty} A_N = \bigcap_{N=1}^{\infty} B_N . \end{equation*}

Either expression is measurable by Lemma 1.2.13(vi),(vii).

(ii) The \(A_N\) increase to \(E\), so \(m(A_N) \to m(E)\) (Exercise 1.2.11(i)); the \(B_N\) decrease to \(E\) with \(m(B_1) \le m(F) < \infty\), so the finiteness hypothesis of Exercise 1.2.11(ii) holds and \(m(B_N) \to m(E)\). Monotonicity and \(A_N \subset E_N \subset B_N\) squeeze \(m(A_N) \le m(E_N) \le m(B_N)\), whence \(m(E_N) \to m(E)\).

(iii) Take \(E_n := [n,n+1] \times [0,1]^{d-1}\), a box, hence measurable with \(m(E_n) = 1\) for every \(n\) (Lemma 1.2.6). Since \(x \notin E_n\) once \(n > x_1\), the \(E_n\) converge pointwise to \(\emptyset\), yet \(m(E_n) = 1 \ne 0 = m(\emptyset)\).

Take \(A := \bigcap_{n \ge 1} U_n\), a \(G_\delta\) hull of \(E\), where the open \(U_n \supset E\) satisfy \(m^*(U_n) \le m^*(E) + 1/n\).

If \(m^*(E) = +\infty\), take instead \(A := \mathbb{R}^d\), which is open, hence measurable (Lemma 1.2.13(i)), and contains boxes of arbitrarily large volume, so \(m(A) = +\infty = m^*(E)\) by monotonicity (Exercise 1.2.3) and Lemma 1.2.6.

If \(m^*(E) < \infty\), such \(U_n\) exist by outer regularity (Lemma 1.2.12), and \(A\) is measurable by Lemma 1.2.13(i),(vii) and contains \(E\). Monotonicity (Exercise 1.2.3) applied to \(E \subset A \subset U_n\) gives

\begin{equation*} m^*(E) \leq m(A) \leq m^*(U_n) \leq m^*(E) + \frac{1}{n} \end{equation*}

for every \(n\); let \(n \to \infty\).

Write \(S\) for the supremum. Every compact \(K \subset E\) is closed, hence measurable (Lemma 1.2.13(ii)), and monotonicity (Exercise 1.2.3) gives \(m(K) \le m(E)\); so \(S \le m(E)\).

For the reverse it suffices that \(S > t\) for every real \(t < m(E)\), which also covers \(m(E) = +\infty\). Fix such a \(t\). The sets \(E_n := E \cap \overline{B}(0,n)\) are measurable (Lemma 1.2.13(ii),(vii)) and increase to \(E\), so \(m(E_n) \to m(E)\) by Exercise 1.2.11(i) and some \(m(E_n) > t\); and \(E_n\) lies in a box, so \(m(E_n) < \infty\) by monotonicity and Lemma 1.2.6. Put \(\varepsilon := m(E_n) - t > 0\). Inner approximation by closed sets (Exercise 1.2.7(iv), applicable since \(E_n\) is measurable) supplies a closed \(F \subset E_n \subset E\) with \(m^*(E_n \setminus F) \le \varepsilon/2\), and \(F \subset \overline{B}(0,n)\) is bounded, hence compact by Heine-Borel. Finite subadditivity (Exercise 1.2.3) applied to \(E_n \subset F \cup (E_n \setminus F)\) gives

\begin{equation*} m(F) \ \ge \ m(E_n) - m^*(E_n \setminus F) \ \ge \ t + \varepsilon/2 \ > \ t , \end{equation*}

so \(S \ge m(F) > t\).

Two facts are used throughout: \(A \Delta C \subset (A \Delta B) \cup (B \Delta C)\), so \(m^*(A \Delta C) \le m^*(A \Delta B) + m^*(B \Delta C)\) by subadditivity (Exercise 1.2.3); and a bounded set lies in a box, hence has finite outer measure (monotonicity and Lemma 1.2.6). We run the cycles (i) \(\Rightarrow\) (ii) \(\Rightarrow\) (iii) \(\Rightarrow\) (ix) \(\Rightarrow\) (viii) \(\Rightarrow\) (vii) \(\Rightarrow\) (i) and (i) \(\Rightarrow\) (iv) \(\Rightarrow\) (v) \(\Rightarrow\) (vi) \(\Rightarrow\) (vii).

• (i) \(\Rightarrow\) (ii): outer regularity (Lemma 1.2.12) gives open \(U \supset E\) with \(m(U) \le m(E) + \varepsilon < \infty\); \(U\) and \(U \setminus E\) are measurable (Lemma 1.2.13(i),(v),(vii)) and \(U = E \sqcup (U \setminus E)\), so finite additivity (Lemma 1.2.15) and the finiteness of \(m(E)\) give \(m^*(U \setminus E) = m(U) - m(E) \le \varepsilon\).
• (ii) \(\Rightarrow\) (iii): take such a \(U\) with \(m^*(U \setminus E) \le \varepsilon/2\) and put \(U_R := U \cap B(0,R)\), open and bounded, increasing to \(U\); since \(m(U) < \infty\), Exercise 1.2.11(i) fixes \(R\) with \(m(U \setminus U_R) = m(U) - m(U_R) \le \varepsilon/2\). As \(E \subset U\), we get \(E \Delta U_R \subset (U \setminus E) \cup (U \setminus U_R)\), so \(m^*(E \Delta U_R) \le \varepsilon\).
• (iii) \(\Rightarrow\) (ix): take bounded open \(U\) with \(m^*(E \Delta U) \le \varepsilon/2\) and let \(F_n\) be the union of the closed dyadic cubes of sidelength \(2^{-n}\) contained in \(U\), a finite union since \(U\) is bounded; \(F_n \subset F_{n+1}\) since the children of a cube inside \(U\) stay inside \(U\), and \(\bigcup_n F_n = U\) because (as in the proof of Lemma 1.2.11) every point of the open set \(U\) lies in a contained dyadic cube, hence in one of every sufficiently small sidelength. Each \(F_n\) is closed, hence measurable, and \(m(U) < \infty\), so Exercise 1.2.11(i) gives an \(n\) with \(m(U \setminus F_n) = m(U) - m(F_n) \le \varepsilon/2\); since \(U \Delta F_n = U \setminus F_n\),

\begin{equation*} m^*(E \Delta F_n) \le m^*(E \Delta U) + m(U \setminus F_n) \le \varepsilon . \end{equation*}

• (ix) \(\Rightarrow\) (viii): a finite union of closed dyadic cubes is elementary.
• (viii) \(\Rightarrow\) (vii): an elementary set is Jordan measurable, hence Lebesgue measurable (Exercise 1.2.8), with finite measure equal to its elementary measure (Lemma 1.2.6).
• (vii) \(\Rightarrow\) (i): \(E\) differs from a measurable set by arbitrarily small outer measure, so is measurable by Exercise 1.2.7, (vi) \(\Rightarrow\) (i); and taking \(S\) measurable of finite measure with \(m^*(E \Delta S) \le 1\), subadditivity on \(E \subset S \cup (E \Delta S)\) gives \(m(E) \le m(S) + 1 < \infty\).
• (i) \(\Rightarrow\) (iv): Exercise 1.2.7(iv) gives closed \(F^{\prime} \subset E\) with \(m^*(E \setminus F^{\prime}) \le \varepsilon/2\), and as \(E \cap \overline{B}(0,R)\) increases to \(E\) with \(m(E) < \infty\), Exercise 1.2.11(i) fixes \(R\) with \(m(E \setminus \overline{B}(0,R)) \le \varepsilon/2\); then \(F := F^{\prime} \cap \overline{B}(0,R) \subset E\) is closed and bounded, hence compact, and \(E \setminus F \subset (E \setminus F^{\prime}) \cup (E \setminus \overline{B}(0,R))\) has \(m^* \le \varepsilon\).
• (iv) \(\Rightarrow\) (v): \(F \subset E\) gives \(E \Delta F = E \setminus F\).
• (v) \(\Rightarrow\) (vi): a compact set is closed, hence measurable (Lemma 1.2.13(ii)), and bounded.
• (vi) \(\Rightarrow\) (vii): a bounded measurable set has finite measure.

(i) \(\Rightarrow\) (ii). An elementary \(A\) is measurable (Exercise 1.2.8) with \(m^*(A) = m(A) < \infty\) (Lemma 1.2.6), and \(A \cap E\), \(A \setminus E\) are measurable (Lemma 1.2.13(v),(vii)), disjoint, with union \(A\), so finite additivity (Lemma 1.2.15) gives

\begin{equation*} m(A) = m(A \cap E) + m(A \setminus E) = m^*(A \cap E) + m^*(A \setminus E). \end{equation*}

(ii) \(\Rightarrow\) (iii). A box is elementary with elementary measure \(|B|\).

(iii) \(\Rightarrow\) (i). Since \(E = \bigcup_N (E \cap B_N)\) with \(B_N := [-N,N]^d\), and countable unions of measurable sets are measurable (Lemma 1.2.13(vi)), it suffices to treat \(F := E \cap B\) for \(B := B_N\) fixed. As \(B \setminus E = B \setminus F\), the hypothesis at \(B\) reads \(|B| = m^*(F) + m^*(B \setminus F)\), both terms being at most \(|B| < \infty\) by monotonicity.

Given \(\varepsilon > 0\), outer regularity (Lemma 1.2.12) supplies open \(U \supset F\) and \(V \supset B \setminus F\) with \(m(U) \le m^*(F) + \varepsilon\) and \(m(V) \le m^*(B \setminus F) + \varepsilon\), both measurable (Lemma 1.2.13(i)) of finite measure. From \(B \subset U \cup V\) we get \(m(U \cup V) \ge |B|\) (monotonicity, Lemma 1.2.6), while decomposing \(U \cup V = U \sqcup (V \setminus U)\) and \(V = (V \cap U) \sqcup (V \setminus U)\) gives \(m(U \cup V) + m(U \cap V) = m(U) + m(V)\), all terms finite; hence

\begin{equation*} m(U \cap V) \le \bigl(m^*(F) + \varepsilon\bigr) + \bigl(m^*(B \setminus F) + \varepsilon\bigr) - |B| = 2\varepsilon . \end{equation*}

Trim \(U\) outside \(B\): by continuity of \(\delta \mapsto (2N+2\delta)^d\), fix \(\delta > 0\) so that the open box \(W := (-N-\delta, N+\delta)^d\) has \(m(W \setminus B) = (2N+2\delta)^d - (2N)^d \le \varepsilon\), and put \(\tilde U := U \cap W\), open with \(F \subset U \cap B \subset \tilde U\). A point of \(\tilde U \setminus F\) lies either in \(B\), hence in \(U \cap (B \setminus F) \subset U \cap V\), or outside \(B\), hence in \(W \setminus B\); so

\begin{equation*} m^*(\tilde U \setminus F) \le m(U \cap V) + m(W \setminus B) \le 3\varepsilon . \end{equation*}

As \(\varepsilon\) was arbitrary, \(F\) is measurable by Definition 1.2.2.

Elementary sets \(A \supset E\) exist since \(E\) is bounded, and are Lebesgue measurable (Exercise 1.2.8) with \(m^*(A) = m(A) < \infty\) (Lemma 1.2.6).

(i) Take first elementary \(A^{\prime} \subset A\), both containing \(E\), and note \(A \setminus E = (A^{\prime} \setminus E) \sqcup (A \setminus A^{\prime})\). We claim

\begin{equation*} m^*(A \setminus E) = m^*(A^{\prime} \setminus E) + m(A \setminus A^{\prime}). \end{equation*}

For “\(\le\)”, Exercise 1.2.14 gives measurable \(S \supset A^{\prime} \setminus E\) with \(m(S) = m^*(A^{\prime} \setminus E)\), and \(A \setminus E \subset S \cup (A \setminus A^{\prime})\), so subadditivity applies. For “\(\ge\)”, Exercise 1.2.14 gives measurable \(T \supset A \setminus E\) with \(m(T) = m^*(A \setminus E)\), which we may intersect with \(A\) so that \(m(T) \le m(A) < \infty\); since the measurable set \(A \setminus A^{\prime}\) lies in \(T\), finite additivity gives \(m(T) = m(A \setminus A^{\prime}) + m(T \setminus (A \setminus A^{\prime}))\), and \(T \setminus (A \setminus A^{\prime}) \supset A^{\prime} \setminus E\) (the latter lies in \(T\) and misses \(A \setminus A^{\prime}\)), so monotonicity gives \(m^*(A^{\prime} \setminus E) \le m^*(A \setminus E) - m(A \setminus A^{\prime})\).

Combining with \(m(A) = m(A^{\prime}) + m(A \setminus A^{\prime})\), all quantities finite,

\begin{equation*} m(A) - m^*(A \setminus E) = m(A^{\prime}) - m^*(A^{\prime} \setminus E) . \end{equation*}

For general elementary \(A, A^{\prime} \supset E\), apply this twice through the elementary set \(A \cap A^{\prime} \supset E\).

(ii) Fix elementary \(A \supset E\). Subadditivity on \(A = E \cup (A \setminus E)\) gives \(m(A) \le m^*(E) + m^*(A \setminus E)\), and \(m^*(A \setminus E) \le m(A) < \infty\) permits rearranging to \(m_*(E) \le m^*(E)\).

If \(E\) is measurable, then \(A = E \sqcup (A \setminus E)\) with both parts measurable of finite measure, so finite additivity (Lemma 1.2.15) gives \(m_*(E) = m(A) - m(A \setminus E) = m(E) = m^*(E)\).

Conversely suppose \(m(A) = m^*(E) + m^*(A \setminus E)\), all three finite, and argue as in Exercise 1.2.17. Given \(\varepsilon > 0\), outer regularity (Lemma 1.2.12) supplies open \(U \supset E\) and \(V \supset A \setminus E\) of finite measure with \(m(U) \le m^*(E) + \varepsilon\) and \(m(V) \le m^*(A \setminus E) + \varepsilon\); from \(A \subset U \cup V\) and \(m(U \cup V) + m(U \cap V) = m(U) + m(V)\),

\begin{equation*} m(U \cap V) \le \bigl(m^*(E) + \varepsilon\bigr) + \bigl(m^*(A \setminus E) + \varepsilon\bigr) - m(A) = 2\varepsilon . \end{equation*}

Outer regularity again gives open \(O \supset A\) with \(m(O \setminus A) = m(O) - m(A) \le \varepsilon\), and \(\tilde U := U \cap O\) is open with \(E \subset \tilde U\); a point of \(\tilde U \setminus E\) lies either in \(A\), hence in \(U \cap (A \setminus E) \subset U \cap V\), or outside \(A\), hence in \(O \setminus A\). So \(m^*(\tilde U \setminus E) \le 3\varepsilon\), and \(E\) is measurable by Definition 1.2.2.

(ii) \(\Rightarrow\) (i). If \(E = G \setminus N\) with \(G = \bigcap_n U_n\) a \(G_\delta\) and \(N\) null, then \(G\) is measurable by Lemma 1.2.13(i),(vii) and \(N\) by Lemma 1.2.13(iii), so \(E = G \cap (\mathbb{R}^d \setminus N)\) is measurable by Lemma 1.2.13(v),(vii).

(iii) \(\Rightarrow\) (i). If \(E = F \cup N\) with \(F = \bigcup_n F_n\) an \(F_\sigma\) and \(N\) null, then \(F\) is measurable by Lemma 1.2.13(ii),(vi) and \(N\) by Lemma 1.2.13(iii), so \(E\) is measurable by Lemma 1.2.13(vi).

(i) \(\Rightarrow\) (ii). Definition 1.2.2 supplies open \(U_n \supset E\) with \(m^*(U_n \setminus E) \le 1/n\); then \(G := \bigcap_n U_n\) is a \(G_\delta\) containing \(E\), and \(G \setminus E \subset U_n \setminus E\) for every \(n\) forces \(m^*(G \setminus E) = 0\), so \(N := G \setminus E\) is null and \(E = G \setminus N\).

(i) \(\Rightarrow\) (iii). Inner approximation by closed sets (Exercise 1.2.7(iv)) supplies closed \(F_n \subset E\) with \(m^*(E \setminus F_n) \le 1/n\); then \(F := \bigcup_n F_n \subset E\) is an \(F_\sigma\), and \(E \setminus F \subset E \setminus F_n\) for every \(n\) forces \(N := E \setminus F\) to be null, so \(E = F \cup N\).

Lebesgue outer measure is translation invariant: a box \(B = \prod_i I_i\) translates to the box \(B + x = \prod_i (I_i + x_i)\) of the same volume, so any countable box cover of \(S\) translates to one of \(S + x\) of the same total volume, giving \(m^*(S + x) \le m^*(S)\); applying this to \(S + x\) and the translation by \(-x\) gives equality.

Now let \(\varepsilon > 0\). Definition 1.2.2 supplies an open \(U \supset E\) with \(m^*(U \setminus E) \le \varepsilon\); translation being a homeomorphism, \(U + x\) is open and contains \(E + x\), and translation commutes with set difference, so

\begin{equation*} m^*\bigl((U + x) \setminus (E + x)\bigr) = m^*\bigl((U \setminus E) + x\bigr) = m^*(U \setminus E) \le \varepsilon . \end{equation*}

Hence \(E + x\) is Lebesgue measurable by Definition 1.2.2, with

\begin{equation*} m(E + x) = m^*(E + x) = m^*(E) = m(E). \end{equation*}

Write \(\|T\|\) for the operator norm and use the convention \(0 \cdot \infty = 0\). Three preliminaries:

(a) \(T\) maps null sets to null sets. Assume \(T \ne 0\). A closed cube \(Q\) of sidelength \(r\) has \(\operatorname{diam} T(Q) \le \|T\| r \sqrt{d}\), and a set of diameter \(\delta\) lies in a closed cube of sidelength \(\delta\), so \(m^*(T(Q)) \le C|Q|\) with \(C := (\|T\|\sqrt d)^d\). Given null \(N\) and \(\varepsilon > 0\), outer regularity (Lemma 1.2.12) gives open \(U \supset N\) with \(m^*(U) \le \varepsilon\), Lemma 1.2.11 writes \(U = \bigcup_n Q_n\) as almost disjoint closed cubes, and Lemma 1.2.9 gives \(\sum_n |Q_n| = m^*(U) \le \varepsilon\); countable subadditivity on \(T(N) \subset \bigcup_n T(Q_n)\) gives \(m^*(T(N)) \le C\varepsilon\) for every \(\varepsilon\).

(b) A proper subspace \(V \subsetneq \mathbb{R}^d\) is null, hence so is every affine hyperplane (translation invariance, Exercise 1.2.20). Pick a unit \(v \perp V\), so \(V \subset H := \{y : y \cdot v = 0\}\); put \(H_R := H \cap [-R,R]^d\) and \(c := m^*(H_R) \le (2R)^d < \infty\). The translates \(S_k := H_R + (k/N)v\), \(k = 1,\dots,N\), each have outer measure \(c\) (Exercise 1.2.20) and satisfy \(\operatorname{dist}(S_1 \cup \dots \cup S_k, S_{k+1}) \ge 1/N > 0\), since \(|y - y^{\prime}| \ge |(y-y^{\prime}) \cdot v|\) by Cauchy-Schwarz; so repeated use of Lemma 1.2.5 gives \(m^*(S_1 \cup \dots \cup S_N) = Nc\). This union lies in \([-R-1,R+1]^d\), so \(Nc \le (2R+2)^d\) for all \(N\) and \(c = 0\); now \(H = \bigcup_R H_R\) is null by countable subadditivity.

(c) \(T(E)\) is measurable whenever \(E\) is. By Exercise 1.2.19, (i) \(\Rightarrow\) (iii), \(E = F \cup N\) with \(F\) an \(F_\sigma\) and \(N\) null; each closed \(F_n\) is the countable union of the compacts \(F_n \cap \overline{B}(0,k)\), whose continuous images are compact, hence measurable (Lemma 1.2.13(ii)), while \(T(N)\) is null by (a), hence measurable (Lemma 1.2.13(iii)); apply Lemma 1.2.13(vi).

(i) \(\det T = 0\). Then \(T(\mathbb{R}^d)\) is a proper subspace, null by (b), so \(T(E)\) is null and \(m(T(E)) = 0 = |\det T| \, m(E)\).

(ii) \(\det T \ne 0\), so \(T\) is a bijection and \(T(E)\) is measurable by (c). Put \(c := m(T([0,1)^d)) < \infty\), a bounded measurable set. Two facts recur: \(T(E+x) = T(E) + Tx\), so \(m(T(E+x)) = m(T(E))\) by Exercise 1.2.20; and \(T\) injective carries disjoint sets to disjoint sets and commutes with unions and differences, making \(E \mapsto m(T(E))\) countably additive (Lemma 1.2.15).

• Dyadic cubes. \([0,1)^d\) is the disjoint union of the \(2^{nd}\) translates of \(Q_0 := [0,2^{-n})^d\), so \(c = 2^{nd} m(T(Q_0))\) and \(m(T(Q)) = c|Q|\) for every half-open dyadic \(Q\) of sidelength \(2^{-n}\). The same holds for the closed dyadic cube \(Q\): its difference from the half-open cube with the same corner lies in the \(2d\) affine hyperplanes carrying its faces, null by (b), and \(T\) of a null set is null by (a).
• Open sets. Write \(U = \bigcup_n Q_n\) with \(Q_n\) almost disjoint closed dyadic cubes and \(m(U) = \sum_n |Q_n|\) (Lemmas 1.2.11, 1.2.9). The interiors \(Q_n^\circ\) are disjoint, \(Z := U \setminus \bigcup_n Q_n^\circ \subset \bigcup_n \partial Q_n\) is null by (b), and \(m(T(Q_n^\circ)) = m(T(Q_n)) = c|Q_n|\) by (a); as \(T(U)\) is the disjoint union of the \(T(Q_n^\circ)\) with the null set \(T(Z)\), countable additivity gives \(m(T(U)) = c\,m(U)\).
• \(c > 0\): otherwise each \(T([0,1)^d + k)\), \(k \in \mathbb{Z}^d\), is null, and surjectivity makes \(\mathbb{R}^d\) their countable union, forcing \(m(\mathbb{R}^d) = 0\) against \(m([0,1]^d) = 1\) (Lemma 1.2.6).
• All measurable \(E\). If \(m(E) < \infty\), Exercise 1.2.16, (i) \(\Rightarrow\) (ii), gives open \(U_n \supset E\) of finite measure with \(m(U_n \setminus E) \le 1/n\), which we may take decreasing; then \(G := \bigcap_n U_n \supset E\) has \(G \setminus E\) null, so \(m(G) = m(E)\) and, by (a) with injectivity, \(m(T(G)) = m(T(E))\). The \(T(U_n)\) decrease to \(T(G)\) with \(m(T(U_1)) = c\,m(U_1) < \infty\), so Exercise 1.2.11(ii) applies to both sequences:

\begin{equation*} m(T(E)) = \lim_{n \to \infty} m(T(U_n)) = \lim_{n \to \infty} c\, m(U_n) = c\, m(G) = c\, m(E). \end{equation*}

If \(m(E) = \infty\), apply this to \(E \cap B(0,k)\) and let \(k \to \infty\), using Exercise 1.2.11(i) and \(c > 0\).

The constant is \(|\det T|\). Writing \(c(T)\) for it and applying the above twice, \(m(ST(E)) = c(S)c(T)m(E) = c(ST)m(E)\), so \(c(ST) = c(S)c(T)\) on taking \(E = [0,1]^d\) of measure \(1\). An orthogonal \(T\) satisfies \(|Ty| = |y|\), so it maps the closed unit ball \(B\) onto itself and \(m(B) = c(T)m(B)\) with \(0 < d^{-d/2} \le m(B) \le 2^d < \infty\), whence \(c(T) = 1 = |\det T|\); and \(T = \operatorname{diag}(\sigma_1,\dots,\sigma_d)\) with \(\sigma_i > 0\) sends \([0,1)^d\) to a box of volume \(\prod_i \sigma_i\), so \(c(T) = \prod_i \sigma_i = |\det T|\). For invertible \(T\), the singular value decomposition \(T = U \Sigma V^{\mathsf T}\) with \(U, V\) orthogonal and all \(\sigma_i > 0\) gives

\begin{equation*} c(T) = c(U)\, c(\Sigma)\, c(V^{\mathsf T}) = \prod_{i=1}^d \sigma_i = |\det T| . \end{equation*}

Throughout, \(a := (m^d)^*(E)\), \(b := (m^{d^{\prime}})^*(F)\), and \(0 \cdot \infty = 0\).

(i) Suppose first \(a, b < \infty\), and given \(\varepsilon > 0\) take box covers \(E \subset \bigcup_n B_n\) with \(\sum_n |B_n| \le a + \varepsilon\) and \(F \subset \bigcup_m B_m^{\prime}\) with \(\sum_m |B_m^{\prime}| \le b + \varepsilon\). The products \(B_n \times B_m^{\prime}\) are boxes of volume \(|B_n| \, |B_m^{\prime}|\) covering \(E \times F\), so summing the non-negative double series in any order,

\begin{equation*} (m^{d+d^{\prime}})^*(E \times F) \leq \Big(\sum_n |B_n|\Big)\Big(\sum_m |B_m^{\prime}|\Big) \leq (a+\varepsilon)(b+\varepsilon); \end{equation*}

let \(\varepsilon \to 0\). If one of \(a, b\) is infinite and the other positive, then \(ab = +\infty\). In the only remaining case, say \(a = 0\) and \(b = +\infty\), we have \(E \times F \subset \bigcup_k (E \times [-k,k]^{d^{\prime}})\), and the finite case gives each term outer measure \(0 \cdot (2k)^{d^{\prime}} = 0\), so countable subadditivity gives \((m^{d+d^{\prime}})^*(E \times F) = 0 = ab\).

(ii) Reduce to bounded \(E, F\): the sets \(E_k := E \cap [-k,k]^d\) and \(F_k := F \cap [-k,k]^{d^{\prime}}\) are bounded and measurable (Lemma 1.2.13), with \(E_k \times F_k\) increasing to \(E \times F\), so the bounded case plus Lemma 1.2.13(vi) makes \(E \times F\) measurable and Exercise 1.2.11(i), applied three times, gives

\begin{equation*} m^{d+d^{\prime}}(E \times F) = \lim_{k \to \infty} m^d(E_k) \, m^{d^{\prime}}(F_k) = m^d(E) \, m^{d^{\prime}}(F) , \end{equation*}

the last step because \(\alpha_k \uparrow \alpha\) and \(\beta_k \uparrow \beta\) in \([0,+\infty]\) force \(\alpha_k\beta_k \to \alpha\beta\) under our convention (if \(\alpha = 0\) every \(\alpha_k = 0\); if \(\alpha, \beta > 0\), fix \(k_0\) with \(\alpha_{k_0}, \beta_{k_0} > 0\), and when \(\alpha = \infty\) note \(\alpha_k\beta_k \ge \alpha_k \beta_{k_0} \to \infty\)).

So let \(E, F\) be bounded and measurable, inside open boxes \(W, W^{\prime}\), and put \(C := |W| + |W^{\prime}| < \infty\). Given \(\varepsilon > 0\), Definition 1.2.2 (after intersecting with \(W\), resp. \(W^{\prime}\)) gives open \(U, V\) with \(E \subset U \subset W\), \(F \subset V \subset W^{\prime}\), \(m^*(U \setminus E) \le \varepsilon\) and \(m^*(V \setminus F) \le \varepsilon\). Then \(U \times V\) is open, contains \(E \times F\), and a point of \((U \times V) \setminus (E \times F)\) has either first coordinate outside \(E\) or second outside \(F\), so

\begin{equation*} (U \times V) \setminus (E \times F) \subset \big((U \setminus E) \times V\big) \cup \big(U \times (V \setminus F)\big) , \end{equation*}

and subadditivity with (i) gives \(m^*((U \times V) \setminus (E \times F)) \le \varepsilon|W^{\prime}| + |W|\varepsilon = C\varepsilon\). Hence \(E \times F\) is measurable by Definition 1.2.2.

For the identity, take closed boxes \(B \supset E\) and \(B^{\prime} \supset F\). The sets \(E \times F\), \((B \setminus E) \times F\) and \(B \times (B^{\prime} \setminus F)\) are measurable by the previous paragraph (\(B \setminus E\), \(B^{\prime} \setminus F\) being bounded measurable), pairwise disjoint (the first two differ in the first coordinate, the third has second coordinate off \(F\)), with union \(B \times B^{\prime}\), a box of volume \(|B||B^{\prime}|\) (Lemma 1.2.6). Finite additivity (Lemma 1.2.15(ii)) with all terms finite, the bounds of (i), and \(m^d(B \setminus E) = |B| - m^d(E)\), \(m^{d^{\prime}}(B^{\prime} \setminus F) = |B^{\prime}| - m^{d^{\prime}}(F)\) then give

\begin{equation*} \begin{aligned} m^{d+d^{\prime}}(E \times F) &\geq |B||B^{\prime}| - \big(|B| - m^d(E)\big) m^{d^{\prime}}(F) - |B|\big(|B^{\prime}| - m^{d^{\prime}}(F)\big) \\ &= m^d(E)\, m^{d^{\prime}}(F), \end{aligned} \end{equation*}

and the reverse inequality is (i).

Uniqueness is the whole content: \(m\) obeys (i),(ii) by Lemma 1.2.15, (iii) by Exercise 1.2.20, and (iv) by Lemma 1.2.6. Let \(\mu\) obey (i)-(iv). Padding a finite disjoint family with copies of \(\emptyset\) in (ii) makes \(\mu\) finitely additive, hence monotone and countably subadditive (Exercise 1.2.12).

Put \(c := \mu([0,1)^d)\); then \(c \in [2^{-d},1]\). Indeed \(c \leq \mu([0,1]^d) = 1\) by monotonicity, while sorting \(x \in [0,1]^d\) by the set \(S\) of coordinates with \(x_i = 1\) covers \([0,1]^d\) by the \(2^d\) sets \(P_S \subset e_S + [0,1)^d\), \(e_S := \sum_{i \in S} e_i\), so subadditivity and (iii) give

\begin{equation*} 1 = \mu([0,1]^d) \leq \sum_{S \subset \{1,\ldots,d\}} \mu(e_S + [0,1)^d) = 2^d c . \end{equation*}

Next, \(\mu(B) = c|B|\) for every box \(B\). The \(2^{nd}\) half-open dyadic cubes \(\prod_i [j_i2^{-n},(j_i+1)2^{-n})\), \(0 \leq j_i < 2^n\), partition \([0,1)^d\) and are translates of \([0,2^{-n})^d\), so finite additivity and (iii) give \(\mu([0,2^{-n})^d) = c\,2^{-nd}\); a half-open box \(B\) with all endpoints in \(2^{-n}\mathbb{Z}\) is a disjoint union of \(|B|2^{nd}\) such translates, whence \(\mu(B) = c|B|\). For a general box \(B = \prod_i I_i\) with \((a_i,b_i) \subset I_i \subset [a_i,b_i]\), fix \(\delta > 0\) and \(n\) with \(2^{-n} \leq \delta\). Rounding each endpoint outward to \(2^{-n}\mathbb{Z}\) puts \(B\) inside a half-open dyadic box with \(i\)-th side \(< b_i - a_i + 3\delta\); if \(b_i - a_i > 4\delta\) for all \(i\), rounding \(a_i + \delta\) up and \(b_i - \delta\) down puts a nonempty half-open dyadic box with \(i\)-th side \(> b_i - a_i - 4\delta\) inside \(B\). So by monotonicity and the dyadic case,

\begin{equation*} \begin{aligned} c \prod_{i=1}^d (b_i - a_i - 4\delta) \ \leq\ \mu(B) \ \leq\ c \prod_{i=1}^d (b_i - a_i + 3\delta), \end{aligned} \end{equation*}

and \(\delta \to 0\) gives \(\mu(B) = c|B|\) when every \(b_i > a_i\); if some \(b_i = a_i\), the upper bound alone forces \(\mu(B) = 0 = c|B|\).

Hence \(c = 1\): every elementary \(A\) is a finite disjoint union of boxes (Lemma 1.1.2), so \(\mu(A) = c\,m(A)\), and \(A = [0,1]^d\) with (iv) gives \(c = 1\). In particular \(\mu(B) = |B|\) for every box.

Therefore \(\mu \leq m\) on measurable sets: for any countable box cover \(E \subset \bigcup_n B_n\), monotonicity and countable subadditivity give \(\mu(E) \leq \sum_n |B_n|\), and infimising over covers yields \(\mu(E) \leq m^*(E) = m(E)\).

Equality now follows by a squeeze. If \(E \subset B\) with \(B\) a half-open box, then \(B \setminus E\) is measurable (Lemma 1.2.13) and finite additivity of \(\mu\) and of \(m\) (Lemma 1.2.15) gives

\begin{equation*} \mu(E) + \mu(B \setminus E) = |B| = m(E) + m(B\setminus E), \end{equation*}

all four terms being at most \(|B| < \infty\); since \(\mu \leq m\) on each of \(E\) and \(B \setminus E\), neither inequality can be strict, so \(\mu(E) = m(E)\). For general measurable \(E\), sum over the disjoint cubes \(Q_k := k + [0,1)^d\), \(k \in \mathbb{Z}^d\), using (ii) for \(\mu\) and for \(m\):

\begin{equation*} \begin{aligned} \mu(E) &= \sum_{k \in \mathbb{Z}^d} \mu(E \cap Q_k) = \sum_{k \in \mathbb{Z}^d} m(E \cap Q_k) = m(E). \end{aligned} \end{equation*}

Everything rests on two facts about null sets (subsets and countable unions of null sets are null) and on \(\Delta\) being an associative, commutative group operation on \(2^A\), since \(1_{E\Delta F} = 1_E + 1_F \bmod 2\). (In (ii) the printed parenthetical should read \(m^*(E\Delta E^{\prime}) = m^*(F \Delta F^{\prime})\): the sets involved need not be measurable.)

(i) Reflexivity: \(E \Delta E = \emptyset\). Symmetry: \(E \Delta F = F \Delta E\). Transitivity: \(E \Delta G = (E \Delta F) \Delta (F \Delta G) \subset (E \Delta F) \cup (F \Delta G)\), a subset of a union of two null sets.

(ii) Well-definedness. Subadditivity applied to \(X \subset Y \cup (X \Delta Y)\) gives \(m^*(X) \leq m^*(Y) + m^*(X \Delta Y)\) for all \(X,Y \subset \mathbb{R}^d\). If \(E \Delta F\) and \(E^{\prime}\Delta F^{\prime}\) are null then

\begin{equation*} \begin{aligned} (E \Delta E^{\prime}) \Delta (F \Delta F^{\prime}) &= (E \Delta F) \Delta (E^{\prime} \Delta F^{\prime}) \\ &\subset (E \Delta F) \cup (E^{\prime} \Delta F^{\prime}) \end{aligned} \end{equation*}

is null, so that inequality with \(X = E \Delta E^{\prime}\), \(Y = F \Delta F^{\prime}\) and then reversed gives \(m^*(E \Delta E^{\prime}) = m^*(F \Delta F^{\prime})\).

Metric axioms: \(d([E],[E^{\prime}]) \leq m^*(A) = m(A) < \infty\) since \(A\) is bounded, so \(d\) is \(\mathbb{R}^+\)-valued; it vanishes exactly when \(E \Delta E^{\prime}\) is null, i.e. when \([E] = [E^{\prime}]\); it is symmetric; and \(E \Delta E^{\prime\prime} \subset (E \Delta E^{\prime}) \cup (E^{\prime} \Delta E^{\prime\prime})\) with subadditivity gives the triangle inequality.

Completeness: a Cauchy sequence with a convergent subsequence converges, so we may assume \(d([E_n],[E_{n+1}]) \leq 2^{-n}\). Put \(F_n := E_n \Delta E_{n+1}\), \(G_N := \bigcup_{n \geq N} F_n\), \(G := \bigcap_{N} G_N\); then \(m^*(G_N) \leq 2^{-N+1}\) by countable subadditivity, so \(G\) is null. For \(x \in A \setminus G\) we have \(x \notin F_n\) for all large \(n\), so \(1_{E_n}(x)\) is eventually constant, and

\begin{equation*} E := \{x \in A \setminus G : x \in E_n \text{ for all sufficiently large } n\} \subset A \end{equation*}

is defined. If \(x \in E \Delta E_n\) lay in no \(F_k\) with \(k \geq n\) then \(1_{E_k}(x) = 1_{E_n}(x)\) for all \(k \geq n\), forcing \(1_E(x) = 1_{E_n}(x)\); so \(E \Delta E_n \subset G \cup G_n\) and

\begin{equation*} d([E_n],[E]) \leq m^*(G) + m^*(G_n) \leq 2^{-n+1} \to 0 . \end{equation*}

(iii) \(\mathcal{L}\) is a union of classes: if \(E \in \mathcal{L}\) and \(E \sim F\) then \(F = E \Delta (E \Delta F)\) differs from \(E\) by a null, hence measurable, set (Lemma 1.2.13(iii)).

Density: given \(E \in \mathcal{L}\) and \(\varepsilon > 0\), \(m(E) \leq m(A) < \infty\), so (i)\(\Leftrightarrow\)(viii) of Exercise 1.2.16 supplies elementary \(F\) with \(m^*(E \Delta F) \leq \varepsilon\); as \(E \subset A\) we have \(E \Delta (F \cap A) \subset E \Delta F\), so \(F \cap A \subset A\) is elementary with \(d([E],[F\cap A]) \leq \varepsilon\).

Closedness: if \(E_n \in \mathcal{L}\) and \([E_n] \to [E]\), then \(E\) differs from a measurable set by outer measure at most \(\varepsilon\) for every \(\varepsilon > 0\), so \(E \in \mathcal{L}\) by Exercise 1.2.7(vi).

Elementary sets are Jordan, hence Lebesgue, measurable (Exercise 1.2.8), so \(\mathcal{E}/\sim \,\subset \mathcal{L}/\sim\); closedness bounds the closure above and density bounds it below, giving \(\overline{\mathcal{E}/\sim} = \mathcal{L}/\sim\), complete as a closed subset of \(2^A/\sim\).

(iv) For all \(E,F \in \mathcal{L}\), \(E \subset F \cup (E \Delta F)\) and its mirror give

\begin{equation*} |m([E]) - m([F])| \leq m^*(E \Delta F) = d([E],[F]), \end{equation*}

all terms finite since \(m(E) \leq m(A) < \infty\). Taking \(E \sim F\) shows \(m\) descends to \(\mathcal{L}/\sim\); the inequality itself makes it \(1\)-Lipschitz, hence continuous, and it restricts on \(\mathcal{E}/\sim\) to elementary measure. If \(f,g\) are continuous and agree on \(\mathcal{E}/\sim\), pick elementary \(F_n \subset A\) with \([F_n] \to [E]\) by (iii): then \(f([E]) = \lim_n f([F_n]) = \lim_n g([F_n]) = g([E])\).

(i) Cover \(\Gamma := \gamma([a,b])\) by \(n\) cubes of sidelength \(O(1/n)\); total volume \(O(n^{1-d}) \to 0\). In detail, assume \(a < b\) (else \(\Gamma\) is a point) and put \(M := \sup_{[a,b]}|\gamma^{\prime}| < \infty\), finite since \(\gamma^{\prime}\) is continuous on a compact interval. Coordinatewise, \(\gamma(t) - \gamma(s) = \int_s^t \gamma^{\prime}\), so \(|\gamma(t)-\gamma(s)| \leq M(t-s)\) for \(s \leq t\). Split \([a,b]\) into \(n\) intervals \(I_k\) of length \((b-a)/n\) with left endpoints \(t_k\); then \(\gamma(I_k)\) lies in the ball of radius \(M(b-a)/n\) about \(\gamma(t_k)\), hence in the concentric closed cube \(Q_k\) of sidelength \(2M(b-a)/n\). As \(\Gamma = \bigcup_k \gamma(I_k)\), subadditivity and Lemma 1.2.6 give

\begin{equation*} m^*(\Gamma) \leq \sum_{k=1}^n |Q_k| = n \left(\frac{2M(b-a)}{n}\right)^{d} = \big(2M(b-a)\big)^d \, n^{1-d} . \end{equation*}

With \(d \geq 2\) the exponent \(1-d\) is negative, so \(n \to \infty\) forces \(m^*(\Gamma) = 0\), and \(\Gamma\) is measurable with \(m(\Gamma) = 0\) by Lemma 1.2.13(iii). The condition \(d \geq 2\) is needed because for \(d=1\) the bound is \(m^*(\Gamma) \leq 2M(b-a)\), and indeed \(\gamma(t) := t\) on \([0,1]\) has image of measure \(1\).

(ii) If \([0,1]^d \subset \bigcup_n \Gamma_n\) with each \(\Gamma_n\) such a curve, then by (i) and countable subadditivity

\begin{equation*} 1 = m^*\big([0,1]^d\big) \leq \sum_{n=1}^\infty m^*(\Gamma_n) = 0, \end{equation*}

a contradiction.

Take \(A_n := V + q_n\), where \(V \subset [0,1]\) is the Vitali set of Proposition 1.2.18 and \(q_1,\ldots,q_N \in \mathbb{Q}\cap[-1,1]\) are distinct with \(N > 3/m^*(V)\); additivity must fail somewhere along \(A_1, A_1\cup A_2, \ldots\), and the first failure is the required pair.

That proof supplies the inclusions

\begin{equation*} [0,1] \subset \bigcup_{q \in \mathbb{Q}\cap[-1,1]} (V+q), \qquad \bigcup_{q \in \mathbb{Q}\cap[-1,1]} (V+q) \subset [-1,2] \end{equation*}

(its (1.4) and (1.5)), and the translates \(V+q\), \(q \in \mathbb{Q}\), are pairwise disjoint: \(x+q = y+q^{\prime}\) with \(q \neq q^{\prime}\) makes \(x - y\) a nonzero rational, so \(x \neq y\) lie in one coset, against the construction. Also \(m^*\) is translation invariant on arbitrary sets, since \(B \mapsto B+x\) is a volume-preserving bijection between countable box covers of \(A\) and of \(A+x\).

Hence \(c := m^*(V) > 0\): by (1.4), countable subadditivity and translation invariance,

\begin{equation*} 1 = m^*([0,1]) \leq \sum_{q \in \mathbb{Q}\cap[-1,1]} m^*(V+q) = \sum_{q \in \mathbb{Q}\cap[-1,1]} c , \end{equation*}

which is \(0\) if \(c = 0\).

Suppose now \(m^*(A \cup B) = m^*(A) + m^*(B)\) for all disjoint bounded \(A,B \subset \mathbb{R}\). Induction on \(N\) (the union of the first \(N\) is bounded and disjoint from the next) extends this to \(m^*(A_1\cup\cdots\cup A_N) = \sum_{n\leq N} m^*(A_n)\) for pairwise disjoint bounded sets. The \(A_n = V+q_n\) are bounded, pairwise disjoint and each of outer measure \(c\), so with (1.5) and monotonicity,

\begin{equation*} Nc = \sum_{n=1}^N m^*(A_n) = m^*\Big(\bigcup_{n=1}^N (V+q_n)\Big) \leq m^*([-1,2]) = 3 , \end{equation*}

contradicting \(N > 3/c\). So let \(k\) be the largest index with \(m^*(A_1\cup\cdots\cup A_k) = \sum_{n\leq k} m^*(A_n)\); it exists (\(k=1\) works) and \(k < N\) by the display. Then \(E := A_1\cup\cdots\cup A_k\) and \(F := A_{k+1}\) are disjoint bounded sets with \(m^*(E) + m^*(F) = \sum_{n \leq k+1} m^*(A_n) \neq m^*(E \cup F)\).

Take \(E := V \times \{0\}\), with \(V \subset [0,1]\) the non-measurable set of Proposition 1.2.18. Then \(E\) lies in the degenerate box \([0,1]\times\{0\}\), of volume \(1 \cdot 0 = 0\), so \(m^*(E) = 0\) by Lemma 1.2.6 and monotonicity, and \(E\) is measurable with \(m(E) = 0\) by Lemma 1.2.13(iii). But \(\pi(E) = V\) is not measurable.

Everything runs off the canonical form. If \(f = c_1 1_{E_1} + \dots + c_k 1_{E_k}\), the sets \(E_i\) cut \(\mathbb{R}^d\) into at most \(2^k\) disjoint measurable atoms on which \(f\) is constant (the text preceding Definition 1.3.3), so \(f\) has at most \(2^k\) distinct values \(a_1,\dots,a_m\), each level set \(A_p := f^{-1}(\{a_p\})\) is a finite union of atoms and hence measurable, the \(A_p\) are disjoint with union \(\mathbb{R}^d\), and by Lemma 1.3.4 the representations \(f = \sum_p a_p 1_{A_p}\) and \(f = \sum_i c_i 1_{E_i}\) agree:

\begin{equation*} \mathrm{Simp}\int_{\mathbb{R}^d} f(x)\ dx = \sum_{p=1}^m a_p m(A_p). \end{equation*}

Running the same atom construction on the sets of several simple functions at once yields a common refinement: one partition \(\mathbb{R}^d = A_1 \cup \dots \cup A_m\) on which all are simultaneously constant. Arithmetic in \([0,+\infty]\) uses \(0 \cdot \infty = 0\), under which it stays commutative, associative, distributive and monotone.

(i) Concatenating representations, \(f+g = \sum_i c_i 1_{E_i} + \sum_j d_j 1_{F_j}\) is again simple, so Definition 1.3.3 (legitimate by Lemma 1.3.4) gives

\begin{equation*} \begin{aligned} \mathrm{Simp}\int_{\mathbb{R}^d} f + g &= \sum_{i=1}^k c_i m(E_i) + \sum_{j=1}^l d_j m(F_j) \\ &= \mathrm{Simp}\int_{\mathbb{R}^d} f + \mathrm{Simp}\int_{\mathbb{R}^d} g . \end{aligned} \end{equation*}

Likewise \(cf = \sum_i (cc_i) 1_{E_i}\), so by distributivity \(\mathrm{Simp}\int cf = c\sum_i c_i m(E_i) = c\,\mathrm{Simp}\int f\).

(ii) In canonical form the integral \(\sum_p a_p m(A_p)\) is finite iff every term is; moreover the support is the disjoint union \(\bigcup_{p : a_p > 0} A_p\), so by finite additivity

\begin{equation*} m(\{f \ne 0\}) = \sum_{p : a_p > 0} m(A_p), \qquad \{f = \infty\} = \bigcup_{p : a_p = \infty} A_p . \end{equation*}

(\(\Rightarrow\)) If the integral is finite, then \(a_p = \infty\) forces \(m(A_p) = 0\) (as \(\infty \cdot t = \infty\) for \(t>0\)), so \(f\) is finite a.e.; and \(a_p > 0\) forces \(m(A_p) < \infty\), so the support has finite measure. (\(\Leftarrow\)) If \(f\) is finite a.e. with \(m(\{f \ne 0\}) < \infty\), then \(a_p = \infty\) gives \(A_p \subset \{f=\infty\}\) null and \(a_p m(A_p) = \infty\cdot 0 = 0\); \(0 < a_p < \infty\) gives \(m(A_p) \le m(\{f\ne0\}) < \infty\); \(a_p = 0\) gives \(0\).

(iii) \(\sum_p a_p m(A_p) = 0\) iff \(m(A_p) = 0\) for every \(p\) with \(a_p > 0\), iff \(m(\{f \ne 0\}) = \sum_{p : a_p>0} m(A_p) = 0\).

(iv) Let \(N := \{f \ne g\}\), null and hence measurable. With \(f = \sum_p a_p 1_{A_p}\) canonical, \(f 1_{\mathbb{R}^d\setminus N} = \sum_p a_p 1_{A_p \setminus N}\) is simple and \(m(A_p) = m(A_p\setminus N)\) by finite additivity, so

\begin{equation*} \mathrm{Simp}\int_{\mathbb{R}^d} f = \sum_p a_p m(A_p \setminus N) = \mathrm{Simp}\int_{\mathbb{R}^d} f 1_{\mathbb{R}^d \setminus N}, \end{equation*}

and likewise for \(g\). But \(f 1_{\mathbb{R}^d\setminus N}\) and \(g 1_{\mathbb{R}^d\setminus N}\) are the same function.

(v) With \(N := \{f > g\}\) null, (iv) lets us replace \(f,g\) by \(f1_{\mathbb{R}^d\setminus N}, g1_{\mathbb{R}^d\setminus N}\) without changing either integral, so assume \(f \le g\) everywhere. On a common refinement, \(f \equiv \alpha_p\) and \(g \equiv \beta_p\) on \(A_p\), with \(\alpha_p \le \beta_p\) for nonempty \(A_p\) (evaluate at a point; empty ones contribute \(0\)), so by monotonicity

\begin{equation*} \mathrm{Simp}\int_{\mathbb{R}^d} f = \sum_p \alpha_p m(A_p) \le \sum_p \beta_p m(A_p) = \mathrm{Simp}\int_{\mathbb{R}^d} g. \end{equation*}

(vi) \(1_E = 1 \cdot 1_E\), so Definition 1.3.3 gives \(\mathrm{Simp}\int 1_E = 1 \cdot m(E) = m(E)\).

Uniqueness uses only (i) and (vi): for any representation \(f = \sum_i c_i 1_{E_i}\), iterating additivity and then homogeneity gives

\begin{equation*} I(f) = \sum_{i=1}^k c_i I(1_{E_i}) = \sum_{i=1}^k c_i m(E_i) = \mathrm{Simp}\int_{\mathbb{R}^d} f(x)\ dx . \end{equation*}

For \(k = 0\), homogeneity with \(c = 0\) gives \(I(0) = 0 \cdot I(0) = 0 = \mathrm{Simp}\int 0\).

Everything follows from one subtraction-free lemma: if \(f : \mathbb{R}^d \to \mathbb{R}\) is absolutely integrable simple, \(h_1,h_2\) are unsigned simple with finite integrals, and \(f = h_1 - h_2\) wherever both \(h_i\) are finite, then \(\mathrm{Simp}\int f = \mathrm{Simp}\int h_1 - \mathrm{Simp}\int h_2\).

Preliminaries. Writing a complex simple \(h\) over its finitely many level sets, \(h = \sum_p a_p 1_{A_p}\) with \(A_p := h^{-1}(\{a_p\})\) measurable (the atom construction before Lemma 1.3.4), any \(\Phi\) with \(\Phi(0) = 0\) gives \(\Phi \circ h = \sum_p \Phi(a_p)1_{A_p}\), again simple; taking \(\Phi = |\cdot|,\mathrm{Re},\mathrm{Im},(\cdot)_\pm\) shows \(|f|, \mathrm{Re}\,f, \mathrm{Im}\,f, (\mathrm{Re}\,f)_\pm, (\mathrm{Im}\,f)_\pm\) are simple. Each of the last four is dominated by \(|f|\), so by Exercise 1.3.1(v) its unsigned integral is at most \(\mathrm{Simp}\int|f| < \infty\), and Definition 1.3.6 defines \(\mathrm{Simp}\int f \in \mathbb{C}\).

Proof of the lemma: by Exercise 1.3.1(ii) both \(h_i\) are finite off a null set \(N\), and for \(x \notin N\) the four reals \(f_\pm(x), h_1(x), h_2(x)\) satisfy \(f_+ - f_- = h_1 - h_2\), i.e.

\begin{equation*} f_+(x) + h_2(x) = h_1(x) + f_-(x). \end{equation*}

So the unsigned simple functions \(f_+ + h_2\) and \(h_1 + f_-\) agree a.e., and Exercise 1.3.1(iv),(i) give \(\mathrm{Simp}\int f_+ + \mathrm{Simp}\int h_2 = \mathrm{Simp}\int h_1 + \mathrm{Simp}\int f_-\); all four are finite reals, so subtracting and using Definition 1.3.6 gives the claim.

(i) Real case. \(f+g\) is simple with \(|f+g| \le |f|+|g|\), hence absolutely integrable by Exercise 1.3.1(v),(i); and \(f+g = (f_++g_+) - (f_-+g_-)\), so the lemma with \(h_1 := f_++g_+\), \(h_2 := f_-+g_-\) (finite integrals by Exercise 1.3.1(i)) and regrouping finite reals gives \(\mathrm{Simp}\int(f+g) = \mathrm{Simp}\int f + \mathrm{Simp}\int g\). For \(c \in \mathbb{R}\), \(|cf| = |c||f|\) makes \(cf\) absolutely integrable, and: for \(c>0\), \((cf)_\pm = cf_\pm\) gives \(\mathrm{Simp}\int cf = c\,\mathrm{Simp}\int f\) by Exercise 1.3.1(i); for \(c=0\), both sides vanish (\(\mathrm{Simp}\int f\) being a finite real); for \(c=-1\), \((-f)_\pm = f_\mp\) gives \(\mathrm{Simp}\int(-f) = -\mathrm{Simp}\int f\); for \(c<0\), compose the cases \(|c|\) and \(-1\).

Complex case. \(|f+g| \le |f|+|g|\) and \(|cf| = |c||f|\) keep everything absolutely integrable. From \(\mathrm{Re}(f+g) = \mathrm{Re}\,f + \mathrm{Re}\,g\), the same for \(\mathrm{Im}\), Definition 1.3.6 and real additivity, \(\mathrm{Simp}\int(f+g) = \mathrm{Simp}\int f + \mathrm{Simp}\int g\). For (1.11) with \(c \in \mathbb{R}\), \(\mathrm{Re}(cf) = c\,\mathrm{Re}\,f\) and \(\mathrm{Im}(cf) = c\,\mathrm{Im}\,f\) with real homogeneity give \(\mathrm{Simp}\int cf = c\,\mathrm{Simp}\int f\); for \(c = i\), \(\mathrm{Re}(if) = -\mathrm{Im}\,f\) and \(\mathrm{Im}(if) = \mathrm{Re}\,f\) with the case \(c=-1\) give

\begin{equation*} \mathrm{Simp}\int if = -\mathrm{Simp}\int \mathrm{Im}\,f + i\,\mathrm{Simp}\int \mathrm{Re}\,f = i\,\mathrm{Simp}\int f ; \end{equation*}

and general \(c = a+bi\) follows from \(cf = af + b(if)\) with complex additivity. Finally \(\mathrm{Re}(\overline f) = \mathrm{Re}\,f\), \(\mathrm{Im}(\overline f) = -\mathrm{Im}\,f\) give

\begin{equation*} \begin{aligned} \mathrm{Simp}\int \overline f &= \mathrm{Simp}\int \mathrm{Re}\,f - i\,\mathrm{Simp}\int \mathrm{Im}\,f \\ &= \overline{\mathrm{Simp}\int f}, \end{aligned} \end{equation*}

those two unsigned-difference integrals being real.

(ii) If \(f = g\) a.e. then the four pairs \((\mathrm{Re}\,f)_\pm,(\mathrm{Re}\,g)_\pm\) and \((\mathrm{Im}\,f)_\pm,(\mathrm{Im}\,g)_\pm\) agree a.e., so their unsigned integrals agree by Exercise 1.3.1(iv), and Definition 1.3.6 assembles both integrals from these four numbers by the same formula.

(iii) \(|1_E| = 1_E\) has unsigned integral \(m(E)\) (Exercise 1.3.1(vi)), so \(1_E \in \mathrm{Simp}_{\mathrm{abs}}(\mathbb{R}^d)\) exactly when \(m(E) < \infty\), which is how the claim is to be read. For such \(E\), \((1_E)_+ = 1_E\) and \((1_E)_- = \mathrm{Im}\,1_E = 0\), so Definition 1.3.6 gives \(\mathrm{Simp}\int 1_E = m(E) - 0 + i\cdot 0 = m(E)\).

Uniqueness uses only (i) and (iii). Write \(f = \sum_{p=1}^m a_p 1_{A_p}\) over its nonzero values, \(A_p := f^{-1}(\{a_p\})\) disjoint and measurable. Since \(|f| \ge |a_p|1_{A_p}\), Exercise 1.3.1(v),(vi),(i) give \(|a_p| m(A_p) \le \mathrm{Simp}\int|f| < \infty\), so \(m(A_p) < \infty\) and \(1_{A_p} \in \mathrm{Simp}_{\mathrm{abs}}(\mathbb{R}^d)\). Iterating additivity and then (1.11),

\begin{equation*} I(f) = \sum_{p=1}^m a_p I(1_{A_p}) = \sum_{p=1}^m a_p m(A_p), \end{equation*}

and the identical computation for the genuine integral (which obeys (i),(iii)) gives \(\mathrm{Simp}\int f = \sum_p a_p m(A_p)\). For \(m = 0\), (1.11) with \(c = 0\) gives \(I(0) = 0 = \mathrm{Simp}\int 0\).

Give \([0,+\infty]\) its usual topology: \(W\) is open iff \(W \cap [0,+\infty)\) is relatively open and, when \(+\infty \in W\), \(W\) contains a ray \((\lambda,+\infty]\); in particular relatively open subsets of \([0,+\infty)\) are open. Clauses of Lemma 1.3.9 are cited by number, and sup/inf/limsup/liminf are over an at most countable family. One preliminary: if \(f\) is measurable and \(W\) is open then \(f^{-1}(W)\) is measurable, since \(f^{-1}(W \cap [0,+\infty))\) is measurable by Lemma 1.3.9(xi) and \(f^{-1}(\{+\infty\}) = \{f \ge +\infty\}\) by Lemma 1.3.9(vi).

(i) For \(\lambda = +\infty\), \(\{f > \lambda\} = \emptyset\); for \(\lambda < +\infty\), \(\{f > \lambda\} = f^{-1}((\lambda,+\infty])\) is open by continuity, hence measurable. So Lemma 1.3.9(v) applies.

(ii) The constant sequence \(f_n := f\) consists of unsigned simple functions converging pointwise to \(f\), so Definition 1.3.8 applies.

(iii) For \(F := \sup_n f_n\) we have \(F(x) > \lambda\) iff \(f_n(x) > \lambda\) for some \(n\), so \(\{F > \lambda\} = \bigcup_n \{f_n > \lambda\}\) is measurable and Lemma 1.3.9(v) gives \(F\) measurable. For \(G := \inf_n f_n\), \(G(x) < \lambda\) iff \(f_n(x) < \lambda\) for some \(n\), so \(\{G < \lambda\} = \bigcup_n \{f_n < \lambda\}\) is measurable and Lemma 1.3.9(vii) gives \(G\) measurable. Then

\begin{equation*} \limsup_{n \to \infty} f_n = \inf_{N \ge 1} \sup_{n \ge N} f_n , \qquad \liminf_{n \to \infty} f_n = \sup_{N \ge 1} \inf_{n \ge N} f_n \end{equation*}

are measurable by the two cases in turn.

(iv) If \(f = g\) off a null set \(N\), then

\begin{equation*} \begin{aligned} \{f > \lambda\} = \big( \{g > \lambda\} \setminus N \big) \cup \big( \{f > \lambda\} \cap N \big), \end{aligned} \end{equation*}

measurable as the difference of measurable sets (Lemma 1.3.9(v); null sets are measurable) united with a subset of a null set. Lemma 1.3.9(v) again.

(v) \(g := \limsup_n f_n\) is measurable by (iii), and on the co-null set where \(f_n \to f\) the limit equals the limit superior, so \(f = g\) a.e. and (iv) applies.

(vi) If \(U \subset [0,+\infty)\) is relatively open then \(U\) is open in \([0,+\infty]\), so \(W := \varphi^{-1}(U)\) is open by continuity and \((\varphi \circ f)^{-1}(U) = f^{-1}(W)\) is measurable by the preliminary claim. Lemma 1.3.9(xi) then gives \(\varphi \circ f\) measurable.

(vii) By Lemma 1.3.9(iv) pick unsigned simple \(f_n \uparrow f\) and \(g_n \uparrow g\) pointwise. Sums: \(f_n + g_n\) is simple, and for increasing sequences in \([0,+\infty]\), \(\lim(a_n+b_n) = \lim a_n + \lim b_n\) (the usual law when both limits are finite; and \(a_n \to +\infty\) forces \(a_n + b_n \ge a_n \to +\infty\)), so \(f_n + g_n \to f+g\) and Definition 1.3.8 applies. Products: in canonical form

\begin{equation*} f_n g_n = \sum_{p,q} a_p b_q 1_{A_p \cap B_q} \end{equation*}

is simple, and with \(a_n \uparrow a\), \(b_n \uparrow b\) and \(0 \cdot \infty = 0\): (i) if \(a = 0\) then every \(a_n = 0\), so \(a_n b_n = 0 = ab\), and symmetrically for \(b = 0\); (ii) if \(a,b \in (0,\infty)\) this is the product limit law; (iii) if \(a,b > 0\) with \(a = +\infty\), fix \(N\) with \(\beta := b_N > 0\), so \(a_n b_n \ge \beta a_n \to +\infty = ab\) for \(n \ge N\), and symmetrically for \(b = +\infty\). Hence \(f_n g_n \to fg\) and Definition 1.3.8 applies.

Take \(f_n := \frac{1}{n}\lfloor n f\rfloor\), the largest multiple of \(1/n\) below \(f\). If \(0 \le f \le M < \infty\) is measurable, then \(f_n\) has the finitely many values \(k/n\), \(0 \le k \le \lfloor nM\rfloor\), with

\begin{equation*} f_n^{-1}\left(\left\{ \tfrac{k}{n}\right\}\right) = f^{-1}\left( \left[ \tfrac{k}{n}, \tfrac{k+1}{n}\right)\right) \end{equation*}

measurable by Lemma 1.3.9(x), so \(f_n = \sum_k \frac{k}{n} 1_{f^{-1}([k/n,(k+1)/n))}\) is simple (Definition 1.3.2) and bounded by \(M\); and \(0 \le f - f_n < 1/n\) pointwise, so \(\sup_x|f_n - f| \le 1/n \to 0\).

Conversely, let bounded unsigned simple \(f_n \to f\) uniformly (with \(|a-b| := +\infty\) when exactly one of \(a,b\) is \(+\infty\)). Then \(f\) is finite everywhere, since \(f(x_0) = +\infty\) would give \(|f_n(x_0) - f(x_0)| = +\infty\) for all \(n\); it is bounded, since choosing \(n_0\) with \(\sup_x|f_{n_0}-f| \le 1\) and \(M_{n_0}\) bounding \(f_{n_0}\) gives \(f \le M_{n_0}+1\); and it is measurable by Definition 1.3.8, uniform convergence being pointwise.

(\(\Rightarrow\)) If \(f = c_1 1_{E_1} + \dots + c_k 1_{E_k}\) is simple, it is measurable by Exercise 1.3.3(ii), and the \(E_i\) cut \(\mathbb{R}^d\) into at most \(2^k\) atoms on which \(f\) is constant with value \(\sum_{i \in S} c_i\) (the text before Lemma 1.3.4), so \(f\) has at most \(2^k\) values.

(\(\Leftarrow\)) If \(f\) is measurable with distinct values \(a_1,\dots,a_m\), put \(A_p := f^{-1}(\{a_p\})\). Each is measurable, since

\begin{equation*} A_p = \{f \ge a_p\} \cap \{f \le a_p\} \end{equation*}

with \(\{f \ge a_p\}\) measurable by Lemma 1.3.9(vi) and \(\{f \le a_p\}\) the complement of \(\{f > a_p\}\) (Lemma 1.3.9(v)). The \(A_p\) are disjoint and cover \(\mathbb{R}^d\), so \(f = a_1 1_{A_1} + \dots + a_m 1_{A_m}\) pointwise, which is simple by Definition 1.3.2.

Writing \(R(g) := \{(x,t) : 0 \le t < g(x)\}\) for the open region under \(g\), the claim is that

\begin{equation*} E := \{(x,t) : 0 \le t \le f(x)\} = \bigcap_{n=1}^\infty R\!\left(f + \tfrac{1}{n}\right), \end{equation*}

a countable intersection of measurable sets.

For unsigned simple \(g = \sum_{p=1}^m a_p 1_{A_p}\) in canonical form, the slice of \(R(g)\) over \(x \in A_p\) is \([0,a_p)\), so \(R(g) = \bigcup_p A_p \times [0,a_p)\), measurable by Exercise 1.2.22(ii) since each \([0,a_p)\) is empty, a bounded interval, or \([0,+\infty) = \bigcup_{k}[0,k)\). For unsigned measurable \(g\), take simple \(g_n \uparrow g\) (Lemma 1.3.9(iv)): then \(R(g) = \bigcup_n R(g_n)\), because \(t < g_n(x)\) gives \(t < g(x)\), and \(t < \sup_n g_n(x)\) means \(t\) is not an upper bound, so \(t < g_n(x)\) for some \(n\). Hence \(R(g)\) is measurable.

The constant \(\frac1n\) is simple, hence measurable (Exercise 1.3.3(ii)), so \(f + \frac1n\) is unsigned measurable (Exercise 1.3.3(vii)) and \(R(f+\frac1n)\) is measurable. For the displayed identity: if \(0 \le t \le f(x)\) then \(t < f(x) + \frac1n\) for every \(n\) (with \(t < +\infty\) when \(f(x) = +\infty\), as \(t\) is real); conversely \(t < f(x) + \frac1n\) for all \(n\) forces \(t \le f(x)\), trivially if \(f(x) = +\infty\) and by \(n \to \infty\) otherwise.

Extend \(f\) by \(0\) on the null set where it is undefined; no statement is affected, since (i),(ii) ignore null sets, (iii) is invariant under change of extension by Exercise 1.3.3(iv), and in (iv),(v) each preimage changes only inside a null set. Write \(u := (\mathrm{Re}\,f)_+\), \(v := (\mathrm{Re}\,f)_-\), \(p := (\mathrm{Im}\,f)_+\), \(q := (\mathrm{Im}\,f)_-\), so \(\mathrm{Re}\,f = u-v\) and \(\mathrm{Im}\,f = p-q\) are finite-valued. Statements (i) and (ii) coincide by Definition 1.3.11, so it suffices to prove (ii)\(\Rightarrow\)(iii)\(\Rightarrow\)(ii), (iii)\(\Rightarrow\)(iv)\(\Rightarrow\)(iii), and (iv)\(\Leftrightarrow\)(v).

(ii)\(\Rightarrow\)(iii). If \(f_n\) are simple with \(f_n \to f\) a.e., then \(u_n := (\mathrm{Re}\,f_n)_+\) is unsigned simple (compose with \(\Phi(z) = (\mathrm{Re}\,z)_+\), which kills \(0\), as in Exercise 1.3.2), and \(u_n \to u\) a.e. since \(s \mapsto s_+\) is continuous; so \(u\) is unsigned measurable by Lemma 1.3.9(iii). Likewise for \(v,p,q\).

(iii)\(\Rightarrow\)(ii). By Lemma 1.3.9(iv) take bounded unsigned simple \(u_n \uparrow u\), and likewise \(v_n, p_n, q_n\); bounded means finite-valued, so each lies in the complex vector space \(\mathrm{Simp}(\mathbb{R}^d)\) and \(f_n := (u_n - v_n) + i(p_n - q_n)\) is complex simple. All eight values are finite reals, so \(f_n \to (u-v) + i(p-q) = f\) pointwise.

(iii)\(\Rightarrow\)(iv). For \(\lambda \ge 0\), \(\mathrm{Re}\,f(x) > \lambda\) iff \(u(x) > \lambda\) (both force \(u = \mathrm{Re}\,f\)); for \(\lambda < 0\), \(\mathrm{Re}\,f(x) > \lambda\) iff \(v(x) < -\lambda\). Symmetrically \(\{\mathrm{Re}\,f < \lambda\}\) is \(\{v > -\lambda\}\) for \(\lambda \le 0\) and \(\{u < \lambda\}\) for \(\lambda > 0\). All four are measurable by Lemma 1.3.9(v),(vii), and the same holds for \(\mathrm{Im}\,f\) with \(p,q\). Hence for an open rectangle \(Q\),

\begin{equation*} f^{-1}(Q) = \{\mathrm{Re}\,f > a\} \cap \{\mathrm{Re}\,f < b\} \cap \{\mathrm{Im}\,f > c\} \cap \{\mathrm{Im}\,f < d\} \end{equation*}

is measurable; and any open \(U \subset \mathbb{C}\) is the union of the countably many rational-cornered open rectangles \(Q_j\) it contains, so \(f^{-1}(U) = \bigcup_j f^{-1}(Q_j)\) is measurable.

(iv)\(\Rightarrow\)(iii). \(\{u > +\infty\} = \emptyset\) as \(u\) is finite-valued, and for \(0 \le \lambda < \infty\), \(\{u > \lambda\} = \{\mathrm{Re}\,f > \lambda\} = f^{-1}(\{\mathrm{Re}\,z > \lambda\})\) is a preimage of an open half-plane; so \(u\) is measurable by Lemma 1.3.9(v). Likewise \(\{v > \lambda\} = f^{-1}(\{\mathrm{Re}\,z < -\lambda\})\), and \(p,q\) via \(\{\mathrm{Im}\,z > \lambda\}\) and \(\{\mathrm{Im}\,z < -\lambda\}\).

(iv)\(\Leftrightarrow\)(v). \(f^{-1}(\mathbb{C}\setminus S) = \mathbb{R}^d \setminus f^{-1}(S)\), and the Lebesgue measurable sets are closed under complementation.

Each part is an application of Exercise 1.3.7, cited by clause number.

(i) \(f^{-1}(U)\) is open, hence measurable (Lemma 1.2.13(i)), so 1.3.7(iv) applies.

(ii) If \(f = c_1 1_{E_1} + \dots + c_k 1_{E_k}\) is simple, the \(E_i\) cut \(\mathbb{R}^d\) into at most \(2^k\) atoms on which \(f\) is constant, so \(f\) has finitely many values; and \(f\) is the limit of the constant sequence \(f,f,\dots\), hence measurable by Definition 1.3.11. Conversely, if \(f\) is measurable with distinct values \(c_1,\dots,c_k\), then each \(E_i := f^{-1}(\{c_i\})\) is measurable by 1.3.7(v) (singletons are closed), and the \(E_i\) are disjoint and cover \(\mathbb{R}^d\), so \(f = c_1 1_{E_1} + \dots + c_k 1_{E_k}\) is simple (Definition 1.3.2).

(iii) If simple \(g_n \to g\) off a null set \(N^{\prime}\) and \(f = g\) off a null set \(N\), then \(g_n \to f\) off the null set \(N \cup N^{\prime}\).

(iv) Off a null set \(\mathrm{Re}\,f_n \to \mathrm{Re}\,f\) and \(\mathrm{Im}\,f_n \to \mathrm{Im}\,f\), hence so do their positive and negative parts, \(t \mapsto \max(\pm t, 0)\) being continuous. Those parts are unsigned measurable by 1.3.7(iii), so their a.e. limits are unsigned measurable by Exercise 1.3.3(v), and 1.3.7(iii) read backwards gives \(f\) measurable.

(v) \(\varphi^{-1}(U)\) is open by continuity, so \((\varphi \circ f)^{-1}(U) = f^{-1}(\varphi^{-1}(U))\) is measurable by 1.3.7(iv), and 1.3.7(iv) again gives \(\varphi \circ f\) measurable. (Preimages are read inside the co-null set where \(f\) is defined.)

(vi) Take simple \(f_n \to f\) and \(g_n \to g\) a.e., the exceptional set being a union of two null sets. \(\mathrm{Simp}(\mathbb{R}^d)\) is a commutative \(*\)-algebra, so \(f_n + g_n\) and \(f_n g_n\) are simple, and continuity of addition and multiplication on \(\mathbb{C}\) gives \(f_n + g_n \to f+g\) and \(f_n g_n \to fg\) a.e.

Sandwich \(f\) between monotone sequences of simple functions whose gap has vanishing integral. By Exercise 1.1.22 (Riemann = Darboux), for each \(n\) there are piecewise constant \(g_n \le f \le h_n\) on \([a,b]\) with

\begin{equation*} \mathrm{p.c.}\!\int_a^b h_n(x)\,dx - \mathrm{p.c.}\!\int_a^b g_n(x)\,dx \le \frac{1}{n}; \end{equation*}

extend all by zero, which preserves \(g_n \le f \le h_n\) since \(f\) vanishes off \([a,b]\). Each extension is \(\sum_j c_j 1_{I_j}\) over the finitely many intervals of a partition, hence simple (Definition 1.3.2). Put \(G_n := \max(g_1,\dots,g_n)\) and \(H_n := \min(h_1,\dots,h_n)\); refining partitions keeps these piecewise constant, so simple, with \(G_n \le f \le H_n\), \(G_n\) increasing, \(H_n\) decreasing, and all bounded (\(G_n \ge g_1\), \(H_n \le h_1\)). Monotone bounded sequences converge, so \(G := \lim_n G_n\) and \(H := \lim_n H_n\) exist with \(G \le f \le H\), and both are measurable by Definition 1.3.11.

It remains to prove \(G = H\) a.e. Fix \(\varepsilon > 0\) and put \(\phi_n := H_n - G_n \ge 0\), piecewise constant; as \(G_n \ge g_n\) and \(H_n \le h_n\), monotonicity of the piecewise constant integral (Exercise 1.1.21) gives \(\mathrm{p.c.}\!\int \phi_n \le 1/n\). The set \(A_{n,\varepsilon} := \{x \in [a,b] : \phi_n(x) \ge \varepsilon\}\) is a finite union of intervals, hence elementary, and \(\varepsilon 1_{A_{n,\varepsilon}} \le \phi_n\), so by Exercise 1.1.21 again

\begin{equation*} \begin{aligned} \varepsilon \, m(A_{n,\varepsilon}) &= \mathrm{p.c.}\!\int_a^b \varepsilon 1_{A_{n,\varepsilon}}(x)\,dx \\ &\le \mathrm{p.c.}\!\int_a^b \phi_n(x)\,dx \le \frac{1}{n}. \end{aligned} \end{equation*}

Since \(H \le H_n\) and \(G \ge G_n\), we have \(0 \le H - G \le \phi_n\), so \(\{H-G \ge \varepsilon\} \subset A_{n,\varepsilon}\) (the left-hand set lies in \([a,b]\), as \(H = G = 0\) outside), and monotonicity of outer measure with Lemma 1.2.6 gives \(m^*(\{H-G\ge\varepsilon\}) \le 1/(\varepsilon n)\) for every \(n\), hence \(0\). Therefore \(\{H > G\} = \bigcup_k \{H - G \ge 1/k\}\) is null, so \(f = G\) a.e. and \(f\) is measurable by Exercise 1.3.8(iii).

All ten parts are moves on Definition 1.3.12,

\begin{equation*} \begin{aligned} \underline{\int} f = \sup_{0 \le \varphi \le f} \mathrm{Simp}\int \varphi, \qquad \overline{\int} f = \inf_{h \ge f} \mathrm{Simp}\int h , \end{aligned} \end{equation*}

together with two facts from the Venn argument before Lemma 1.3.4: any simple \(\varphi\) is \(\sum_{i=1}^k c_i 1_{A_i}\) over a measurable partition, with \(\mathrm{Simp}\int \varphi = \sum_i c_i m(A_i)\); and over a common such partition \(\varphi 1_E\) (for measurable \(E\)), pointwise minima and maxima, and \(v - u\) for finite simple \(u \le v\), are all simple.

(i) \(f\) competes with itself, giving \(\underline{\int} f \ge \mathrm{Simp}\int f \ge \overline{\int} f\); monotonicity of the simple integral (Exercise 1.3.1(v)) reverses both inequalities.

(ii) \(N := \{f > g\}\) is null, hence measurable (Lemma 1.2.13(iii),(v)). For simple \(\varphi \le f\), the simple function \(\varphi 1_{\mathbb{R}^d \setminus N}\) is \(\le g\) everywhere and equals \(\varphi\) off \(N\), so \(\mathrm{Simp}\int\varphi = \mathrm{Simp}\int \varphi 1_{\mathbb{R}^d\setminus N} \le \underline{\int} g\) by Exercise 1.3.1(iv); take the supremum. Dually, for simple \(h \ge g\), the simple function \(h + \infty\cdot 1_N\) is \(\ge f\) everywhere and equals \(h\) off \(N\), so \(\overline{\int} f \le \mathrm{Simp}\int h\); take the infimum.

(iii) For \(c = 0\), \(cf \equiv 0\) has only the zero minorant, so both sides vanish (\(0 \cdot \infty = 0\)). For \(c > 0\), \(\varphi \mapsto c\varphi\) is a bijection between the simple minorants of \(f\) and those of \(cf\), \(\mathrm{Simp}\int c\varphi = c\,\mathrm{Simp}\int\varphi\) by Exercise 1.3.1(i), and \(t \mapsto ct\) is an increasing bijection of \([0,+\infty]\), hence commutes with suprema.

(iv) Apply (ii) in both directions.

(v) For simple \(\varphi \le f\) and \(\psi \le g\), \(\varphi+\psi \le f+g\) is simple, so Exercise 1.3.1(i) gives \(\underline{\int}(f+g) \ge \mathrm{Simp}\int\varphi + \mathrm{Simp}\int\psi\); take suprema, using \(\sup(A+B) = \sup A + \sup B\) in \([0,+\infty]\).

(vi) Dually with simple \(h_1 \ge f\), \(h_2 \ge g\), using \(\inf(A+B) = \inf A + \inf B\).

(vii) The inequality \(\ge\) is (v) applied to \(f = f1_E + f1_{\mathbb{R}^d\setminus E}\). For \(\le\), a simple \(\varphi \le f\) gives simple \(\varphi 1_E \le f1_E\) and \(\varphi 1_{\mathbb{R}^d\setminus E} \le f1_{\mathbb{R}^d\setminus E}\) (here \(E\) measurable is used), so by Exercise 1.3.1(i)

\begin{equation*} \begin{aligned} \mathrm{Simp}\int \varphi &= \mathrm{Simp}\int \varphi 1_E + \mathrm{Simp}\int \varphi 1_{\mathbb{R}^d \setminus E} \\ &\le \underline{\int} f 1_E + \underline{\int} f 1_{\mathbb{R}^d \setminus E}; \end{aligned} \end{equation*}

take the supremum.

(viii) The \(\min(f,n)\) increase and are \(\le f\), so by (ii) their lower integrals increase to some \(L \le \underline{\int} f\). Conversely, for simple \(\varphi = \sum_i c_i 1_{A_i} \le f\), \(\min(\varphi,n) = \sum_i \min(c_i,n)1_{A_i}\) is simple and \(\le \min(f,n)\), so

\begin{equation*} \begin{aligned} \underline{\int} \min(f,n) \ge \mathrm{Simp}\int \min(\varphi,n) = \sum_{i=1}^k \min(c_i,n)\, m(A_i). \end{aligned} \end{equation*}

Each term increases to \(c_i m(A_i)\): eventually constant if \(c_i < \infty\); and if \(c_i = \infty\), either \(m(A_i) = 0\) and every term is \(0\), or \(n\,m(A_i) \to \infty\). Summing the finitely many terms gives \(L \ge \mathrm{Simp}\int\varphi\), hence \(L \ge \underline{\int} f\).

(ix) Same argument with \(B_n := \{|x| \le n\}\): the \(f1_{B_n}\) increase to \(f\), giving \(L \le \underline{\int} f\); and \(\varphi 1_{B_n} = \sum_i c_i 1_{A_i \cap B_n} \le f1_{B_n}\) is simple with \(m(A_i \cap B_n) \nearrow m(A_i)\) by Exercise 1.2.11(i). Hence \(c_i m(A_i\cap B_n) \nearrow c_i m(A_i)\) (clear for \(c_i < \infty\); for \(c_i = \infty\) the terms all vanish if \(m(A_i) = 0\) and are eventually \(+\infty\) otherwise), so \(L \ge \mathrm{Simp}\int\varphi\) and \(L \ge \underline{\int} f\).

(x) Write \(s := f+g\), bounded by \(M\) and supported in \(S\) with \(m(S) < \infty\); then \(f,g \le s\), and any simple \(\psi \le M\) vanishing off \(S\) has \(\mathrm{Simp}\int\psi \le M\,m(S) < \infty\) (Exercise 1.3.1(v)), so every quantity below is finite and subtraction is legitimate. For \(\le\): given simple \(\varphi \le f\), the function \(h := s - \varphi\) is unsigned simple with \(h = f+g-\varphi \ge g\), and \(h + \varphi = s\) with Exercise 1.3.1(i) gives \(\mathrm{Simp}\int h = \mathrm{Simp}\int s - \mathrm{Simp}\int\varphi\); so \(\mathrm{Simp}\int \varphi + \overline{\int} g \le \mathrm{Simp}\int s\), and take the supremum over \(\varphi\). For \(\ge\): given simple \(h \ge g\), replace \(h\) by \(\min(h,s)\) – still simple, still \(\ge g\) since \(g \le s\), and of no larger integral by Exercise 1.3.1(v) – so assume \(g \le h \le s\). Then \(\varphi := s-h\) is unsigned simple with \(\varphi = f+g-h \le f\), and \(\mathrm{Simp}\int\varphi + \mathrm{Simp}\int h = \mathrm{Simp}\int s\) gives

\begin{equation*} \begin{aligned} \underline{\int} f + \mathrm{Simp}\int h \ge \mathrm{Simp}\int \varphi + \mathrm{Simp}\int h = \mathrm{Simp}\int s ; \end{aligned} \end{equation*}

the infimum over such \(h\) computes \(\overline{\int} g\), since that reduction shows restricting to \(h \le s\) does not change it.

Squeeze \(f\) between simple functions differing by \(2\varepsilon_n\) on the support. Let \(0 \le f \le M\) vanish off a measurable \(E\) with \(m(E) < \infty\). Exercise 1.3.4 gives bounded simple \(f_n \to f\) uniformly; put \(\varepsilon_n := \sup_x |f_n - f| \to 0\) (finite for large \(n\)) and

\begin{equation*} g_n := \max(f_n - \varepsilon_n, 0)\,1_E, \qquad h_n := (f_n + \varepsilon_n)\,1_E , \end{equation*}

both unsigned simple. All three functions vanish off \(E\); on \(E\), \(|f_n - f| \le \varepsilon_n\) gives \(f \ge f_n - \varepsilon_n\) and \(f \ge 0\), so \(f \ge g_n\), while \(f \le f_n + \varepsilon_n = h_n\). Also \(0 \le h_n - g_n \le 2\varepsilon_n 1_E\) (on \(E\) the difference is \(2\varepsilon_n\) where \(f_n \ge \varepsilon_n\) and is \(f_n + \varepsilon_n \le 2\varepsilon_n\) elsewhere), so by Exercise 1.3.1(i),(v) and Definition 1.3.12,

\begin{equation*} \overline{\int} f \le \mathrm{Simp}\int h_n \le \mathrm{Simp}\int g_n + 2\varepsilon_n m(E) \le \underline{\int} f + 2\varepsilon_n m(E). \end{equation*}

Every term is finite, since \(\underline{\int} f \le \overline{\int} f \le \mathrm{Simp}\int M1_E = M\,m(E) < \infty\), so \(n \to \infty\) gives \(\overline{\int} f \le \underline{\int} f\); the reverse always holds.

Both hypotheses are needed. (i) Unbounded with finite-measure support: in \(d=1\) take disjoint \(I_k \subset (0,1]\) with \(m(I_k) = 4^{-k}\) and \(f := \sum_k 2^k 1_{I_k}\), measurable as the pointwise limit of \(\sum_{k \le n} 2^k 1_{I_k}\). If \(h \ge f\) is simple with largest finite value \(C\), pick \(k\) with \(2^k > C\): then \(h = +\infty\) on \(I_k\), so \(\mathrm{Simp}\int h \ge \infty \cdot 4^{-k} = \infty\) and \(\overline{\int} f = \infty\). But a simple \(\varphi \le f\) is finite-valued, say \(\varphi \le C\): if \(C < 2\) then \(\varphi \le C1_{(0,1]}\) gives \(\mathrm{Simp}\int\varphi \le 2\); otherwise take \(2^K \le C < 2^{K+1}\) and \(U_K := \bigcup_{k>K} I_k\), of measure \(4^{-K}/3\), so that \(\varphi \le \sum_{k \le K} 2^k 1_{I_k} + C 1_{U_K}\) and

\begin{equation*} \mathrm{Simp}\int \varphi \le \sum_{k=1}^{\infty} 2^{-k} + 2^{K+1}\frac{4^{-K}}{3} \le 1 + \tfrac{2}{3} \le 3 . \end{equation*}

Hence \(\underline{\int} f \le 3 < \infty = \overline{\int} f\).

(ii) Bounded with infinite-measure support: \(f := \sum_n 2^{-n}1_{[n,n+1)}\) is bounded by \(1/2\) and supported on \([1,+\infty)\). Any simple \(h \ge f\) is positive on \([1,+\infty)\), which its finitely many level sets partition, so one of them has infinite measure and positive value, forcing \(\mathrm{Simp}\int h = \infty\) and \(\overline{\int} f = \infty\). For simple \(\varphi \le f\) with smallest positive value \(c\), \(\varphi \le 2^{-n}\) on \([n,n+1)\) forces \(\varphi = 0\) there once \(2^{-n} < c\), so \(\varphi\) is supported in \([1,N+1)\) and \(\mathrm{Simp}\int\varphi \le \sum_{n \le N} 2^{-n} \le 1\). Hence \(\underline{\int} f \le 1 < \infty = \overline{\int} f\).

For \(\le\), Exercise 1.2.14 gives a measurable \(A \supset E\) with \(m(A) = m^*(E)\), and \(1_A \ge 1_E\) is simple, so \(\overline{\int} 1_E \le \mathrm{Simp}\int 1_A = m^*(E)\) by Exercise 1.3.1(vi). For \(\ge\), a simple \(h \ge 1_E\) written over a measurable partition has \(A := \{h \ge 1\}\) measurable and containing \(E\), with \(h \ge 1_A\), so \(\mathrm{Simp}\int h \ge m(A) \ge m^*(E)\) by Exercise 1.3.1(v); take the infimum.

Dually \(\underline{\int} 1_E = m_*(E) := \sup\{m(A) : A \subset E \text{ measurable}\}\): each such \(1_A\) is a simple minorant, and conversely a simple \(\varphi = \sum_i c_i 1_{A_i} \le 1_E\) over a partition into nonempty atoms has, for each \(i\) with \(c_i > 0\), \(c_i \le 1\) and \(A_i \subset E\) (evaluate at any point of \(A_i\)), so with \(A := \bigcup_{c_i > 0} A_i \subset E\),

\begin{equation*} \mathrm{Simp}\int \varphi = \sum_i c_i m(A_i) \le \sum_{i : c_i > 0} m(A_i) = m(A) \le m_*(E). \end{equation*}

Failure of additivity: take \(E \subset [0,1]\) non-measurable (Proposition 1.2.18) and \(F := [0,1]\setminus E\), also non-measurable. Then \(1_E + 1_F = 1_{[0,1]}\) is simple, so Exercise 1.3.10(i) gives \(\underline{\int}(1_E+1_F) = \overline{\int}(1_E+1_F) = 1\).

For complementary subsets of \([0,1]\), \(m_*(E) = 1 - m^*(F)\): a measurable hull \(B \supset F\) inside \([0,1]\) with \(m(B) = m^*(F)\) (Exercise 1.2.14, intersected with \([0,1]\)) makes \([0,1]\setminus B \subset E\) measurable, giving \(\ge\); and any measurable \(A \subset E\) has \([0,1]\setminus A \supset F\) measurable, so \(m^*(F) \le 1 - m(A)\), giving \(\le\).

Moreover \(m^*(E) + m^*(F) > 1\). The inequality \(\ge\) is subadditivity on \([0,1] = E \cup F\); and if equality held, then with hulls \(A \supset E\), \(B \supset F\) in \([0,1]\),

\begin{equation*} [0,1] \setminus B \subset E \subset A , \qquad m(A) - m([0,1] \setminus B) = m^*(E) - (1 - m^*(F)) = 0 , \end{equation*}

so \(E \setminus ([0,1]\setminus B)\) would be a subset of a measurable null set, hence null and measurable, making \(E\) measurable. Therefore

\begin{equation*} \overline{\int} 1_E + \overline{\int} 1_F = m^*(E) + m^*(F) > 1 = \overline{\int} (1_E + 1_F), \end{equation*}

\begin{equation*} \begin{aligned} \underline{\int} 1_E + \underline{\int} 1_F = 2 - (m^*(E) + m^*(F)) < 1 = \underline{\int} (1_E + 1_F). \end{aligned} \end{equation*}

Write \(\Omega_f := \{(x,t) : 0 \le t \le f(x)\}\), measurable by Exercise 1.3.6, and recall \(\int f = \underline{\int} f\) (Definition 1.3.13).

For simple \(\varphi = \sum_{i=1}^k c_i 1_{A_i}\) over a measurable partition, the fibre above \(x \in A_i\) is \(J_i := [0,c_i]\) (or \([0,+\infty)\) if \(c_i = \infty\)), so \(\Omega_\varphi = \bigcup_i A_i \times J_i\) disjointly, and Exercise 1.2.22(ii) with finite additivity gives

\begin{equation*} m^{d+1}(\Omega_\varphi) = \sum_{i=1}^k m^d(A_i)\, m^1(J_i) = \sum_{i=1}^k c_i\, m(A_i) = \mathrm{Simp}\int_{\mathbb{R}^d} \varphi \end{equation*}

(with \(0\cdot\infty = 0\), the convention of Definition 1.3.3). Every simple \(\varphi \le f\) has \(\Omega_\varphi \subset \Omega_f\), so \(m^{d+1}(\Omega_f) \ge \mathrm{Simp}\int\varphi\), and the supremum over \(\varphi\) gives \(m^{d+1}(\Omega_f) \ge \int f\).

For the reverse, the graph \(G := \Omega_f \setminus \Omega_f^{<}\) is null, where \(\Omega_f^{<} := \{(x,t) : 0 \le t < f(x)\}\): the two fibres differ by the single point \(f(x)\) when \(f(x) < \infty\) and coincide otherwise, so \(G = \{(x,f(x)) : f(x) < \infty\}\). Fix \(R, n\), put \(B_R := \{|x| \le R\}\) and \(E_{j,n} := \{j2^{-n} \le f < (j+1)2^{-n}\}\), measurable by Lemma 1.3.9, disjoint, with union \(\{f < \infty\}\); then

\begin{equation*} G \cap (B_R \times \mathbb{R}) \subset \bigcup_{j=0}^{\infty} \big( (E_{j,n} \cap B_R) \times [j2^{-n}, (j+1)2^{-n}] \big), \end{equation*}

so countable subadditivity, Exercise 1.2.22(ii) and countable additivity of \(m^d\) give \(m^{d+1}(G \cap (B_R\times\mathbb{R})) \le 2^{-n} m(B_R)\), which is \(0\) as \(n \to \infty\); and \(G = \bigcup_R G \cap (B_R\times\mathbb{R})\).

Now take simple \(f_n \uparrow f\) (Lemma 1.3.9(iv)). If \(0 \le t < f(x) = \sup_n f_n(x)\) then \(t < f_n(x)\) for some \(n\), so \(\Omega_f^{<} \subset \bigcup_n \Omega_{f_n}\), an increasing union. Hence by Exercise 1.2.11(i) and the simple case,

\begin{equation*} \begin{aligned} m^{d+1}(\Omega_f) &\le m^{d+1}(\Omega_f^{<}) + m^{d+1}(G) \\ &\le \lim_{n \to \infty} \mathrm{Simp}\int_{\mathbb{R}^d} f_n \le \underline{\int_{\mathbb{R}^d}} f . \end{aligned} \end{equation*}

Uniqueness is the content: the Lebesgue integral obeys (i) by Definition 1.3.13 with Exercise 1.3.10(i), (ii) by Corollary 1.3.14, and (iii),(iv) by Exercise 1.3.10(viii),(ix). Let \(I\) obey (i)-(iv), and put \(B_n := \{|x| \le n\}\); note \(\min(f,n)\) is measurable by Exercise 1.3.3(vi) and \(f1_{B_n}\) by Exercise 1.3.3(vii).

Two tools. If \(u \le v\) are finite-valued unsigned measurable, then \(v-u\) is unsigned measurable with \(u + (v-u) = v\): on \([0,+\infty)\)-valued functions unsigned and complex measurability agree (the remark after Exercise 1.3.7), so \(v-u\) is complex measurable by Exercise 1.3.8(v),(vi) and unsigned measurable by Exercise 1.3.7(iii). Hence (ii) gives \(I(v) = I(u) + I(v-u) \ge I(u)\): \(I\) is monotone on finite-valued functions.

It suffices to treat \(f\) bounded and vanishing off some \(B_m\). Granting that, \(\min(f,n)1_{B_m}\) is bounded by \(n\) and supported in \(B_m\), so letting \(m \to \infty\) with (iv) for \(I\) and Exercise 1.3.10(ix) for the integral gives \(I(\min(f,n)) = \int \min(f,n)\); then \(n \to \infty\) with (iii) and Exercise 1.3.10(viii) gives \(I(f) = \int f\).

So let \(0 \le f \le M\) vanish off \(E := B_m\). As in Exercise 1.3.11, Exercise 1.3.4 supplies bounded simple \(f_k \to f\) uniformly, and with \(\varepsilon_k := \sup_x|f_k - f| \to 0\) the functions

\begin{equation*} g_k := \max(f_k - \varepsilon_k, 0)1_E, \qquad h_k := (f_k + \varepsilon_k)1_E \end{equation*}

are simple, bounded, supported in \(E\), with \(g_k \le f \le h_k\) and \(0 \le h_k - g_k \le 2\varepsilon_k 1_E\). Since \(g_k \le f\) are finite-valued, the difference tool with (ii),(i) gives \(I(f) = \mathrm{Simp}\int g_k + I(f-g_k)\), while \(0 \le f - g_k \le 2\varepsilon_k 1_E\) with monotonicity and (i) gives \(0 \le I(f-g_k) \le 2\varepsilon_k m(E)\); hence

\begin{equation*} \begin{aligned} \mathrm{Simp}\int_{\mathbb{R}^d} g_k \ \le\ I(f) \ \le\ \mathrm{Simp}\int_{\mathbb{R}^d} g_k + 2\varepsilon_k m(E). \end{aligned} \end{equation*}

The same bracket holds for \(\int f\): Definition 1.3.12 gives \(\mathrm{Simp}\int g_k \le \int f \le \mathrm{Simp}\int h_k\), and \(h_k = g_k + (h_k-g_k)\) with Exercise 1.3.1(i),(v) gives \(\mathrm{Simp}\int h_k \le \mathrm{Simp}\int g_k + 2\varepsilon_k m(E)\). All terms are finite, since \(\mathrm{Simp}\int g_k \le M\,m(E) < \infty\), so \(|I(f) - \int f| \le 2\varepsilon_k m(E) \to 0\).

Translation is a simple-integral-preserving bijection between the two classes of minorants. Write \(\tau_y g(x) := g(x+y)\), so \(\tau_y\) is a bijection of unsigned functions with inverse \(\tau_{-y}\), preserving pointwise order.

It preserves measurability: \(\{\tau_y g > \lambda\} = E_\lambda - y\) where \(E_\lambda := \{g > \lambda\}\), and \(E_\lambda\) measurable (Lemma 1.3.9(v)) makes \(E_\lambda - y\) measurable (Exercise 1.2.20); read Lemma 1.3.9(v) backwards. In particular \(\tau_y f\) is measurable. It also preserves simplicity and the simple integral: \(1_E(x+y) = 1_{E-y}(x)\) gives \(\tau_y g = \sum_i c_i 1_{E_i - y}\), simple by Exercise 1.2.20, and by Definition 1.3.3 with Lemma 1.3.4,

\begin{equation*} \begin{aligned} \mathrm{Simp}\!\int_{\mathbb{R}^d} \tau_y g &= \sum_{i=1}^k c_i\, m(E_i - y) = \sum_{i=1}^k c_i\, m(E_i) \\ &= \mathrm{Simp}\!\int_{\mathbb{R}^d} g . \end{aligned} \end{equation*}

Hence \(g \mapsto \tau_y g\) maps \(\mathcal{G} := \{g \text{ simple} : 0 \le g \le f\}\) into \(\mathcal{H} := \{h \text{ simple} : 0 \le h \le \tau_y f\}\), and \(h \mapsto \tau_{-y}h\) maps \(\mathcal{H}\) into \(\mathcal{G}\) for the same reasons; the two are mutually inverse and preserve simple integrals, so the suprema in Definitions 1.3.12-1.3.13 agree: \(\int f(x+y)\,dx = \int f(x)\,dx\).

Composition with \(T^{-1}\) is a bijection between the two classes of minorants that multiplies simple integrals by \(D := |\det T| \in (0,\infty)\). It preserves measurability: \(\{g \circ T^{-1} > \lambda\} = T(E_\lambda)\) with \(E_\lambda := \{g > \lambda\}\) measurable (Lemma 1.3.9(v)), and \(T(E_\lambda)\) is measurable by Exercise 1.2.21. And since \(1_E \circ T^{-1} = 1_{T(E)}\), a simple \(g = \sum_i c_i 1_{E_i}\) has \(g \circ T^{-1} = \sum_i c_i 1_{T(E_i)}\) simple, with

\begin{equation*} \mathrm{Simp}\!\int_{\mathbb{R}^d} g \circ T^{-1} = \sum_{i=1}^k c_i\, m(T(E_i)) = D \sum_{i=1}^k c_i\, m(E_i) = D\, \mathrm{Simp}\!\int_{\mathbb{R}^d} g \end{equation*}

by Definition 1.3.3, Lemma 1.3.4 and Exercise 1.2.21 (legitimate in \([0,+\infty]\) as \(0 < D < \infty\)). As \(T^{-1}\) is a bijection, \(g \le f\) iff \(g \circ T^{-1} \le f \circ T^{-1}\), so \(g \mapsto g\circ T^{-1}\) and \(h \mapsto h \circ T\) are mutually inverse maps between \(\mathcal{G} := \{g \text{ simple} : 0 \le g \le f\}\) and \(\mathcal{H} := \{h \text{ simple} : 0 \le h \le f\circ T^{-1}\}\). Multiplication by \(D\) commutes with suprema in \([0,+\infty]\), so Definitions 1.3.12-1.3.13 give \(\int f(T^{-1}x)\,dx = D\int f(x)\,dx\).

Applying this to \(S := T^{-1}\), with \(S^{-1} = T\) and \(|\det S| = 1/|\det T|\), gives the equivalent form \(\int f(Tx)\,dx = |\det T|^{-1}\int f(x)\,dx\).

Zero-extension carries piecewise constant functions to simple functions with the same integral, squeezing \(\int_{\mathbb{R}} f\) between the two Darboux integrals. Riemann integrability forces \(f\) bounded, so the extension is a genuine unsigned function, measurable by Exercise 1.3.9; and by Exercise 1.1.22 the Riemann integral equals both Darboux integrals. If \(g = \sum_{i=1}^n c_i 1_{I_i}\) is unsigned piecewise constant on \([a,b]\), each \(I_i\) is elementary with \(m(I_i) = |I_i|\) (Lemma 1.2.6), so its zero-extension \(\tilde g\) is simple with

\begin{equation*} \mathrm{Simp}\!\int_{\mathbb{R}} \tilde g(x)\,dx = \sum_{i=1}^n c_i\, m(I_i) = \mathrm{p.c.}\!\int_a^b g(x)\,dx \end{equation*}

by Definition 1.3.3 and Lemma 1.3.4.

For \(\ge\): a piecewise constant \(g \le f\) may be replaced by \(\max(g,0)\), still piecewise constant, still \(\le f\) (as \(f \ge 0\)) and of no smaller integral (Exercise 1.1.21(2)), so the lower Darboux integral is unchanged by restricting to unsigned minorants; for such \(g\), \(0 \le \tilde g \le f\) on \(\mathbb{R}\), so \(\mathrm{p.c.}\!\int_a^b g = \mathrm{Simp}\!\int_{\mathbb{R}}\tilde g \le \int_{\mathbb{R}} f\), and the supremum gives \(\int_a^b f \le \int_{\mathbb{R}} f\).

For \(\le\): a piecewise constant \(h \ge f\) is automatically unsigned, and \(\tilde h \ge f\) on \(\mathbb{R}\) is simple, so Exercise 1.3.10(i),(ii) give \(\int_{\mathbb{R}} f \le \mathrm{Simp}\!\int_{\mathbb{R}}\tilde h = \mathrm{p.c.}\!\int_a^b h\); the infimum gives \(\int_{\mathbb{R}} f \le \int_a^b f\).

Both parts are Markov’s inequality (Lemma 1.3.15) applied to \(I := \int_{\mathbb{R}^d} f\): the sets \(\{f \ge \lambda\}\) are measurable by Lemma 1.3.9(vi), and \(m(\{f \ge \lambda\}) \le I/\lambda\) for \(0 < \lambda < \infty\).

(i) If \(I < \infty\), then \(A := \{f = +\infty\} = \bigcap_n \{f \ge n\}\) is measurable with \(m(A) \le m(\{f \ge n\}) \le I/n \to 0\), so \(f\) is finite a.e. The converse fails for \(f \equiv 1\), which is simple and finite everywhere with

\begin{equation*} \begin{aligned} \int_{\mathbb{R}^d} 1\,dx = \mathrm{Simp}\!\int_{\mathbb{R}^d} 1_{\mathbb{R}^d}(x)\,dx = m(\mathbb{R}^d) = +\infty \end{aligned} \end{equation*}

by Exercise 1.3.10(i) and Exercise 1.3.1(vi).

(ii) If \(f = 0\) a.e., then Exercise 1.3.10(iv),(i) give \(I = \int 0 = \mathrm{Simp}\!\int 0 = 0\). Conversely, if \(I = 0\), Markov with \(\lambda = 1/n\) makes each \(E_n := \{f \ge 1/n\}\) null, and \(\{f \ne 0\} = \bigcup_n E_n\) is then null by countable subadditivity.

Rearranging into a subtraction-free identity reduces everything to Corollary 1.3.14; all pointwise identities below need only hold a.e., since the integral ignores null sets (Exercise 1.3.10(iv)).

The operations stay in \(L^1\): \(f+g\) and \(cf\) are measurable by Exercise 1.3.8(vi) (the constant \(c\) being measurable by Exercise 1.3.8(i)) and \(\overline f\) by Exercise 1.3.8(v); and \(|f+g| \le |f|+|g|\) with Exercise 1.3.10(ii) and Corollary 1.3.14, \(|cf| = |c||f|\) with Exercise 1.3.10(iii), and \(|\overline f| = |f|\) keep the \(L^1\) norms finite. For real absolutely integrable \(f\), \(\int f_\pm \le \int|f| < \infty\), so every difference below is of finite reals.

Real additivity. With \(h := f+g\), rearrange \(h_+ - h_- = (f_+-f_-)+(g_+-g_-)\) into the unsigned identity

\begin{equation*} h_+ + f_- + g_- = h_- + f_+ + g_+ , \end{equation*}

all six functions unsigned measurable (Exercise 1.3.7(iii)). Corollary 1.3.14 turns this into the same identity for the integrals; all six are finite, so rearranging back gives \(\int(f+g) = \int f + \int g\) by (1.12).

Real homogeneity. For \(c \ge 0\), \((cf)_\pm = cf_\pm\), so Exercise 1.3.10(iii) with (1.12) gives \(\int cf = c\int f\); for \(c = -1\), \((-f)_\pm = f_\mp\) gives \(\int(-f) = -\int f\); a general \(c < 0\) is \((-1)\cdot|c|\).

Complex case. \(\mathrm{Re}\,f\) and \(\mathrm{Im}\,f\) are measurable (Exercise 1.3.7(iii)) and absolutely integrable, being dominated by \(|f|\). Additivity follows from \(\mathrm{Re}(f+g) = \mathrm{Re}\,f + \mathrm{Re}\,g\), the same for \(\mathrm{Im}\), Definition 1.3.17 and the real case. For homogeneity write \(c = a+ib\) and \(X := \int \mathrm{Re}\,f\), \(Y := \int\mathrm{Im}\,f\); from \(\mathrm{Re}(cf) = a\,\mathrm{Re}\,f - b\,\mathrm{Im}\,f\) and \(\mathrm{Im}(cf) = a\,\mathrm{Im}\,f + b\,\mathrm{Re}\,f\), real linearity gives \(\int\mathrm{Re}(cf) = aX-bY\) and \(\int\mathrm{Im}(cf) = aY+bX\), so Definition 1.3.17 gives

\begin{equation*} \int_{\mathbb{R}^d} cf = (aX - bY) + i(aY + bX) = (a+ib)(X+iY) = c \int_{\mathbb{R}^d} f . \end{equation*}

Finally \(\mathrm{Re}\,\overline f = \mathrm{Re}\,f\) and \(\mathrm{Im}\,\overline f = -\mathrm{Im}\,f\) give \(\int\overline f = X - iY = \overline{\int f}\).

Each of the three substitutions \(\Phi\) – \(f \mapsto f(\cdot+y)\), \(f \mapsto f \circ T^{-1}\), and extension by zero – is a precomposition (or a zero-extension), hence commutes with every pointwise operation fixing \(0\): \(\mathrm{Re}(\Phi f) = \Phi(\mathrm{Re}\,f)\), \((\Phi u)_\pm = \Phi(u_\pm)\), \(|\Phi f| = \Phi|f|\). So with \(u_1 := (\mathrm{Re}\,f)_+\), \(u_2 := (\mathrm{Re}\,f)_-\), \(u_3 := (\mathrm{Im}\,f)_+\), \(u_4 := (\mathrm{Im}\,f)_-\), which are unsigned measurable by Exercise 1.3.7(iii) with finite integrals (each is \(\le |f|\)), Definition 1.3.17 reads

\begin{equation*} \int_{\mathbb{R}^d} f = \Big( \int u_1 - \int u_2 \Big) + i \Big( \int u_3 - \int u_4 \Big) , \end{equation*}

and the unsigned result applies termwise.

Exercise 1.3.15. \(\tau_y f := f(\cdot+y)\) is defined a.e., since \(E - y\) has full measure when \(E\) does (Exercise 1.2.20); the \(\tau_y u_j\) are exactly its four unsigned parts, measurable by Exercise 1.3.15, so \(\tau_y f\) is measurable by Exercise 1.3.7(iii), and \(\int|\tau_y f| = \int \tau_y|f| = \int|f| < \infty\). Applying Exercise 1.3.15 to each \(u_j\) and rearranging the eight finite terms gives \(\int f(x+y)\,dx = \int f(x)\,dx\).

Exercise 1.3.16. \(f \circ T^{-1}\) is defined a.e. (the complement of \(T(E)\) is \(T(\mathbb{R}^d\setminus E)\), null by Exercise 1.2.21), measurable as above, and \(\int|f\circ T^{-1}| = D\int|f| < \infty\) with \(D := |\det T|\). Applying Exercise 1.3.16 to each \(u_j\),

\begin{equation*} \int_{\mathbb{R}^d} f(T^{-1}(x))\,dx = D\Big( \int u_1 - \int u_2 \Big) + iD\Big( \int u_3 - \int u_4 \Big) = D \int_{\mathbb{R}^d} f(x)\,dx , \end{equation*}

all terms finite; replacing \(T\) by \(T^{-1}\) gives \(\int f(Tx)\,dx = |\det T|^{-1}\int f(x)\,dx\).

Exercise 1.3.17. Riemann integrability of a complex \(f\) means that of \(\mathrm{Re}\,f\) and \(\mathrm{Im}\,f\), and zero-extension commutes with \(\mathrm{Re}\) and \(\mathrm{Im}\), so it suffices to take \(u : [a,b] \to \mathbb{R}\) Riemann integrable and prove \(\int_{\mathbb{R}}\tilde u = \int_a^b u\). Here \(\tilde u\) is measurable (Exercise 1.3.9) with \(|\tilde u| \le M1_{[a,b]}\), hence absolutely integrable by Exercise 1.3.10(i),(ii). Both \(u_\pm\) are Riemann integrable: given \(\varepsilon > 0\), Exercise 1.1.22 supplies piecewise constant \(g \le u \le h\) with \(\mathrm{p.c.}\!\int_a^b(h-g) \le \varepsilon\); then \(g_+ \le u_+ \le h_+\) are piecewise constant with \(0 \le h_+ - g_+ \le h-g\) (as \(t \mapsto \max(t,0)\) is \(1\)-Lipschitz), so \(\mathrm{p.c.}\!\int_a^b(h_+-g_+) \le \varepsilon\) by Exercise 1.1.21, and \(\varepsilon \to 0\) gives Darboux, hence Riemann, integrability; apply this also to \(u_- = (-u)_+\), with \(-u\) Riemann integrable by Exercise 1.1.24(1). Since \((\tilde u)_\pm = \widetilde{u_\pm}\), the unsigned Exercise 1.3.17 and (1.12) give

\begin{equation*} \begin{aligned} \int_{\mathbb{R}} \tilde u(x)\,dx = \int_a^b u_+(x)\,dx - \int_a^b u_-(x)\,dx = \int_a^b u(x)\,dx , \end{aligned} \end{equation*}

the last step by Exercise 1.1.24(1).

The integral is the sum, and both sides are controlled by the single identity \(\int_{\mathbb{R}} |f| = \sum_{n \in \mathbb{Z}} |c_n|\) in \([0,+\infty]\).

Put

\begin{equation*} f_N := f 1_{[-N,\,N+1)} = \sum_{|n| \le N} c_n 1_{[n,n+1)}, \qquad g_N := \sum_{|n| \le N} |c_n| 1_{[n,n+1)} . \end{equation*}

Each \(f_N\) is simple and \(f_N \to f\) pointwise, so \(f\) (hence \(|f|\)) is measurable by Exercise 1.3.8(iv),(v). Now \(g_N \le |f|\) pointwise, while \(|f| 1_{\{|x| \le N\}} \le g_N\) because \(|x| \le N\) forces \(\lfloor x \rfloor \in \{-N,\dots,N\}\); so by Definition 1.3.12, monotonicity, and compatibility with the simple integral (Exercise 1.3.10(i),(ii)),

\begin{align*} \sum_{|n| \le N} |c_n| = \mathrm{Simp}\!\int_{\mathbb{R}} g_N &\le \int_{\mathbb{R}} |f| , \\ \int_{\mathbb{R}} |f(x)| 1_{\{|x| \le N\}}\, dx &\le \sum_{|n| \le N} |c_n| . \end{align*}

Letting \(N \to \infty\) in the second line by vertical truncation (Exercise 1.3.10(ix)) gives \(\int_{\mathbb{R}} |f| = \sum_{n \in \mathbb{Z}} |c_n|\). By Definition 1.3.17, \(f\) is absolutely integrable exactly when \(\sum_{n} |c_n| < \infty\), i.e. exactly when \(\sum_n c_n\) converges absolutely.

Assume that. Then \(f_N \in L^1\) with \(\int_{\mathbb{R}} f_N = \sum_{|n| \le N} c_n\), by linearity (Exercise 1.3.19) and \(\int_{\mathbb{R}} 1_{[n,n+1)} = m([n,n+1)) = 1\). The tail \(h_N := f - f_N \in L^1\) has the same form as \(f\) with coefficients \(c_n 1_{|n| > N}\), so the identity above applied to it gives \(\int_{\mathbb{R}} |h_N| = \sum_{|n| > N} |c_n|\). Splitting a complex \(w \in L^1\) into real and imaginary parts and each of those into positive and negative parts gives \(|\int w| \le 2\int |w|\) (Corollary 1.3.14 and (1.12)); hence

\begin{align*} \Big| \int_{\mathbb{R}} f(x)\,dx - \sum_{|n| \le N} c_n \Big| &= \Big| \int_{\mathbb{R}} h_N(x)\,dx \Big| \\ &\le 2 \sum_{|n| > N} |c_n| \longrightarrow 0 . \end{align*}

So the symmetric partial sums converge to \(\int_{\mathbb{R}} f(x)\,dx\), and absolute convergence makes \(\sum_{n \in \mathbb{Z}} c_n\) that same limit, independently of ordering.

Both identities reduce to the pointwise splitting \(h = h 1_E + h 1_F\), where \(h : \mathbb{R}^d \to \mathbb{C}\) is the extension of \(f\) by zero, so that \(\int_{E \cup F} f = \int_{\mathbb{R}^d} h\) by the definition of the localised integral.

By hypothesis \(h\) is measurable with \(\int_{\mathbb{R}^d} |h| < \infty\); since \(1_E\) is measurable and \(|h 1_E| \le |h|\), Exercise 1.3.8(vi) and monotonicity (Exercise 1.3.10(ii)) place \(h 1_E\) and \(h 1_F\) in \(L^1(\mathbb{R}^d \to \mathbb{C})\).

First identity. The extension by zero of \(f|_E\) is \(h 1_E\) (on \(E \subset E \cup F\) we have \(h = f\)); the extension by zero of \(f 1_E : E \cup F \to \mathbb{C}\) is also \(h 1_E\) (as \(h\) vanishes off \(E \cup F\)). Both sides are thus \(\int_{\mathbb{R}^d} h 1_E\), and in particular \(f|_E\) is absolutely integrable.

Second identity. Disjointness gives \(h = h 1_E + h 1_F\) pointwise (check the three cases \(x \in E\), \(x \in F\), \(x \notin E \cup F\)), so by linearity of the absolutely convergent integral (Exercise 1.3.19),

\begin{equation*} \int_{E \cup F} f = \int_{\mathbb{R}^d} h 1_E + \int_{\mathbb{R}^d} h 1_F = \int_E f(x)\,dx + \int_F f(x)\,dx , \end{equation*}

the last step by the first identity and its analogue with \(F\) in place of \(E\).

Lusin’s theorem (Theorem 1.3.28) survives both relaxations: patch it across a partition of \(\mathbb{R}^d\) into open shells, then conjugate a merely measurable \(f\) by the bounded homeomorphism \(\Phi(z) := z/(1+|z|^2)^{1/2}\).

(i) \(f\) locally absolutely integrable. Fix \(\varepsilon > 0\), put \(Q_0 := \emptyset\), \(Q_n := [-n,n]^d\), and

\begin{equation*} U_n := (-n,n)^d \setminus Q_{n-1} \qquad (n \ge 1), \end{equation*}

open, bounded and pairwise disjoint. The leftover \(S := \mathbb{R}^d \setminus \bigcup_n U_n\) lies in \(\bigcup_n \partial Q_n\) (if \(n\) is least with \(x \in Q_n\) then \(x \notin U_n\) forces \(x \in \partial Q_n\)), and each \(\partial Q_n\) is a finite union of boxes with a degenerate side, so \(S\) is null. Each \(f 1_{U_n}\) is absolutely integrable by hypothesis, so Theorem 1.3.28 with parameter \(\varepsilon/2^n\) supplies measurable \(E_n\), \(m(E_n) \le \varepsilon/2^n\), off which \(f 1_{U_n}\) is continuous. Put

\begin{equation*} E := S \cup \bigcup_{n \ge 1} (E_n \cap U_n), \qquad m(E) \le \varepsilon . \end{equation*}

Given \(x \notin E\), let \(n\) be the unique index with \(x \in U_n\) and pick \(\rho > 0\) with \(B(x,\rho) \subset U_n\). Then \(f = f 1_{U_n}\) on \(B(x,\rho)\), and \(B(x,\rho) \setminus E \subset \mathbb{R}^d \setminus E_n\) (a point of \(U_n \cap E_n\) lies in \(E\)); since \(x \notin E_n\), continuity of \((f 1_{U_n})|_{\mathbb{R}^d \setminus E_n}\) at \(x\) therefore gives continuity of \(f|_{\mathbb{R}^d \setminus E}\) at \(x\).

(ii) \(f\) measurable. In polar form \(r \mapsto r/(1+r^2)^{1/2}\) is an increasing bijection of \([0,\infty)\) onto \([0,1)\), so \(\Phi\) is a continuous bijection of \(\mathbb{C}\) onto the open disc \(D\) with continuous inverse \(\Psi(w) = w/(1-|w|^2)^{1/2}\). Then \(g := \Phi \circ f\) is measurable (Exercise 1.3.8(v)) and \(|g| < 1\), so \(\int_{\mathbb{R}^d} |g| 1_B \le m(B) < \infty\) for every bounded measurable \(B\) (Exercise 1.3.10(i),(ii)): \(g\) is locally absolutely integrable. By (i) there is \(E\) with \(m(E) \le \varepsilon\) and \(g|_{\mathbb{R}^d \setminus E}\) continuous, whence

\begin{equation*} f|_{\mathbb{R}^d \setminus E} = \Psi \circ \big( g|_{\mathbb{R}^d \setminus E} \big) \end{equation*}

is continuous.

(iii) \(f\) finite only a.e. Let \(Z\) be the null set where \(f\) is undefined or infinite and \(f_0 := f 1_{\mathbb{R}^d \setminus Z}\), measurable (Exercise 1.3.8(iii)) and finite everywhere. Apply (ii) to get \(E_0\) with \(m(E_0) \le \varepsilon\); then \(E := E_0 \cup Z\) has \(m(E) \le \varepsilon\) and \(f = f_0\) off \(E\), so \(f|_{\mathbb{R}^d \setminus E}\) is a restriction of the continuous \(f_0|_{\mathbb{R}^d \setminus E_0}\).

Take \(f_n\) continuous with \(\|h_n^{\prime} - f_n\|_{L^1} \le 1/(n2^{n+1})\), where \(h_n^{\prime}\) is \(f\) truncated to \(B(0,n)\) and to the set where it is bounded.

(\(\Leftarrow\)) Continuous functions are measurable (Exercise 1.3.8(i)) and a pointwise a.e. limit of measurable functions is measurable (Exercise 1.3.8(iv)), so \(f\) is measurable.

(\(\Rightarrow\)) Fix \(n \ge 1\) and put \(h := f 1_{B(0,n)}\), measurable (Exercise 1.3.8(vi)) and supported in a ball of finite measure. By Exercise 1.3.25(ii) with parameter \(2^{-n-1}\) there are measurable \(E\) with \(m(E) \le 2^{-n-1}\) and \(M < \infty\) with \(|h| \le M\) on \(B(0,n) \setminus E\). Then \(h^{\prime} := h 1_{\mathbb{R}^d \setminus E}\) satisfies \(|h^{\prime}| \le M 1_{B(0,n)}\), so

\begin{equation*} \int_{\mathbb{R}^d} |h^{\prime}(x)|\, dx \le M\, m(B(0,n)) < \infty \end{equation*}

by Exercise 1.3.10(i),(ii),(iii): \(h^{\prime} \in L^1\). Theorem 1.3.20(iii) supplies a continuous compactly supported \(f_n\) with \(\|h^{\prime} - f_n\|_{L^1(\mathbb{R}^d)} \le 1/(n2^{n+1})\), and Markov’s inequality (Lemma 1.3.15) at level \(\lambda = 1/n\) gives

\begin{equation*} m\big(\{ |h^{\prime} - f_n| \ge 1/n \}\big) \le n \|h^{\prime} - f_n\|_{L^1} \le 2^{-n-1} . \end{equation*}

On \(B(0,n) \setminus E\) we have \(f = h = h^{\prime}\), so \(F_n := \{x \in B(0,n) : |f(x) - f_n(x)| \ge 1/n\}\) satisfies \(F_n \subset E \cup \{|h^{\prime} - f_n| \ge 1/n\}\) and hence \(m(F_n) \le 2^{-n}\), the claim of the hint.

Put \(G := \bigcap_{N \ge 1} \bigcup_{n \ge N} F_n\). Countable subadditivity gives \(m(G) \le \sum_{n \ge N} 2^{-n} = 2^{-N+1}\) for every \(N\), so \(m(G) = 0\). If \(x \notin G\), then \(x \notin F_n\) for all \(n \ge\) some \(N\), and \(x \in B(0,n)\) once \(n > |x|\); for such \(n\) the definition of \(F_n\) forces \(|f(x) - f_n(x)| < 1/n\), so \(f_n(x) \to f(x)\). Thus \(f_n \to f\) pointwise off the null set \(G\).

Method (2) for Lusin. Let \(f\) be measurable and \(\varepsilon > 0\). Take continuous \(f_n \to f\) a.e. as above; Egorov’s theorem (Theorem 1.3.26) gives measurable \(A\) with \(m(A) \le \varepsilon\) on whose complement the convergence is locally uniform. Fix \(x_0 \notin A\) and set \(B := B(x_0,1) \cap (\mathbb{R}^d \setminus A)\), a bounded subset of \(\mathbb{R}^d \setminus A\); by Definition 1.3.21 the convergence is uniform on \(B\), so \(f|_B\), a uniform limit of the continuous \(f_n|_B\), is continuous. Since \(B\) is a relatively open neighbourhood of \(x_0\) in \(\mathbb{R}^d \setminus A\), \(f|_{\mathbb{R}^d \setminus A}\) is continuous at \(x_0\), and \(x_0\) was arbitrary.

(i) Take \(R := n+1\) with \(n\) chosen below. Write \(A_n := \{|x| \le n\}\). The function \(|f|\) is unsigned measurable (Exercise 1.3.8(v)) with \(\int_{\mathbb{R}^d} |f| < \infty\), so divisibility (Exercise 1.3.10(vii)) gives

\begin{equation*} \int_{\mathbb{R}^d} |f| = \int_{\mathbb{R}^d} |f| 1_{A_n} + \int_{\mathbb{R}^d} |f| 1_{\mathbb{R}^d \setminus A_n} , \end{equation*}

all three terms being finite by monotonicity (Exercise 1.3.10(ii)). Vertical truncation (Exercise 1.3.10(ix)) sends \(\int |f| 1_{A_n} \to \int |f|\), so subtraction (legitimate, all terms finite) gives \(\int |f| 1_{\mathbb{R}^d \setminus A_n} \to 0\). Fix \(n\) with this tail at most \(\varepsilon\); since \(\mathbb{R}^d \setminus B(0,n+1) \subset \mathbb{R}^d \setminus A_n\), monotonicity gives \(\int_{\mathbb{R}^d \setminus B(0,R)} |f| \le \varepsilon\).

(ii) Take \(E := \{x : |f(x)| > n_0\}\) for a suitable \(n_0\); then \(|f| \le n_0\) on all of \(\mathbb{R}^d \setminus E\), so \(M := n_0\) serves for every \(R\). Let \(S\) be a set of finite measure off which \(f\) vanishes, and put \(S_n := \{|f| > n\} = f^{-1}(\{|z| > n\})\), measurable by Exercise 1.3.7(iv). These decrease, and

\begin{equation*} S_1 \subset \{ f \ne 0 \} \subset S, \qquad \bigcap_{n \ge 1} S_n = \emptyset , \end{equation*}

the first giving \(m(S_1) \le m^*(S) < \infty\) and the second holding because \(f\) is complex-valued, hence finite everywhere. Downward monotone convergence (Exercise 1.2.11(ii)), whose finite-measure hypothesis is met by \(m(S_1) < \infty\), yields \(m(S_n) \to m(\emptyset) = 0\); choose \(n_0\) with \(m(S_{n_0}) \le \varepsilon\).

The three axioms of Definition 1.4.1 hold, the union axiom by de Morgan.

Empty set: \(\emptyset\) is the empty union of boxes, hence elementary, hence in \(\mathcal{E}[\mathbb{R}^d]\).

Complement: if \(E\) is elementary then \(E^c\) is co-elementary; if \(E = F^c\) with \(F\) elementary then \(E^c = F\) is elementary. Either way \(E^c \in \mathcal{E}[\mathbb{R}^d]\).

Finite unions: let \(E, F \in \mathcal{E}[\mathbb{R}^d]\) and use Exercise 1.1.1 (elementary sets are closed under \(\cup\), \(\cap\), \(\setminus\)).

• (i) \(E, F\) elementary: \(E \cup F\) is elementary.
• (ii) \(E\) elementary, \(F = G^c\) with \(G\) elementary: \(E \cup F = (E^c \cap G)^c = (G \setminus E)^c\), co-elementary.
• (iii) \(E = G^c\), \(F = H^c\) with \(G, H\) elementary: \(E \cup F = (G \cap H)^c\), co-elementary.

In each case \(E \cup F \in \mathcal{E}[\mathbb{R}^d]\), so \(\mathcal{E}[\mathbb{R}^d]\) is a Boolean algebra on \(\mathbb{R}^d\).

Everything follows from \(Y \setminus (E \cap Y) = E^c \cap Y\) and \((E \cap Y) \cup (F \cap Y) = (E \cup F) \cap Y\), complements being taken in \(Y\) on the left and in \(X\) on the right.

Members of \(\mathcal{B}\upharpoonright_Y\) are subsets of \(Y\), and \(\emptyset = \emptyset \cap Y \in \mathcal{B}\upharpoonright_Y\). For \(A = E \cap Y\) with \(E \in \mathcal{B}\), the first identity (distribute: \(Y \cap (E^c \cup Y^c) = Y \cap E^c\)) exhibits \(Y \setminus A\) as \(E^c \cap Y \in \mathcal{B}\upharpoonright_Y\), since \(E^c \in \mathcal{B}\); the second identity exhibits a union of two members as \((E \cup F) \cap Y \in \mathcal{B}\upharpoonright_Y\). So \(\mathcal{B}\upharpoonright_Y\) is a Boolean algebra on \(Y\) (no measurability of \(Y\) used).

Now suppose \(Y \in \mathcal{B}\). The two right-hand descriptions are the same collection by definition, so it suffices to prove \(\mathcal{B}\upharpoonright_Y = \mathcal{B} \cap 2^Y\).

• (\(\subset\)) For \(A = E \cap Y\) with \(E \in \mathcal{B}\): a Boolean algebra is closed under finite intersections (\(E \cap Y = (E^c \cup Y^c)^c\)), so \(A \in \mathcal{B}\), and \(A \subset Y\).
• (\(\supset\)) For \(A \in \mathcal{B}\) with \(A \subset Y\): \(A = A \cap Y \in \mathcal{B}\upharpoonright_Y\).

The atoms are recoverable from the algebra alone, as its minimal non-empty members, so only the labelling can differ.

Two facts, each immediate from disjointness. (A) Distinct indices give distinct atoms: \(A_\alpha = A_\beta\) with \(\alpha \ne \beta\) would force \(A_\alpha = A_\alpha \cap A_\beta = \emptyset\). (B) If \(E = \bigcup_{\alpha \in J} A_\alpha\) meets \(A_\beta\), then any \(x \in E \cap A_\beta\) lies in some \(A_\alpha\) with \(\alpha \in J\), whence \(\alpha = \beta\) and \(A_\beta \subset E\). Both hold for the primed family too.

(\(\Leftarrow\)) If \(A^{\prime}_{\phi(\alpha)} = A_\alpha\) for a bijection \(\phi\), the two families have exactly the same members, so \(\bigcup_{\alpha \in J} A_\alpha = \bigcup_{\alpha^{\prime} \in \phi(J)} A^{\prime}_{\alpha^{\prime}}\) for every \(J \subset I\), and the two algebras coincide.

(\(\Rightarrow\)) Suppose \(\mathcal{A} := \mathcal{A}((A_\alpha)) = \mathcal{A}((A^{\prime}_{\alpha^{\prime}}))\) and fix \(\alpha \in I\). Then \(A_\alpha = \bigcup_{\alpha^{\prime} \in J^{\prime}} A^{\prime}_{\alpha^{\prime}}\) with \(J^{\prime} \ne \emptyset\) (as \(A_\alpha \ne \emptyset\)); pick \(\alpha^{\prime} \in J^{\prime}\), so \(A^{\prime}_{\alpha^{\prime}} \subset A_\alpha\). Reversing roles, \(A^{\prime}_{\alpha^{\prime}} = \bigcup_{\beta \in J} A_\beta\) with \(J \ne \emptyset\), and every \(\beta \in J\) has \(\emptyset \ne A_\beta \subset A_\alpha\), forcing \(\beta = \alpha\) by disjointness; hence \(J = \{\alpha\}\) and

\begin{equation*} A^{\prime}_{\alpha^{\prime}} = A_\alpha . \end{equation*}

By (A) this \(\alpha^{\prime}\) is unique, so \(\phi(\alpha) := \alpha^{\prime}\) is well defined and satisfies \(A^{\prime}_{\phi(\alpha)} = A_\alpha\). It is injective by (A) again, and surjective because the same argument run from a fixed \(\alpha^{\prime} \in I^{\prime}\) produces \(\alpha\) with \(A_\alpha = A^{\prime}_{\alpha^{\prime}}\), i.e. \(\phi(\alpha) = \alpha^{\prime}\).

The atoms are

\begin{equation*} A_x := \bigcap \{ E \in \mathcal{B} : x \in E \} \qquad (x \in X), \end{equation*}

the smallest \(\mathcal{B}\)-measurable set containing \(x\).

Since \(X = \emptyset^c \in \mathcal{B}\) the family intersected is non-empty, and it is finite, so \(A_x \in \mathcal{B}\) by closure under finite intersections (Definition 1.4.1 with de Morgan); also \(x \in A_x\), and \(A_x \subset E\) for every \(E \in \mathcal{B}\) containing \(x\).

If \(y \in A_x\) then \(A_y = A_x\). Indeed every \(E \in \mathcal{B}\) with \(x \in E\) has \(A_x \subset E\), hence \(y \in E\), so the family defining \(A_y\) contains that defining \(A_x\) and \(A_y \subset A_x\). Conversely let \(E \in \mathcal{B}\) with \(y \in E\); were \(x \notin E\) we would get \(x \in E^c \in \mathcal{B}\), so \(A_x \subset E^c\) and \(y \in E^c\), a contradiction. Thus \(x \in E\) and \(A_x \subset E\); intersecting over such \(E\) gives \(A_x \subset A_y\).

Consequently any two of the \(A_x\) are equal or disjoint (a common point \(z\) gives \(A_x = A_z = A_y\)), and since each \(x\) lies in \(A_x\), the distinct ones \(A_1,\dots,A_n\) are a partition of \(X\) into non-empty sets, finite in number because \(\mathcal{B}\) is. Write \(\mathcal{A} := \mathcal{A}((A_i)_{i=1}^n)\). Then \(\mathcal{A} \subset \mathcal{B}\) by closure under finite unions, and \(\mathcal{B} \subset \mathcal{A}\) because any \(E \in \mathcal{B}\) satisfies \(E = \bigcup_{x \in E} A_x\) (each \(A_x \subset E\) by minimality, each \(x \in A_x\)). So \(\mathcal{B} = \mathcal{A}\) is atomic.

For the cardinality, \(\Phi : 2^{\{1,\dots,n\}} \to \mathcal{B}\), \(\Phi(J) := \bigcup_{i \in J} A_i\), is surjective by the last display and injective: if \(i \in J \setminus J^{\prime}\) then \(\emptyset \ne A_i \subset \Phi(J)\) while \(A_i \cap \Phi(J^{\prime}) = \emptyset\) by disjointness. Hence \(|\mathcal{B}| = 2^n\).

An atomic algebra containing every singleton is the discrete algebra; each of the four contains every singleton but omits some set, so none is atomic.

The lemma: if \(\mathcal{B} = \mathcal{A}((A_\alpha)_{\alpha \in I})\) with non-empty atoms (discard the empty ones, Example 1.4.7) and \(\{x\} \in \mathcal{B}\), then the atom \(A_\beta \ni x\) meets \(\{x\}\), so \(A_\beta \subset \{x\}\) by observation (B) of Exercise 1.4.3, i.e. \(A_\beta = \{x\}\). Every atom is thus a singleton and \(\mathcal{B} = 2^X\). It remains, for each algebra, to exhibit a singleton and a non-member.

(i) \(\mathcal{E}[\mathbb{R}^d]\) (Exercise 1.4.1). Singletons are degenerate boxes. The half-space \(H := \{x_1 > 0\}\) and \(H^c\) are both unbounded, hence neither is elementary (a finite union of boxes is bounded), so \(H \notin \mathcal{E}[\mathbb{R}^d]\).

(ii) \(\mathcal{J}[\mathbb{R}^d]\) (Example 1.4.4). Singletons are elementary, hence Jordan measurable. For \(E := [0,1]^d \cap \mathbb{Q}^d\) we have \(\overline{E} = [0,1]^d\) and \(E^\circ = \emptyset\) (every ball holds a point with all coordinates irrational), so \(\partial E = [0,1]^d\) has Jordan outer measure \(1 \ne 0\) and \(E\) is not Jordan measurable by Exercise 1.1.18(3); \(E^c\) is unbounded, hence not Jordan measurable either.

(iii) \(\mathcal{L}[\mathbb{R}^d]\) (Example 1.4.5). Singletons are null, hence measurable. Run Proposition 1.2.18 with the dense countable subgroup \(\mathbb{Q}^d\): pick \(x_C \in C \cap [0,1]^d\) from each coset \(C \in \mathbb{R}^d/\mathbb{Q}^d\) and let \(V\) be the set of representatives. The translates \(V+q\), \(q \in Q := \mathbb{Q}^d \cap [-1,1]^d\), are pairwise disjoint (each coset is met once) and, as \(y \in [0,1]^d\) gives \(y - x_C \in Q\), they squeeze

\begin{equation*} [0,1]^d \subset \bigcup_{q \in Q} (V+q) \subset [-1,2]^d . \end{equation*}

Were \(V\) measurable, translation invariance (Exercise 1.2.20) with countable additivity and monotonicity would force \(1 \le \sum_{q \in Q} m(V) \le 3^d\), impossible since that sum is \(0\) when \(m(V) = 0\) and \(+\infty\) when \(m(V) > 0\) (\(Q\) is countably infinite).

(iv) \(\mathcal{N}(\mathbb{R}^d)\) (Example 1.4.6). Singletons are null. But \(E := [0,1]^d\) has \(m(E) = 1\), and \(E^c \supset [2,3]^d\) gives \(m(E^c) \ge 1\), so \(E\) is neither null nor co-null.

Each axiom of Definition 1.4.1 is inherited from the \(\mathcal{B}_\alpha\) one \(\alpha\) at a time, so \(\mathcal{B} := \bigcap_{\alpha \in I} \mathcal{B}_\alpha\) is a Boolean algebra on \(X\).

Indeed \(\emptyset \in \mathcal{B}_\alpha\) for every \(\alpha\), hence \(\emptyset \in \mathcal{B}\); if \(E \in \mathcal{B}\) then \(E^c \in \mathcal{B}_\alpha\) for every \(\alpha\), hence \(E^c \in \mathcal{B}\); and likewise \(E, F \in \mathcal{B}\) give \(E \cup F \in \mathcal{B}_\alpha\) for every \(\alpha\), hence \(E \cup F \in \mathcal{B}\). (Check!) For \(I = \emptyset\) the condition is vacuous, so the intersection is \(2^X\), the stated convention, and a Boolean algebra by Example 1.4.3.

\(\mathcal{B} \subset \mathcal{B}_\alpha\) for each \(\alpha\), i.e. \(\mathcal{B}\) is coarser than every \(\mathcal{B}_\alpha\); and if \(\mathcal{C}\) is any Boolean algebra with \(\mathcal{C} \subset \mathcal{B}_\alpha\) for all \(\alpha\), then every \(E \in \mathcal{C}\) lies in every \(\mathcal{B}_\alpha\), so \(\mathcal{C} \subset \mathcal{B}\). Hence \(\mathcal{B}\) is the finest Boolean algebra coarser than all the \(\mathcal{B}_\alpha\), and is unique as such.

\(\langle \mathcal{F}\rangle_{\mathrm{bool}} = \mathcal{E}[\mathbb{R}^d]\), where \(\mathcal{F}\) is the collection of boxes.

(\(\subset\)) \(\mathcal{E}[\mathbb{R}^d]\) is a Boolean algebra (Exercise 1.4.1) containing every box, so it is one of the algebras intersected in Definition 1.4.10.

(\(\supset\)) Let \(\mathcal{C}\) be any Boolean algebra containing \(\mathcal{F}\). An elementary set is a finite union \(B_1 \cup \dots \cup B_k\) of boxes (Definition 1.1.1), hence lies in \(\mathcal{C}\) by closure under finite unions (the case \(k=0\) being \(\emptyset \in \mathcal{C}\)); a co-elementary set \(F^c\) then lies in \(\mathcal{C}\) by closure under complements. So \(\mathcal{E}[\mathbb{R}^d] \subset \mathcal{C}\), and intersecting over all such \(\mathcal{C}\) gives \(\mathcal{E}[\mathbb{R}^d] \subset \langle \mathcal{F}\rangle_{\mathrm{bool}}\).

\(\langle \mathcal{F}\rangle_{\mathrm{bool}}\) is the atomic algebra on the \(2^n\) cells cut out by \(\mathcal{F} = \{A_1,\dots,A_n\}\): writing \(A^{(1)} := A\) and \(A^{(0)} := X \setminus A\), set

\begin{equation*} A_\omega := \bigcap_{i=1}^{n} A_i^{(\omega_i)} \qquad (\omega \in \{0,1\}^n). \end{equation*}

These partition \(X\): they are disjoint because \(\omega \ne \omega^{\prime}\) differ in some digit \(i\), where \(A_i^{(\omega_i)}\) and \(A_i^{(\omega_i^{\prime})}\) are complementary; and they cover \(X\) because \(x\) lies in \(A_\omega\) for \(\omega_i := 1_{x \in A_i}\) (some cells may be empty, which Example 1.4.7 permits).

Put \(\mathcal{A} := \mathcal{A}((A_\omega)_{\omega})\). Each \(A_i = \bigcup_{\omega : \omega_i = 1} A_\omega\), since a point of \(A_i\) lies in the cell of its own digit string, whose \(i\)-th digit is \(1\); so \(\mathcal{F} \subset \mathcal{A}\) and \(\langle \mathcal{F}\rangle_{\mathrm{bool}} \subset \mathcal{A}\) by minimality (Definition 1.4.10). Conversely each \(A_\omega\) is a finite intersection of members of \(\mathcal{F}\) and their complements, hence lies in \(\langle \mathcal{F}\rangle_{\mathrm{bool}}\) (Boolean algebras are closed under finite intersections, Remark 1.4.1), and so does every union of the \(2^n\) cells. Thus \(\langle \mathcal{F}\rangle_{\mathrm{bool}} = \mathcal{A}\).

The map \(J \mapsto \bigcup_{\omega \in J} A_\omega\) from the power set of \(\{0,1\}^n\) onto \(\mathcal{A}\) is surjective, so

\begin{equation*} \#\langle \mathcal{F}\rangle_{\mathrm{bool}} = \#\mathcal{A} \le 2^{2^n} < \infty . \end{equation*}

Sharpness: take \(X := \{0,1\}^n\) and \(A_i := \{x \in X : x_i = 1\}\). Then \(A_\omega = \{\omega\}\), every cell is a non-empty singleton, and the map above becomes \(J \mapsto J\), a bijection; so \(\langle\mathcal{F}\rangle_{\mathrm{bool}} = 2^X\) has cardinality exactly \(2^{2^n}\).

\(\mathcal{G} := \bigcup_{n \ge 0} \mathcal{F}_n\) is itself a Boolean algebra containing \(\mathcal{F}\), and is contained in every such algebra.

The hierarchy increases: \(E \in \mathcal{F}_{n-1}\) is the union of the one-element family \(\{E\} \subset \mathcal{F}_{n-1}\), so \(E \in \mathcal{F}_n\) by clause (ii). Hence any two members of \(\mathcal{G}\) lie in a common \(\mathcal{F}_n\).

\(\mathcal{G}\) is a Boolean algebra containing \(\mathcal{F} = \mathcal{F}_0\). Empty set: \(\emptyset\) is the empty union of sets in \(\mathcal{F}_0\), so \(\emptyset \in \mathcal{F}_1\). Complement: for \(E \in \mathcal{F}_n\) with \(n \ge 1\), clause (ii) makes \(E\) either a finite union \(U\) of sets in \(\mathcal{F}_{n-1}\) or \(X \setminus U\) for such a \(U\), and \(X \setminus E\) is then the other, so \(X \setminus E \in \mathcal{F}_n\); for \(n = 0\), \(X \setminus E\) is the complement of the one-fold union \(E\), so \(X \setminus E \in \mathcal{F}_1\). Finite unions: given \(E, E^{\prime} \in \mathcal{F}_n\), the set \(E \cup E^{\prime}\) is a finite union of sets in \(\mathcal{F}_n\), hence lies in \(\mathcal{F}_{n+1}\) (unions of more than two sets follow by induction). Therefore \(\langle \mathcal{F}\rangle_{\mathrm{bool}} \subset \mathcal{G}\) by Definition 1.4.10.

Conversely \(\mathcal{F}_n \subset \langle \mathcal{F}\rangle_{\mathrm{bool}}\) by induction on \(n\): the case \(n = 0\) is the definition, and if \(\mathcal{F}_{n-1} \subset \langle\mathcal{F}\rangle_{\mathrm{bool}}\) then any finite union \(U\) of its members lies in \(\langle\mathcal{F}\rangle_{\mathrm{bool}}\) (which contains \(\emptyset\), covering the empty union) and so does \(X \setminus U\). Taking the union over \(n\) gives \(\mathcal{G} \subset \langle\mathcal{F}\rangle_{\mathrm{bool}}\).

Membership in \(\mathcal{A} := \mathcal{A}((A_\alpha)_{\alpha \in I}) = \{\bigcup_{\alpha \in J} A_\alpha : J \subset I\}\) is indexed by subsets \(J \subset I\), and the three axioms of Definition 1.4.12 are the corresponding operations on index sets.

Empty set: \(J = \emptyset\). Complement: for \(E = \bigcup_{\alpha \in J} A_\alpha\),

\begin{equation*} X \setminus E = \bigcup_{\alpha \in I \setminus J} A_\alpha , \end{equation*}

since the atoms cover \(X\) (so \(x \notin E\) lies in some \(A_\beta\), necessarily with \(\beta \notin J\)) and are disjoint (so \(x \in A_\beta\) with \(\beta \notin J\) cannot lie in \(E\)). Unions: if \(E_n = \bigcup_{\alpha \in J_n} A_\alpha\) then \(\bigcup_n E_n = \bigcup_{\alpha \in J} A_\alpha\) with \(J := \bigcup_n J_n\), by regrouping; no countability of \(I\) is needed, so \(\mathcal{A}\) is closed under arbitrary unions. Hence \(\mathcal{A}\) is a \(\sigma\)-algebra.

The named cases: the trivial algebra and the discrete algebra are the atomic algebras of the one-atom and the singleton partitions (Example 1.4.7); every finite Boolean algebra is atomic by Exercise 1.4.4; and \(\mathcal{D}_n(\mathbb{R}^d)\) (Example 1.4.8) is atomic because the half-open dyadic cubes of side \(2^{-n}\) partition \(\mathbb{R}^d\), the cube containing \(x\) being the one with indices \(i_r = \lfloor 2^n x_r \rfloor\).

Lebesgue and null are closed under countable unions; elementary and Jordan are broken by a countable union of degenerate boxes.

\(\mathcal{L}[\mathbb{R}^d]\). Lemma 1.2.13(iv),(v),(vi) is exactly the list of axioms in Definition 1.4.12 for the Lebesgue measurable sets.

\(\mathcal{N}(\mathbb{R}^d)\). The class contains \(\emptyset\) and is symmetric under complementation by construction (Example 1.4.6), so only countable unions \(E := \bigcup_n E_n\) need checking.

• (i) Every \(E_n\) null: \(m^*(E) \le \sum_n m^*(E_n) = 0\) by countable subadditivity (Exercise 1.2.3(iii)), so \(E\) is null.
• (ii) Some \(E_{n_0}\) co-null: \(\mathbb{R}^d \setminus E \subset \mathbb{R}^d \setminus E_{n_0}\), so \(m^*(\mathbb{R}^d \setminus E) = 0\) by monotonicity (Exercise 1.2.3(ii)) and \(E\) is co-null.

\(\mathcal{E}[\mathbb{R}^d]\). Each \(E_n := \{1/n\} \times \{0\}^{d-1}\) is a box, hence elementary, but \(S := \bigcup_{n \ge 1} E_n\) is not. Indeed a non-empty box \(B = I_1 \times \dots \times I_d\) inside \(S\) has \(I_1 \subset \{1/n : n \ge 1\}\), a countable set, so \(I_1\) has length \(0\) and is a point, and likewise \(I_r = \{0\}\) for \(r \ge 2\); a finite union of boxes contained in \(S\) is therefore finite, while \(S\) is infinite. And \(\mathbb{R}^d \setminus S\) is unbounded, hence not elementary. So \(S \notin \mathcal{E}[\mathbb{R}^d]\).

\(\mathcal{J}[\mathbb{R}^d]\). Enumerate \(\mathbb{Q} \cap [0,1] = \{q_1, q_2, \dots\}\); each \(E_n := \{q_n\} \times [0,1]^{d-1}\) is a box, hence Jordan measurable, but

\begin{equation*} S := \bigcup_{n \ge 1} E_n = (\mathbb{Q} \cap [0,1]) \times [0,1]^{d-1} \end{equation*}

has \(\overline{S} = [0,1]^d\) and \(S^\circ = \emptyset\) (every ball meets points with irrational first coordinate), so \(\partial S = [0,1]^d\) has Jordan outer measure \(1 \ne 0\) and \(S\) is not Jordan measurable by Exercise 1.1.18(3). Its complement is unbounded, hence not Jordan measurable either, so \(S \notin \mathcal{J}[\mathbb{R}^d]\).

The two identities

\begin{equation*} Y \setminus (E \cap Y) = (X \setminus E) \cap Y, \qquad \bigcup_{n=1}^\infty (E_n \cap Y) = \Big(\bigcup_{n=1}^\infty E_n\Big) \cap Y \end{equation*}

give the complement and countable-union axioms of Definition 1.4.12 for \(\mathcal{B}|_Y = \{E \cap Y : E \in \mathcal{B}\}\), since \(X \setminus E\) and \(\bigcup_n E_n\) lie in \(\mathcal{B}\); and \(\emptyset = \emptyset \cap Y \in \mathcal{B}|_Y\). Complements here are relative to \(Y\), and no measurability of \(Y\) is used. Hence \(\mathcal{B}|_Y\) is a \(\sigma\)-algebra on \(Y\).

As in Exercise 1.4.6, each axiom passes to \(\mathcal{B} := \bigcap_{\alpha \in I} \mathcal{B}_\alpha\) one \(\alpha\) at a time: \(\emptyset \in \mathcal{B}_\alpha\) for all \(\alpha\); \(E \in \mathcal{B}\) gives \(X \setminus E \in \mathcal{B}_\alpha\) for all \(\alpha\); and \(E_1, E_2, \dots \in \mathcal{B}\) give \(\bigcup_n E_n \in \mathcal{B}_\alpha\) for all \(\alpha\), each \(\mathcal{B}_\alpha\) being closed under countable unions. So \(\mathcal{B}\) is a \(\sigma\)-algebra (Definition 1.4.12). For \(I = \emptyset\) the convention of Exercise 1.4.6 makes the intersection \(2^X\), a \(\sigma\)-algebra by Exercise 1.4.10.

By construction \(\mathcal{B} \subset \mathcal{B}_\alpha\) for every \(\alpha\), so \(\mathcal{B}\) is coarser than each; and any \(\sigma\)-algebra \(\mathcal{B}^{\prime}\) with \(\mathcal{B}^{\prime} \subset \mathcal{B}_\alpha\) for all \(\alpha\) satisfies \(\mathcal{B}^{\prime} \subset \mathcal{B}\). Hence \(\mathcal{B}\) is the finest \(\sigma\)-algebra coarser than all the \(\mathcal{B}_\alpha\), and is unique as such.

All six generate \(\mathcal{B}[\mathbb{R}^d] = \langle \mathcal{O}\rangle\), by the criterion

\begin{equation*} \mathcal{F} \subset \langle \mathcal{F}^{\prime}\rangle \ \text{ and }\ \mathcal{F}^{\prime} \subset \langle \mathcal{F}\rangle \quad \Longrightarrow \quad \langle \mathcal{F}\rangle = \langle \mathcal{F}^{\prime}\rangle , \end{equation*}

which holds because \(\langle \mathcal{F}^{\prime}\rangle\) is then a \(\sigma\)-algebra containing \(\mathcal{F}\), so it contains the intersection \(\langle\mathcal{F}\rangle\) of all such (Definition 1.4.14). Write \(\mathcal{O}, \mathcal{C}, \mathcal{K}, \mathcal{U}, \mathcal{X}, \mathcal{E}\) for the six families, and use freely that a \(\sigma\)-algebra is closed under complements and countable (hence finite) unions and intersections.

(i) \(\langle \mathcal{O}\rangle = \mathcal{B}[\mathbb{R}^d]\) is Definition 1.4.16.

(ii) Closed and open sets are each other’s complements, so \(\mathcal{C} \subset \langle\mathcal{O}\rangle\) and \(\mathcal{O} \subset \langle\mathcal{C}\rangle\).

(iii) Compact sets are closed (Heine–Borel), so \(\mathcal{K} \subset \mathcal{C}\); conversely a closed \(C\) is \(\bigcup_{n \ge 1} (C \cap \overline{B}(0,n))\), a countable union of closed bounded, hence compact, sets. So \(\langle\mathcal{K}\rangle = \langle\mathcal{C}\rangle = \langle\mathcal{O}\rangle\) by (ii).

(iv) Balls are open. Conversely, let \(U\) be open and \(x \in U\), and take \(r > 0\) with \(B(x,r) \subset U\). Choose \(q \in \mathbb{Q}^d\) with \(|q - x| < r/4\) and a rational \(s \in (r/4, r/2)\); then \(x \in B(q,s)\), and \(y \in B(q,s)\) gives \(|y - x| < s + r/4 < r\), so \(B(q,s) \subset U\). Hence \(U\) is the union of those balls with rational centre and radius that it contains, a countable family, so \(U \in \langle\mathcal{U}\rangle\).

(v) For a box \(B = I_1 \times \dots \times I_d\) with \(I_r\) of endpoints \(a_r \le b_r\), writing \(\pi_r\) for the \(r\)-th coordinate map,

\begin{equation*} B = \bigcap_{r=1}^d \pi_r^{-1}(I_r), \qquad \pi_r^{-1}(I_r) = \pi_r^{-1}\big((a_r,b_r)\big) \cup \bigcup_{c \in P_r} \pi_r^{-1}(\{c\}) , \end{equation*}

where \(P_r \subset \{a_r,b_r\}\) records the included endpoints; the first piece is open and each \(\pi_r^{-1}(\{c\})\) is a closed hyperplane, so \(B \in \langle\mathcal{O}\rangle\) by (ii) and closure under finite unions and intersections. Conversely, repeating (iv) with rational open boxes: given \(x \in U\) open and \(B(x,r) \subset U\), pick rationals \(a_j < x_j < b_j\) with \(b_j - a_j < r/\sqrt{d}\), so that \(x \in Q := \prod_j (a_j,b_j) \subset B(x,r) \subset U\); thus \(U\) is a countable union of boxes.

(vi) Every box is elementary, and every elementary set is a finite union of boxes (Definition 1.1.1), so \(\mathcal{X} \subset \mathcal{E}\) and \(\mathcal{E} \subset \langle\mathcal{X}\rangle\), giving \(\langle\mathcal{E}\rangle = \langle\mathcal{X}\rangle = \langle\mathcal{O}\rangle\) by (v).

\(G := \bigcup_{\alpha \in \omega_1} \mathcal{F}_\alpha\) is a \(\sigma\)-algebra containing \(\mathcal{F}\) and contained in every such, the point being that a countable sequence of countable ordinals is bounded below \(\omega_1\).

The hierarchy increases. \(\mathcal{F}_\beta \subset \mathcal{F}_{\beta+1}\), since \(E \in \mathcal{F}_\beta\) is the union of the one-element family \(\{E\}\), so clause (ii) applies; then \(\mathcal{F}_\beta \subset \mathcal{F}_\alpha\) for \(\beta \le \alpha < \omega_1\) by transfinite induction on \(\alpha\), the successor step by the previous sentence and the limit step by clause (iii).

\(G \subset \langle\mathcal{F}\rangle\). By transfinite induction, \(\mathcal{F}_\alpha \subset \langle\mathcal{F}\rangle\): the case \(\alpha = 0\) is Definition 1.4.14; at a successor \(\alpha = \beta+1\), an at most countable union of sets in \(\mathcal{F}_\beta \subset \langle\mathcal{F}\rangle\) lies in \(\langle\mathcal{F}\rangle\) (pad a finite or empty union with copies of \(\emptyset\)), as does its complement; at a limit, \(\mathcal{F}_\alpha\) is the union of the earlier stages.

\(G\) is a \(\sigma\)-algebra containing \(\mathcal{F} = \mathcal{F}_0\). Empty set: \(\emptyset \in \mathcal{F}_1\), the empty union. Complement: \(E \in \mathcal{F}_\alpha\) makes \(E^c\) the complement of the one-fold union \(\{E\}\), so \(E^c \in \mathcal{F}_{\alpha+1} \subset G\), as \(\alpha+1 < \omega_1\). Countable unions: given \(E_1, E_2, \dots \in G\), let \(\alpha_n\) be the least ordinal with \(E_n \in \mathcal{F}_{\alpha_n}\) (well defined by well-ordering, no choice needed) and \(\alpha := \sup_n \alpha_n = \bigcup_n \alpha_n\). Each \(\alpha_n\) is an at most countable set and a countable union of such is countable (countable choice), so \(\alpha < \omega_1\); by monotonicity of the hierarchy every \(E_n \in \mathcal{F}_\alpha\), whence \(\bigcup_n E_n \in \mathcal{F}_{\alpha+1} \subset G\).

Hence \(\langle\mathcal{F}\rangle \subset G\) as well, and the two are equal.

Every stage of the hierarchy of Exercise 1.4.15 has size at most \(\lambda := \kappa^{\aleph_0}\), and there are \(\aleph_1 \le \lambda\) stages. Two facts are used throughout:

\begin{equation*} \lambda^{\aleph_0} = \kappa^{\aleph_0 \cdot \aleph_0} = \kappa^{\aleph_0} = \lambda , \qquad \lambda \ge 2^{\aleph_0} \ge \aleph_1 , \end{equation*}

the second because \(\kappa \ge 2\) and \(2^{\aleph_0} > \aleph_0\) (Cantor).

By transfinite induction \(|\mathcal{F}_\alpha| \le \lambda\) for all \(\alpha \in \omega_1\). For \(\alpha = 0\), \(|\mathcal{F}_0| = \kappa \le \lambda\). At \(\alpha = \beta+1\): an at most countable subfamily of \(\mathcal{F}_\beta\) is empty or the range of a map \(\omega \to \mathcal{F}_\beta\), so there are at most \(1 + \lambda^{\aleph_0} = \lambda\) of them, each contributing a union and its complement, giving \(|\mathcal{F}_{\beta+1}| \le 2\lambda = \lambda\). At a limit \(\alpha < \omega_1\) there are at most countably many earlier stages, so \(|\mathcal{F}_\alpha| \le \aleph_0 \cdot \lambda = \lambda\). Hence by Exercise 1.4.15

\begin{equation*} |\langle\mathcal{F}\rangle| = \Big|\bigcup_{\alpha \in \omega_1} \mathcal{F}_\alpha\Big| \le \aleph_1 \cdot \lambda = \lambda = \kappa^{\aleph_0} . \end{equation*}

For the Borel algebra, Exercise 1.4.14(v) lets us take \(\mathcal{F}\) to be the boxes, of which there are \(\kappa = \mathfrak{c}\) (each determined by \(2d\) endpoints and \(2d\) open/closed choices), so \(|\mathcal{B}[\mathbb{R}^d]| \le \mathfrak{c}^{\aleph_0} = 2^{\aleph_0} = \mathfrak{c}\), with equality since the \(\mathfrak{c}\) singletons are Borel.

Now let \(C\) be the middle-thirds Cantor set and \(K := C \times \{0\}^{d-1}\). Covering \(C\) by \(2^n\) intervals of length \(3^{-n}\) and thickening by \([-1,1]^{d-1}\) exhibits \(K\) inside elementary sets of measure \(2^{d-1}(2/3)^n \to 0\), so \(m^{*,(J)}(K) = 0\); hence every \(A \subset K\) has Jordan outer measure \(0\), inner measure squeezed to \(0\), and so is Jordan (hence Lebesgue) measurable. But \(|K| = \mathfrak{c}\), so \(K\) has \(2^{\mathfrak{c}} > \mathfrak{c} = |\mathcal{B}[\mathbb{R}^d]|\) subsets, and some \(A \subset K\) is Jordan measurable but not Borel.

Placement. \(\mathcal{B}[\mathbb{R}^d]\) sits strictly above the elementary algebra and each dyadic algebra \(\mathcal{D}_n(\mathbb{R}^d)\) (elementary sets are finite unions of boxes; a dyadic set is an at most countable union of cubes; and \(\mathbb{Q}^d \cap [0,1]^d\) is Borel but lies in neither), strictly below \(\mathcal{L}[\mathbb{R}^d]\) (witness \(A\)) and strictly below the discrete algebra (\(\mathfrak{c} < 2^{\mathfrak{c}}\)), and strictly above the trivial algebra. It is incomparable with the Jordan and null algebras: \(A\) is Jordan measurable and null but not Borel, while \(\mathbb{Q}^d \cap [0,1]^d\) is Borel and not in \(\mathcal{J}[\mathbb{R}^d]\) (Exercise 1.4.11) and \([0,1]^d\) is Borel and neither null nor co-null.

Two runs of Remark 1.4.15, first freezing a box in the second factor, then freezing a Borel set in the first. Throughout, \(\mathcal{B}[\mathbb{R}^{d_1+d_2}]\) is closed under finite intersections and differences, since \(A \cap B = (A^c \cup B^c)^c\) and \(A \setminus B = A \cap B^c\).

Two ambient sets are Borel: \(U \times \mathbb{R}^{d_2}\) for \(U \subset \mathbb{R}^{d_1}\) open (it is open), and, for a box \(B \subset \mathbb{R}^{d_2}\),

\begin{equation*} \mathbb{R}^{d_1} \times B = \bigcup_{n=1}^\infty \big([-n,n]^{d_1} \times B\big), \end{equation*}

a countable union of boxes, Borel by Exercise 1.4.14(v).

Fix a box \(B\) and let \(P(E)\) say that \(E \times B\) is Borel. Apply Remark 1.4.15 with \(\mathcal{F}\) the open subsets of \(\mathbb{R}^{d_1}\), so that \(\langle\mathcal{F}\rangle = \mathcal{B}[\mathbb{R}^{d_1}]\):

• \(P(\emptyset)\), since \(\emptyset \times B = \emptyset\).
• \(P(U)\) for \(U\) open, since \(U \times B = (U \times \mathbb{R}^{d_2}) \cap (\mathbb{R}^{d_1} \times B)\).
• \(P(E) \Rightarrow P(\mathbb{R}^{d_1} \setminus E)\), since \((\mathbb{R}^{d_1} \setminus E) \times B = (\mathbb{R}^{d_1} \times B) \setminus (E \times B)\).
• \(P(E_n)\) for all \(n\) gives \(P(\bigcup_n E_n)\), since \((\bigcup_n E_n) \times B = \bigcup_n (E_n \times B)\).

So \(E \times B\) is Borel for every Borel \(E\) and every box \(B\).

Now fix such an \(E\) and let \(Q(F)\) say that \(E \times F\) is Borel, applying Remark 1.4.15 with \(\mathcal{F}\) the boxes of \(\mathbb{R}^{d_2}\), which generate \(\mathcal{B}[\mathbb{R}^{d_2}]\) by Exercise 1.4.14(v). Here \(Q(\emptyset)\) is trivial and \(Q(B)\) is the previous paragraph; \(E \times \mathbb{R}^{d_2} = \bigcup_n (E \times [-n,n]^{d_2})\) is Borel by that paragraph, so \(E \times (\mathbb{R}^{d_2} \setminus F) = (E \times \mathbb{R}^{d_2}) \setminus (E \times F)\) gives the complement step; and \(E \times \bigcup_n F_n = \bigcup_n (E \times F_n)\) gives countable unions. Hence \(E \times F\) is Borel whenever \(E\) and \(F\) are.

(i) The slice is a preimage: \(E_{x_1} := \{x_2 : (x_1,x_2) \in E\} = \iota^{-1}(E)\) for the continuous map \(\iota(x_2) := (x_1,x_2)\), and preimages commute with all set operations,

\begin{equation*} \iota^{-1}(\emptyset) = \emptyset, \quad (E^c)_{x_1} = (E_{x_1})^c, \quad \Big(\bigcup_{n} E_n\Big)_{x_1} = \bigcup_{n} (E_n)_{x_1}. \end{equation*}

Let \(P(E)\) say that \(E_{x_1}\) is Borel in \(\mathbb{R}^{d_2}\) and apply Remark 1.4.15 with \(\mathcal{F}\) the open subsets of \(\mathbb{R}^{d_1+d_2}\), so \(\langle\mathcal{F}\rangle = \mathcal{B}[\mathbb{R}^{d_1+d_2}]\): \(P(\emptyset)\) holds by the first identity; \(P(U)\) holds for \(U\) open because \(\iota^{-1}(U)\) is open by continuity of \(\iota\); and the last two identities give the complement and countable-union steps. Hence \(P(E)\) holds for every Borel \(E\). The other direction is the same with \(x_1 \mapsto (x_1,x_2)\).

(ii) Let \(A \subset [0,1]^{d_1}\) be the non-Lebesgue-measurable set of Exercise 1.4.5(iii) and put \(E := A \times \{0\}^{d_2}\). Then \(E\) lies in

\begin{equation*} H := \mathbb{R}^{d_1} \times \{0\}^{d_2} = \bigcup_{n=1}^\infty \big([-n,n]^{d_1} \times \{0\}^{d_2}\big), \end{equation*}

and each term is covered by boxes \([-n,n]^{d_1} \times [-\varepsilon,\varepsilon]^{d_2}\) of volume \((2n)^{d_1}(2\varepsilon)^{d_2} \to 0\), so \(H\) is null and hence so is \(E\). Thus \(E\) is Lebesgue measurable with \(m(E) = 0\), yet its slice at \(x_2 = 0\) is \(A\), which is not Lebesgue measurable. Taking instead \(E^{\prime} := \{0\}^{d_1} \times B\) with \(B \subset [0,1]^{d_2}\) non-measurable gives the same failure in the other variable.

\(\langle \mathcal{B}[\mathbb{R}^d] \cup \mathcal{N}(\mathbb{R}^d)\rangle = \mathcal{L}[\mathbb{R}^d]\), the decisive fact being the decomposition \(E = F \cup N\) of Exercise 1.2.19.

(\(\subset\)) Open sets are Lebesgue measurable, so \(\mathcal{B}[\mathbb{R}^d] \subset \mathcal{L}[\mathbb{R}^d]\) by the minimality in Definition 1.4.14; null sets are Lebesgue measurable and \(\mathcal{L}[\mathbb{R}^d]\) is closed under complements, so \(\mathcal{N}(\mathbb{R}^d) \subset \mathcal{L}[\mathbb{R}^d]\) too. Thus \(\mathcal{L}[\mathbb{R}^d]\) is a \(\sigma\)-algebra containing \(\mathcal{B}[\mathbb{R}^d] \cup \mathcal{N}(\mathbb{R}^d)\), hence contains the coarsest one.

(\(\supset\)) Let \(E\) be Lebesgue measurable. By Exercise 1.2.19 ((i) \(\Leftrightarrow\) (iii)), \(E = F \cup N\) with \(F\) an \(F_\sigma\) set and \(N\) null. An \(F_\sigma\) set is a countable union of closed sets, hence Borel, and \(N \in \mathcal{N}(\mathbb{R}^d)\); a \(\sigma\)-algebra is closed under the union of two of its members, so \(E \in \langle \mathcal{B}[\mathbb{R}^d] \cup \mathcal{N}(\mathbb{R}^d)\rangle\).

All four follow from splitting a set into disjoint measurable pieces and adding; nothing is ever subtracted or cancelled in \([0,+\infty]\). A Boolean algebra is closed under intersections and differences (Remark 1.4.1): \(E \cap F = (E^c \cup F^c)^c\) and \(F \setminus E = F \cap E^c\).

(i) \(F = E \cup (F \setminus E)\) is a disjoint union of \(\mathcal{B}\)-measurable sets, so by finite additivity (Definition 1.4.19(ii))

\begin{equation*} \mu(F) = \mu(E) + \mu(F \setminus E) \ge \mu(E), \end{equation*}

since \(\mu(F \setminus E) \ge 0\) and addition in \([0,+\infty]\) is monotone.

(ii) Induct on \(k\); \(k = 0\) is \(\mu(\emptyset) = 0\) (Definition 1.4.19(i)) and \(k = 1\) is trivial. For the step, \(G := E_1 \cup \dots \cup E_k\) is \(\mathcal{B}\)-measurable and \(G \cap E_{k+1} = \bigcup_{j \le k} (E_j \cap E_{k+1}) = \emptyset\), so two-set additivity and the inductive hypothesis give

\begin{equation*} \mu(E_1 \cup \dots \cup E_{k+1}) = \mu(G) + \mu(E_{k+1}) = \mu(E_1) + \dots + \mu(E_{k+1}) . \end{equation*}

(iii) Disjointify: \(F_1 := E_1\) and \(F_j := E_j \setminus (E_1 \cup \dots \cup E_{j-1})\). Each \(F_j\) is \(\mathcal{B}\)-measurable with \(F_j \subset E_j\); they are pairwise disjoint (for \(i < j\), \(F_i \subset E_i\) while \(F_j \cap E_i = \emptyset\)); and they have the same union as the \(E_j\) (given \(x \in \bigcup_j E_j\), the least \(j\) with \(x \in E_j\) puts \(x \in F_j\)). Hence by (ii) and then (i),

\begin{equation*} \mu(E_1 \cup \dots \cup E_k) = \sum_{j=1}^k \mu(F_j) \le \sum_{j=1}^k \mu(E_j) . \end{equation*}

(iv) The disjoint decompositions \(E \cup F = E \cup (F \setminus E)\) and \(F = (E \cap F) \cup (F \setminus E)\) give

\begin{align*} \mu(E \cup F) + \mu(E \cap F) &= \mu(E) + \big(\mu(F \setminus E) + \mu(E \cap F)\big) \\ &= \mu(E) + \mu(F) . \end{align*}

Take \(c_j := \mu(A_j) \in [0,+\infty]\).

By Exercise 1.4.4 the algebra \(\mathcal{B}\) consists exactly of the unions \(\bigcup_{j \in J}A_j\) with \(J \subset \{1,\ldots,k\}\); as the atoms are disjoint and non-empty, \(A_j \subset E\) holds precisely for \(j \in J\), so every \(E \in \mathcal{B}\) is the disjoint union of the atoms it contains. Finite additivity (Exercise 1.4.20(ii)) then gives

\begin{equation*} \mu(E) = \sum_{1\le j\le k:\, A_j\subset E} \mu(A_j) = \sum_{1\le j\le k:\, A_j\subset E} c_j , \end{equation*}

the empty sum being \(0 = \mu(\emptyset)\).

Fix \(x_j \in A_j\) (possible as \(A_j \ne \emptyset\)). Since \(E\) is a union of the pairwise disjoint atoms, \(x_j \in E\) iff \(A_j \subset E\); so with \(\delta_{x_j}(E) = 1_E(x_j)\) (Example 1.4.22) and \(\infty \cdot 0 = 0\) (Example 1.4.24),

\begin{equation*} \sum_{j=1}^k c_j\delta_{x_j}(E) = \sum_{1\le j\le k:\, x_j \in E} c_j = \sum_{1\le j\le k:\, A_j\subset E} c_j = \mu(E) . \end{equation*}

Uniqueness: evaluating the formula at \(E = A_i\), the only \(j\) with \(A_j \subset A_i\) is \(j = i\) (disjointness forces \(A_j = \emptyset\) otherwise), so \(c_i = \mu(A_i)\) is forced.

Both parts follow from two facts about sums \(\sum_{n=1}^{\infty} a_n := \sup_N \sum_{n \le N} a_n\) of elements of \([0,+\infty]\).

Fact 1. \(c \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} c a_n\) for every \(c \in [0,+\infty]\). Finite distributivity holds in \([0,+\infty]\) (Check! – treat the terms \(0\) and \(\infty\) separately, using \(0\cdot\infty = 0\)), so it suffices that \(c \sup_N s_N = \sup_N (c s_N)\) for non-decreasing \(s_N\): for \(c = 0\) both sides are \(0\); for \(0 < c < \infty\) multiplication by \(c\) is an order-isomorphism of \([0,+\infty]\); for \(c = +\infty\) both sides equal \(+\infty\) exactly when some \(s_N > 0\), and \(0\) otherwise.

Fact 2 (Tonelli for series). For any \(a_{n,m} \in [0,+\infty]\),

\begin{equation*} \begin{aligned} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} a_{n,m} &= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} a_{n,m} = S, \\ S &:= \sup\Big\{ \textstyle\sum_{(n,m) \in F} a_{n,m} : F \subset \mathbb{N}^2 \text{ finite} \Big\}. \end{aligned} \end{equation*}

By symmetry it suffices to prove the first sum is \(S\). Every finite \(F\) lies in some square \(\{1,\ldots,N\}^2\), whence \(S \le \sum_n \sum_m a_{n,m}\). Conversely, fix \(N\); finitely many non-decreasing sequences in \([0,+\infty]\) may be added termwise in the limit, so

\begin{equation*} \sum_{n=1}^{N} \sum_{m=1}^{\infty} a_{n,m} = \lim_{M \to \infty} \sum_{n \le N} \sum_{m \le M} a_{n,m} \le S, \end{equation*}

each inner double sum being a sum over the finite set \(\{1,\ldots,N\}\times\{1,\ldots,M\}\). Let \(N \to \infty\).

(i). \((c\mu)(\emptyset) = c \cdot 0 = 0\), and for disjoint \(E_1,E_2,\ldots \in \mathcal{B}\), countable additivity of \(\mu\) and Fact 1 give

\begin{equation*} c\,\mu\Big( \bigcup_{n=1}^{\infty} E_n \Big) = c \sum_{n=1}^{\infty} \mu(E_n) = \sum_{n=1}^{\infty} c\,\mu(E_n). \end{equation*}

(ii). Put \(\nu(E) := \sum_{n=1}^{\infty} \mu_n(E) \in [0,+\infty]\); then \(\nu(\emptyset) = 0\). For disjoint \(E_1,E_2,\ldots\) with \(E := \bigcup_m E_m\), countable additivity of each \(\mu_n\) and Fact 2 with \(a_{n,m} := \mu_n(E_m)\) give

\begin{equation*} \begin{aligned} \nu(E) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \mu_n(E_m) &= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \mu_n(E_m) \\ &= \sum_{m=1}^{\infty} \nu(E_m). \end{aligned} \end{equation*}

A countably additive measure is finitely additive (pad a finite disjoint union with copies of \(\emptyset\)), hence monotone and finitely subadditive by Exercise 1.4.20.

(i). Disjointify: \(F_n := E_n \setminus (E_1 \cup \cdots \cup E_{n-1}) \in \mathcal{B}\) are pairwise disjoint (for \(m < n\), \(F_n\) misses \(E_m \supset F_m\)) with \(\bigcup_n F_n = \bigcup_n E_n\) (send \(x\) to the least \(n\) with \(x \in E_n\)). By countable additivity and \(F_n \subset E_n\),

\begin{equation*} \mu\Big( \bigcup_{n=1}^{\infty} E_n \Big) = \sum_{n=1}^{\infty} \mu(F_n) \le \sum_{n=1}^{\infty} \mu(E_n). \end{equation*}

(ii). With \(E_0 := \emptyset\), the sets \(F_n := E_n \setminus E_{n-1}\) are disjoint and measurable with \(\bigcup_{n \le N} F_n = E_N\) and \(\bigcup_n F_n = \bigcup_n E_n\), so

\begin{equation*} \mu\Big( \bigcup_{n=1}^{\infty} E_n \Big) = \sum_{n=1}^{\infty} \mu(F_n) = \lim_{N \to \infty} \sum_{n \le N} \mu(F_n) = \lim_{N \to \infty} \mu(E_N), \end{equation*}

and the limit equals the supremum since \(\mu(E_n)\) is non-decreasing.

(iii). Discarding the first \(n_0 - 1\) terms alters neither \(\bigcap_n E_n\) nor \(\lim_n \mu(E_n) = \inf_n \mu(E_n)\) (the sequence is non-increasing by monotonicity), so assume \(\mu(E_1) < \infty\); then every \(\mu(E_n)\) and \(\mu(\bigcap_n E_n)\) is finite. The sets \(E_1 \setminus E_n\) increase to \(E_1 \setminus \bigcap_n E_n\) (De Morgan), so (ii) applies to them; and from the disjoint decomposition \(E_1 = E_n \cup (E_1 \setminus E_n)\) with all terms finite, \(\mu(E_1 \setminus E_n) = \mu(E_1) - \mu(E_n)\), likewise with \(E_n\) replaced by \(\bigcap_n E_n \subset E_1\). Hence

\begin{equation*} \mu(E_1) - \mu\Big( \bigcap_{n=1}^{\infty} E_n \Big) = \lim_{n \to \infty} \big( \mu(E_1) - \mu(E_n) \big), \end{equation*}

and cancelling the finite quantity \(\mu(E_1)\) gives \(\mu(\bigcap_n E_n) = \lim_n \mu(E_n) = \inf_n \mu(E_n)\).

Failure without finiteness. Take \(E_n := [n,+\infty) \subset \mathbb{R}\) with Lebesgue measure: every \(m(E_n) = +\infty\), yet \(\bigcap_n E_n = \emptyset\), so

\begin{equation*} m\Big( \bigcap_{n=1}^{\infty} E_n \Big) = 0 \ne +\infty = \lim_{n \to \infty} m(E_n). \end{equation*}

(i). Indicators take only the values \(0\) and \(1\), so \(1_{E_n}(x) \to 1_E(x)\) forces \(1_{E_n}(x) = 1_E(x)\) for all large \(n\); hence

\begin{equation*} E = \bigcup_{N=1}^{\infty} \bigcap_{n \ge N} E_n \in \mathcal{B}, \end{equation*}

a countable union of countable intersections of \(\mathcal{B}\)-measurable sets.

(ii). By the identity in (i), \(E \subset F\), so \(E\) and every \(E_n \Delta E\) are measurable subsets of \(F\) and have finite measure. Put \(G_N := \bigcup_{n > N}(E_n \Delta E) \in \mathcal{B}\). These decrease, \(\mu(G_1) \le \mu(F) < \infty\), and \(\bigcap_N G_N = \emptyset\): a point in every \(G_N\) would satisfy \(1_{E_n}(x) \ne 1_E(x)\) for infinitely many \(n\), contradicting the eventual equality in (i). Downwards monotone convergence (Exercise 1.4.23(iii), applicable since \(\mu(G_1) < \infty\)) then gives \(\mu(G_N) \to 0\).

For measurable \(A, B \subset F\) we have \(|\mu(A) - \mu(B)| \le \mu(A \Delta B)\): from \(A \subset B \cup (A \setminus B)\) and finite subadditivity (Exercise 1.4.20(iii)), \(\mu(A) \le \mu(B) + \mu(A \setminus B)\), and all terms being finite we may rearrange, then swap \(A\) and \(B\). Taking \(A = E_n\), \(B = E\) and using \(E_n \Delta E \subset G_N\) for \(n > N\),

\begin{equation*} |\mu(E_n) - \mu(E)| \le \mu(E_n \Delta E) \le \mu(G_N) \qquad (n > N), \end{equation*}

so \(\mu(E_n) \to \mu(E)\).

(iii). Take \(E_n := [n,n+1] \subset \mathbb{R}\): each \(x\) lies outside \(E_n\) once \(n > x\), so \(1_{E_n} \to 1_{\emptyset}\) pointwise, yet \(m(E_n) = 1 \ne 0 = m(\emptyset)\).

Take \(c_x := \mu(\{x\})\), where \(\delta_x(E) = 1_E(x)\) (Example 1.4.22).

For an at most countable index set \(I\), the unordered sum \(\sum_{i \in I} a_i := \sup\{\sum_{i \in F} a_i : F \subset I \text{ finite}\}\) agrees with the series along any enumeration of \(I\) (each partial sum is a finite-subset sum; each finite subset lies in an initial segment), so the map \(E \mapsto \sum_{x \in X} c_x 1_E(x) = \sum_{x \in E} c_x\) is unambiguous.

Existence. Let \(E \subset X\), necessarily at most countable, and write it as the disjoint union of its singletons. Countable additivity (padding with copies of \(\emptyset\) when \(E\) is finite) gives

\begin{equation*} \mu(E) = \sum_{x \in E} \mu(\{x\}) = \sum_{x \in E} c_x , \end{equation*}

that is, \(\mu = \sum_{x \in X} c_x \delta_x\).

Uniqueness. Evaluating any such representation at \(E = \{y\}\) forces \(c_y = \mu(\{y\})\).

The completion is

\begin{equation*} \begin{aligned} \overline{\mathcal{B}} &:= \{ A \Delta Z : A \in \mathcal{B}, \ Z \ \mathcal{B}\text{-subnull} \}, \\ \overline{\mu}(A \Delta Z) &:= \mu(A). \end{aligned} \end{equation*}

Throughout, \(\Delta\) makes the subsets of \(X\) a group with identity \(\emptyset\) (immediate from \(1_{A \Delta B} = 1_A + 1_B \bmod 2\)), so \(E = A \Delta Z\) iff \(Z = E \Delta A\).

A \(\sigma\)-algebra containing \(\mathcal{B}\). Taking \(Z = \emptyset\) gives \(\mathcal{B} \subset \overline{\mathcal{B}}\). Complements: \(X \setminus (A \Delta Z) = (X \setminus A) \Delta Z\). Countable unions: let \(E_n = A_n \Delta Z_n\) with \(Z_n \subset N_n\) null, and put \(E := \bigcup_n E_n\), \(A := \bigcup_n A_n \in \mathcal{B}\), \(N := \bigcup_n N_n\), which is null by countable subadditivity (Exercise 1.4.23(i)). Then

\begin{equation*} E \Delta A \subset \bigcup_n (E_n \Delta A_n) = \bigcup_n Z_n \subset N \end{equation*}

(a point of \(E \setminus A\) lies in some \(E_n\) and in no \(A_n\), and symmetrically), so \(E = A \Delta (E \Delta A) \in \overline{\mathcal{B}}\).

Well defined and countably additive. If \(A \Delta Z = B \Delta W\) with \(Z \subset N\), \(W \subset M\) null, then \(A \Delta B = Z \Delta W \subset N \cup M\), which is null by finite subadditivity; since \(A \subset B \cup (A \Delta B)\), monotonicity and subadditivity give \(\mu(A) \le \mu(B)\), and symmetrically, so \(\mu(A) = \mu(B)\). Taking \(Z = \emptyset\) gives \(\overline{\mu}|_{\mathcal{B}} = \mu\), so \(\overline{\mu}(\emptyset) = 0\). For disjoint \(E_n = A_n \Delta Z_n\), set \(N := \bigcup_n N_n\) (null, as above) and \(B_n := A_n \setminus N = E_n \setminus N \in \mathcal{B}\), the two descriptions agreeing because \(E_n\) and \(A_n\) differ only inside \(N\). The \(B_n\) are disjoint with \(E_n \Delta B_n \subset N\), and \(E \Delta B \subset N\) for \(E := \bigcup_n E_n\), \(B := \bigcup_n B_n\); so by countable additivity of \(\mu\),

\begin{equation*} \overline{\mu}(E) = \mu(B) = \sum_{n=1}^{\infty} \mu(B_n) = \sum_{n=1}^{\infty} \overline{\mu}(E_n). \end{equation*}

Complete. If \(\overline{\mu}(A \Delta Z) = 0\) with \(Z \subset N\) null, then \(\mu(A) = 0\) and \(A \Delta Z \subset A \cup N\), a null set; so any \(S \subset A \Delta Z\) is \(\mathcal{B}\)-subnull, whence \(S = \emptyset \Delta S \in \overline{\mathcal{B}}\) with \(\overline{\mu}(S) = \mu(\emptyset) = 0\).

Coarsest. Let \((X,\mathcal{C},\nu)\) be any complete refinement and \(E = A \Delta Z \in \overline{\mathcal{B}}\) with \(Z \subset N\) \(\mathcal{B}\)-null. Then \(N \in \mathcal{B} \subset \mathcal{C}\) with \(\nu(N) = \mu(N) = 0\), so completeness gives \(Z \in \mathcal{C}\) and \(\nu(Z) = 0\), hence \(E \in \mathcal{C}\); and \(E \subset A \cup Z\), \(A \subset E \cup Z\) with finite subadditivity give

\begin{equation*} \nu(E) = \nu(A) = \mu(A) = \overline{\mu}(E). \end{equation*}

So \((X,\mathcal{C},\nu)\) is a refinement of \((X,\overline{\mathcal{B}},\overline{\mu})\).

Uniqueness. Any two coarsest complete refinements each refine the other, hence coincide.

Write \(\mathcal{L} := \mathcal{L}[\mathbb{R}^d]\), \(\mathcal{B} := \mathcal{B}[\mathbb{R}^d]\). By Exercise 1.4.26 it suffices to prove \(\mathcal{L} = \overline{\mathcal{B}}\), where \(\overline{\mathcal{B}}\) is the family of sets \(A \Delta Z\) with \(A\) Borel and \(Z\) inside a Borel null set, and that \(m\) agrees with \(\overline{m}(A \Delta Z) = m(A)\). That \((\mathbb{R}^d,\mathcal{L},m)\) refines \((\mathbb{R}^d,\mathcal{B},m)\) is Lemma 1.2.13(i),(iv),(v),(vi): open sets are Lebesgue measurable and \(\mathcal{L}\) is a \(\sigma\)-algebra, so \(\mathcal{B} \subset \mathcal{L}\), with the same \(m\).

Every Lebesgue null set lies in a Borel null set. If \(m^{\ast}(Z) = 0\), then for each \(k\) cover \(Z\) by boxes of total volume \(\le 1/(2k)\) and enlarge the \(j\)th to an open box, adding at most \(2^{-j}/(2k)\); the union \(V_k\) is open with \(Z \subset V_k\) and \(m^{\ast}(V_k) \le 1/k\). Then \(N := \bigcap_k V_k\) is a Borel \(G_{\delta}\) containing \(Z\) with \(m(N) = 0\).

\(\overline{\mathcal{B}} \subset \mathcal{L}\). If \(Z\) lies in a Borel null set then \(m^{\ast}(Z) = 0\), so \(Z \in \mathcal{L}\) by Lemma 1.2.13(iii); and \(\mathcal{L}\) is a \(\sigma\)-algebra, hence closed under \(A \Delta Z\).

\(\mathcal{L} \subset \overline{\mathcal{B}}\). Let \(E \in \mathcal{L}\). By Exercise 1.2.7 pick open \(U_n \supset E\) with \(m^{\ast}(U_n \setminus E) \le 1/n\); then \(G := \bigcap_n U_n\) is Borel, \(E \subset G\), and \(m^{\ast}(G \setminus E) \le 1/n\) for all \(n\), so \(G \setminus E\) is \(\mathcal{B}\)-subnull by the previous paragraph. Since \(E \subset G\),

\begin{equation*} G \Delta (G \setminus E) = G \cap E = E \in \overline{\mathcal{B}} . \end{equation*}

The measures agree. For \(E = A \Delta Z\) with \(Z\) in a Borel null set we have \(m(Z) = 0\), while \(E \subset A \cup Z\) and \(A \subset E \cup Z\), so finite subadditivity gives \(m(E) = m(A) = \overline{m}(E)\).

Both parts use the triangle inequality \(A \Delta C \subset (A \Delta B) \cup (B \Delta C)\) and the identity \((X \setminus A) \Delta (X \setminus B) = A \Delta B\).

(i). Let

\begin{equation*} \mathcal{C} := \{ E \in \langle \mathcal{A} \rangle : \forall \varepsilon > 0 \ \exists F \in \mathcal{A} \text{ with } \mu(E \Delta F) < \varepsilon \}; \end{equation*}

since \(\mathcal{C} \subset \langle \mathcal{A} \rangle\), it is enough to show \(\mathcal{C}\) is a \(\sigma\)-algebra containing \(\mathcal{A}\).

(a) \(\mathcal{A} \subset \mathcal{C}\): take \(F := E\).

(b) Complements: if \(\mu(E \Delta F) < \varepsilon\) then \(X \setminus F \in \mathcal{A}\) and \(\mu((X \setminus E) \Delta (X \setminus F)) = \mu(E \Delta F) < \varepsilon\).

(c) Countable unions: let \(E := \bigcup_n E_n\) with \(E_n \in \mathcal{C}\), and \(E^{(N)} := \bigcup_{n \le N} E_n\). These increase to \(E\), so \(\mu(E^{(N)}) \to \mu(E)\) by Exercise 1.4.23(ii); all measures are at most \(\mu(X) < \infty\), so \(\mu(E \Delta E^{(N)}) = \mu(E) - \mu(E^{(N)}) \to 0\) and we may fix \(N\) with \(\mu(E \Delta E^{(N)}) < \varepsilon/2\). Choose \(F_n \in \mathcal{A}\) with \(\mu(E_n \Delta F_n) < \varepsilon 2^{-n-1}\) and set \(F := \bigcup_{n \le N} F_n \in \mathcal{A}\); from

\begin{equation*} \Big( \bigcup_{n \le N} E_n \Big) \Delta \Big( \bigcup_{n \le N} F_n \Big) \subset \bigcup_{n \le N} (E_n \Delta F_n) \end{equation*}

and finite subadditivity, \(\mu(E^{(N)} \Delta F) < \varepsilon/2\), whence

\begin{equation*} \mu(E \Delta F) \le \mu(E \Delta E^{(N)}) + \mu(E^{(N)} \Delta F) < \varepsilon . \end{equation*}

(ii). Replacing \(A_n\) by \(A_1 \cup \cdots \cup A_n\) (still in \(\mathcal{A}\), still of finite measure by subadditivity) we may assume \(A_1 \subset A_2 \subset \cdots\) with union \(X\). The sets \(E \cap A_n\) increase to \(E\), so \(\mu(E \cap A_n) \to \mu(E) < \infty\) (Exercise 1.4.23(ii)) and \(\mu(E \setminus A_n) = \mu(E) - \mu(E \cap A_n) \to 0\); fix \(N\) with \(\mu(E \setminus A_N) < \varepsilon/2\) and set \(Y := A_N\), so \(\mu(Y) < \infty\).

Now \(\mathcal{A}|_Y := \{ A \cap Y : A \in \mathcal{A} \}\) is a Boolean algebra on \(Y\) (the complement of \(A \cap Y\) in \(Y\) is \((X \setminus A) \cap Y\)), and \(\langle \mathcal{A} \rangle|_Y = \langle \mathcal{A}|_Y \rangle\): the left side is a \(\sigma\)-algebra on \(Y\) containing \(\mathcal{A}|_Y\), hence contains the right; conversely \(\mathcal{D} := \{ B \subset X : B \cap Y \in \langle \mathcal{A}|_Y \rangle \}\) is a \(\sigma\)-algebra on \(X\) containing \(\mathcal{A}\), so \(\langle \mathcal{A} \rangle \subset \mathcal{D}\). The restriction of \(\mu\) to \(\langle \mathcal{A}|_Y \rangle\) is a measure (Example 1.4.30) of total mass \(\mu(Y) < \infty\) and \(E \cap Y \in \langle \mathcal{A}|_Y \rangle\), so part (i) supplies \(F_0 \in \mathcal{A}\) with \(\mu((E \cap Y) \Delta (F_0 \cap Y)) < \varepsilon/2\).

Put \(F := F_0 \cap Y \in \mathcal{A}\). As \(F \subset Y\), a point of \(E \Delta F\) either lies outside \(Y\), hence in \(E \setminus Y\), or lies in \(Y\), hence in \((E \cap Y) \Delta F\); so by finite subadditivity

\begin{equation*} \mu(E \Delta F) \le \mu(E \setminus A_N) + \mu\big( (E \cap Y) \Delta F \big) < \varepsilon . \end{equation*}

Measurable means \(f^{-1}(U) \in \mathcal{B}\) for every open \(U\) in the target (Definition 1.4.32). Preimages commute with complements and countable unions, and both targets have a countable base – the sets \([0,b)\), \((a,b)\), \((a,+\infty]\) with \(0 \le a < b < \infty\) rational for \([0,+\infty]\), the rational discs for \(\mathbb{C}\) – so it suffices to test basic open sets.

(i). If \(f\) is measurable then \(\{f > \lambda\} = f^{-1}((\lambda,+\infty]) \in \mathcal{B}\) for \(\lambda \ge 0\), and \(\{f > \lambda\} = X\) for \(\lambda < 0\). Conversely, \(\{f \ge b\} = \bigcap_{n}\{f > b - \tfrac1n\} \in \mathcal{B}\), so

\begin{equation*} f^{-1}([0,b)) = X \setminus \{f \ge b\}, \quad f^{-1}((a,+\infty]) = \{f > a\}, \quad f^{-1}((a,b)) = \{f>a\} \cap \{f<b\} \end{equation*}

all lie in \(\mathcal{B}\); every open set is a countable union of these.

(ii). For \(E \in \mathcal{B}\) and \(U\) open, \(1_E^{-1}(U)\) is one of \(\emptyset, E, X \setminus E, X\), all in \(\mathcal{B}\). Conversely \(E = 1_E^{-1}((\tfrac12,+\infty]) \in \mathcal{B}\).

(iii). \(\mathcal{C} := \{E \subset Y : f^{-1}(E) \in \mathcal{B}\}\) is a \(\sigma\)-algebra on the target \(Y\), because preimages commute with complements and countable unions. If \(f\) is measurable then \(\mathcal{C}\) contains every open subset of \(Y\), hence contains the generated \(\sigma\)-algebra \(\mathcal{B}[Y]\). The converse is immediate, open sets being Borel.

(iv). Write \(f = u + iv\) and \(S_{a,b} := \{z \in \mathbb{C} : a < \mathrm{Re}\,z < b\}\), an open strip. If \(f\) is measurable then \(u^{-1}((a,b)) = f^{-1}(S_{a,b}) \in \mathcal{B}\); since \(u\) is real-valued, \(u^{-1}(U) = u^{-1}(U \cap \mathbb{R})\) and \(U \cap \mathbb{R}\) is a countable union of intervals, so \(u^{-1}(U) \in \mathcal{B}\), and likewise for \(v\). Conversely, for an open rectangle \(R\) with \(a < \mathrm{Re}\,z < b\), \(c < \mathrm{Im}\,z < d\),

\begin{equation*} f^{-1}( R) = u^{-1}(S_{a,b}) \cap v^{-1}(S_{c,d}) \in \mathcal{B}, \end{equation*}

and every open \(U \subset \mathbb{C}\) is a countable union of rectangles with rational corners.

(v). Here \(f\) real-valued is read as \(\mathbb{C}\)-valued and \(f_{\pm}\) as \([0,+\infty]\)-valued. If \(f\) is measurable then \(\{f_+ > \lambda\} = f^{-1}(\{\mathrm{Re}\,z > \lambda\}) \in \mathcal{B}\) for \(\lambda \ge 0\) and \(= X\) for \(\lambda < 0\), so \(f_+\) is measurable by (i); applying this to \(-f\) (measurable, as \((-f)^{-1}(U) = f^{-1}(-U)\)) handles \(f_-\). Conversely, \(\{f > \lambda\} = \{f_+ > \lambda\}\) for \(\lambda \ge 0\), while for \(\lambda < 0\), \(\{f > \lambda\} = \{f_- < -\lambda\} = X \setminus \bigcap_n \{f_- > -\lambda - \tfrac1n\} \in \mathcal{B}\); then \(f^{-1}((a,b)) = \{f>a\} \setminus \bigcap_n\{f > b - \tfrac1n\} \in \mathcal{B}\), and as in (iv) every open \(U \subset \mathbb{C}\) meets \(\mathbb{R}\) in a countable union of intervals.

(vi). Unsigned case: for \(\lambda \in [0,\infty)\),

\begin{equation*} \{f > \lambda\} = \bigcup_{m=1}^{\infty} \bigcup_{N=1}^{\infty} \bigcap_{n \ge N} \{ f_n > \lambda + \tfrac1m \} \in \mathcal{B} \end{equation*}

by (i). Indeed if \(f(x) > \lambda\), choose \(m\) with \(\lambda + \tfrac2m < f(x)\) (possible also when \(f(x) = +\infty\)) and use \(f_n(x) \to f(x)\); conversely \(f_n(x) > \lambda + \tfrac1m\) for all \(n \ge N\) forces \(f(x) \ge \lambda + \tfrac1m > \lambda\). So \(f\) is measurable by (i). Complex case: \(u_n \to u\) and \(v_n \to v\) pointwise with \(u_n, v_n\) measurable by (iv); \(t \mapsto t_{\pm}\) is continuous, so \((u_n)_{\pm} \to u_{\pm}\), each \((u_n)_{\pm}\) is measurable by (v), hence \(u_{\pm}\) is measurable by the unsigned case and \(u\) by (v); likewise \(v\), and then \(f\) by (iv).

(vii). \((\varphi \circ f)^{-1}(U) = f^{-1}(\varphi^{-1}(U)) \in \mathcal{B}\), since \(\varphi^{-1}(U)\) is open by continuity. The argument uses nothing about the target, so it applies verbatim for \(\mathbb{C}\).

(viii). Unsigned case: for \(\lambda \in [0,\infty)\),

\begin{equation*} \begin{aligned} \{f+g > \lambda\} &= \bigcup_{q \in \mathbb{Q}} \big( \{f > q\} \cap \{g > \lambda - q\} \big), \\ \{fg > \lambda\} &= \bigcup_{q \in \mathbb{Q},\, q > 0} \big( \{f > q\} \cap \{g > \lambda/q\} \big), \end{aligned} \end{equation*}

with \(\{f > q\} := X\) for \(q < 0\). For the first: \(f(x) > q\) and \(g(x) > \lambda - q\) give \(f(x)+g(x) > \lambda\), and conversely \(f(x) > \lambda - g(x)\) as extended reals, so density of \(\mathbb{Q}\) supplies \(q\) with \(\lambda - g(x) < q < f(x)\). For the second (convention \(0 \cdot \infty = 0\)): \(f(x) > q > 0\) and \(g(x) > \lambda/q\) give \(f(x)g(x) > \lambda\) by monotonicity of multiplication; conversely \(f(x)g(x) > \lambda \ge 0\) forces \(f(x), g(x) > 0\) and \(\lambda/g(x) < f(x)\) (read as \(0\) when \(g(x) = +\infty\)), so pick rational \(q\) with \(\max(0, \lambda/g(x)) < q < f(x)\). Both right-hand sides lie in \(\mathcal{B}\) by (i). Complex case: \(F := (f,g) : X \to \mathbb{C}^2\) satisfies \(F^{-1}(D \times D_1) = f^{-1}(D) \cap g^{-1}(D_1) \in \mathcal{B}\) for discs \(D, D_1\), and every open subset of \(\mathbb{C}^2\) is a countable union of rational disc-products, so \(F^{-1}(U) \in \mathcal{B}\) for all open \(U \subset \mathbb{C}^2\); addition and multiplication being continuous \(\mathbb{C}^2 \to \mathbb{C}\), part (vii) applies.

All three conditions are equivalent, with \(c_\alpha\) the value of \(f\) on \(A_\alpha\); write \(Y\) for the target.

By Example 1.4.7, \(\mathcal{B}\) consists exactly of the unions \(\bigcup_{\beta \in J} A_\beta\) with \(J \subset I\); since the atoms are disjoint, a \(\mathcal{B}\)-measurable set either contains a given atom or is disjoint from it.

(i) Constant on atoms \(\Rightarrow\) measurable. If \(f \equiv c_\alpha\) on \(A_\alpha\), then for any \(E \subset Y\),

\begin{equation*} f^{-1}(E) = \bigcup \{ A_\alpha : \alpha \in I, \ c_\alpha \in E \} \in \mathcal{B}, \end{equation*}

since the atoms partition \(X\); in particular this holds for every open \(E\).

(ii) Measurable \(\Rightarrow\) constant on atoms. If \(f(x) \ne f(y)\) for some \(x,y \in A_\alpha\), choose (both targets being Hausdorff) an open \(U \ni f(x)\) with \(f(y) \notin U\). Then \(f^{-1}(U) \in \mathcal{B}\) meets \(A_\alpha\), hence contains it, forcing \(f(y) \in U\) – a contradiction.

(iii) Equivalence with the representation. Each \(x\) lies in exactly one atom \(A_{\alpha_0}\), so at \(x\) the sum \(\sum_{\alpha \in I} c_\alpha 1_{A_\alpha}\) has at most one non-zero term (using \(0 \cdot \infty = 0\) in the unsigned case) and equals \(c_{\alpha_0}\). The possibly uncountable sum is therefore unambiguous, and \(f = \sum_{\alpha \in I} c_\alpha 1_{A_\alpha}\) holds precisely when \(f \equiv c_\alpha\) on each \(A_\alpha\).

Uniqueness: evaluating at any \(x \in A_\alpha\), which exists as the atoms are non-empty, gives \(c_\alpha = f(x)\).

Take \(E := N \cup \bigcup_{m=1}^{\infty} E_{M_m,m}\), in the notation built below.

Let \(N \in \mathcal{B}\) be a null set containing \(\{x : f_n(x) \not\to f(x)\}\) (Definition 1.4.31), and put

\begin{equation*} \tilde f_n := f_n 1_{X \setminus N}, \qquad \tilde f := f 1_{X \setminus N}, \qquad h_n := \tilde f_n - \tilde f . \end{equation*}

Then \(\tilde f_n \to \tilde f\) pointwise on all of \(X\) (both vanish on \(N\); off \(N\) the original convergence holds), so each \(h_n\) is measurable by Exercise 1.4.29(ii),(vi),(viii), and \(g_n := |h_n| \to 0\) at every point of \(X\). Since \(\{z \in \mathbb{C} : |z| > 1/m\}\) is open,

\begin{equation*} E_{M,m} := \bigcup_{n \ge M} \{ g_n > \tfrac1m \} = \bigcup_{n \ge M} h_n^{-1}\big( \{ |z| > \tfrac1m \} \big) \in \mathcal{B}. \end{equation*}

Fix \(m\). The \(E_{M,m}\) decrease in \(M\) with \(\bigcap_M E_{M,m} = \emptyset\) (a point of the intersection would have \(g_n(x) > 1/m\) for infinitely many \(n\)), and \(\mu(E_{1,m}) \le \mu(X) < \infty\), so downwards monotone convergence (Exercise 1.4.23(iii)) gives \(\mu(E_{M,m}) \to 0\) – this is the only use of finiteness. Choose \(M_m\) with \(\mu(E_{M_m,m}) \le \varepsilon 2^{-m}\); then by countable subadditivity (Exercise 1.4.23(i)),

\begin{equation*} \mu(E) \le \mu(N) + \sum_{m=1}^{\infty} \varepsilon 2^{-m} = \varepsilon . \end{equation*}

For \(x \notin E\) we have \(x \notin N\), so \(g_n(x) = |f_n(x) - f(x)|\), and \(x \notin E_{M_m,m}\), so

\begin{equation*} \sup_{x \in X \setminus E} |f_n(x) - f(x)| \le \tfrac1m \qquad (n \ge M_m); \end{equation*}

as \(m\) is arbitrary, \(f_n \to f\) uniformly off \(E\).

Failure for infinite measure. On \(\mathbb{R}\) with Lebesgue measure take \(f_n := 1_{[n,n+1]}\), so \(f_n \to 0\) pointwise, and \(\varepsilon := 1/2\). If \(|f_n| \le 1/2\) off some measurable \(E\) for all \(n \ge M\), then \([n,n+1] \subset E\) for every \(n \ge M\), so \(E \supset [M,\infty)\) has infinite measure, never at most \(1/2\).

Partition \(X\) into the non-empty atoms \(A_1,\ldots,A_n\) of the finite algebra \(\mathcal{B}\) (Exercise 1.4.4) and write \(f = \sum_{i=1}^n c_i 1_{A_i}\) with \(c_i \in \mathbb{C}\) uniquely determined (Exercise 1.4.30). Call \(f\) absolutely convergent when

\begin{equation*} \mathrm{Simp}\int_X |f|\, d\mu = \sum_{i=1}^n |c_i| \mu(A_i) < \infty , \end{equation*}

equivalently \(c_i = 0\) whenever \(\mu(A_i) = \infty\), and then define

\begin{equation*} \mathrm{Simp}\int_X f\, d\mu := \sum_{i=1}^n c_i \mu(A_i) \in \mathbb{C}, \end{equation*}

with the convention \(0 \cdot \infty = 0\). Absolute convergence makes each term a finite complex number, and relabelling the atoms (Exercise 1.4.3) permutes the sum only.

This agrees with Definition 1.4.34 when \(f\) is unsigned, and with the Section 1.3 recipe \(\mathrm{Simp}\int f := (\mathrm{Simp}\int u_+ - \mathrm{Simp}\int u_-) + i(\mathrm{Simp}\int v_+ - \mathrm{Simp}\int v_-)\): writing \(c_i = a_i + ib_i\), uniqueness in Exercise 1.4.30 gives \(u = \sum_i a_i 1_{A_i}\) and \(u_{\pm} = \sum_i (a_i)_{\pm} 1_{A_i}\), while \(|a_i|, |b_i| \le |c_i|\) makes all four unsigned integrals finite, so the subtraction is legitimate and

\begin{equation*} \mathrm{Simp}\int_X u\, d\mu = \sum_i (a_i)_+ \mu(A_i) - \sum_i (a_i)_- \mu(A_i) = \sum_i a_i \mu(A_i), \end{equation*}

and likewise for \(v\).

Complex linearity, \(\mathrm{Simp}\int_X 1_E\, d\mu = \mu(E)\) for \(E \in \mathcal{B}\) of finite measure, and \(|\mathrm{Simp}\int_X f\, d\mu| \le \mathrm{Simp}\int_X |f|\, d\mu\) all read off from the formula (Check!), using \(f + g = \sum_i (c_i + d_i)1_{A_i}\). Finally, as in Exercise 1.4.33, the value is unchanged under a finite refinement \((X,\mathcal{B}^{\prime},\mu^{\prime})\): each \(A_i\) splits into \(\mathcal{B}^{\prime}\)-atoms of total \(\mu^{\prime}\)-measure \(\mu(A_i)\), on each of which \(f \equiv c_i\), and all terms are finite. So the definition extends, exactly as in Definition 1.4.35, to absolutely convergent complex-valued simple functions on an arbitrary measure space.

By Exercise 1.4.4 both algebras are atomic; let \(A_1,\ldots,A_n\) be the non-empty atoms of \(\mathcal{B}\) and \(A^{\prime}_1,\ldots,A^{\prime}_m\) those of \(\mathcal{B}^{\prime}\).

Each \(A^{\prime}_j\) lies in exactly one \(A_i\): pick \(x \in A^{\prime}_j\) and let \(A_i\) be the \(\mathcal{B}\)-atom containing it; then \(A_i \in \mathcal{B} \subset \mathcal{B}^{\prime}\) meets the \(\mathcal{B}^{\prime}\)-atom \(A^{\prime}_j\), hence contains it (Exercise 1.4.30), and uniqueness follows from disjointness of the \(A_i\). So with \(J_i := \{ j : A^{\prime}_j \subset A_i \}\), the \(J_i\) partition \(\{1,\ldots,m\}\) and \(A_i = \bigcup_{j \in J_i} A^{\prime}_j\) disjointly.

By Exercise 1.4.30, \(f = \sum_i c_i 1_{A_i}\) with \(c_i\) the value of \(f\) on \(A_i\); since \(\mathcal{B} \subset \mathcal{B}^{\prime}\), \(f\) is also \(\mathcal{B}^{\prime}\)-measurable, with \(f = \sum_j c^{\prime}_j 1_{A^{\prime}_j}\) and \(c^{\prime}_j = c_i\) for \(j \in J_i\). Finite additivity of \(\mu^{\prime}\) on the decomposition of \(A_i\) gives \(\sum_{j \in J_i} \mu^{\prime}(A^{\prime}_j) = \mu^{\prime}(A_i) = \mu(A_i)\), so Definition 1.4.34 applied twice yields

\begin{equation*} \begin{aligned} \mathrm{Simp}\int_X f\, d\mu^{\prime} &= \sum_{j=1}^m c^{\prime}_j \mu^{\prime}(A^{\prime}_j) = \sum_{i=1}^n c_i \sum_{j \in J_i} \mu^{\prime}(A^{\prime}_j) \\ &= \sum_{i=1}^n c_i \mu(A_i) = \mathrm{Simp}\int_X f\, d\mu , \end{aligned} \end{equation*}

the factoring of \(c_i\) out of the inner sum being valid because \(c(a+b) = ca + cb\) holds throughout \([0,+\infty]\) under \(0 \cdot \infty = 0\) (Check!).

Common-refinement lemma. Given simple \(f_1,\ldots,f_r : X \to [0,+\infty]\) there is a finite \(\sigma\)-algebra \(\mathcal{B}^{\prime\prime} \subset \mathcal{B}\) making all of them measurable, with non-empty atoms \(A_1,\ldots,A_n\), such that

\begin{equation*} f_s = \sum_{i=1}^n c^{(s)}_i 1_{A_i}, \qquad \mathrm{Simp}\int_X f_s\, d\mu = \sum_{i=1}^n c^{(s)}_i \mu(A_i). \end{equation*}

Indeed, each \(f_s\) is measurable with respect to the finite algebra \(\mathcal{B}_s\) generated by its finitely many level sets \(f_s^{-1}(\{a\}) \in \mathcal{B}\) (Exercise 1.4.29(iii), singletons being Borel); \(\mathcal{B}^{\prime\prime} := \langle \mathcal{B}_1 \cup \cdots \cup \mathcal{B}_r \rangle \subset \mathcal{B}\) is finite by Exercise 1.4.8, hence a \(\sigma\)-algebra (a countable union of its members takes finitely many values, so collapses to a finite one), and is a refinement of each \(\mathcal{B}_s\), so Exercise 1.4.33 lets the integral of Definition 1.4.35 be computed in \(\mathcal{B}^{\prime\prime}\) via Definition 1.4.34. The representations are those of Exercise 1.4.30. Throughout, \([0,+\infty]\) arithmetic with \(0 \cdot \infty = 0\) is commutative, associative, distributive and monotone.

(i). On common atoms \(f = \sum_i c_i 1_{A_i}\), \(g = \sum_i d_i 1_{A_i}\); evaluating at a point of the non-empty \(A_i\) gives \(c_i \le d_i\), hence \(\sum_i c_i \mu(A_i) \le \sum_i d_i \mu(A_i)\).

(ii). For \(\emptyset \ne E \ne X\) the algebra \(\{\emptyset, E, X \setminus E, X\} \subset \mathcal{B}\) has atoms \(E, X \setminus E\) and \(1_E = 1 \cdot 1_E + 0 \cdot 1_{X \setminus E}\), so

\begin{equation*} \mathrm{Simp}\int_X 1_E\, d\mu = 1 \cdot \mu(E) + 0 \cdot \mu(X \setminus E) = \mu(E), \end{equation*}

the last term vanishing even when \(\mu(X \setminus E) = \infty\). For \(E = \emptyset\) or \(E = X\) use \(\{\emptyset, X\}\), whose only atom is \(X\).

(iii). \(cf = \sum_i (c c_i) 1_{A_i}\) is simple and \(\mathcal{B}^{\prime\prime}\)-measurable, so by associativity and distributivity

\begin{equation*} \mathrm{Simp}\int_X cf\, d\mu = \sum_i (c c_i) \mu(A_i) = c \sum_i c_i \mu(A_i) = c \times \mathrm{Simp}\int_X f\, d\mu . \end{equation*}

(iv). On common atoms \(f + g = \sum_i (c_i + d_i) 1_{A_i}\) is simple, so

\begin{equation*} \mathrm{Simp}\int_X (f+g)\, d\mu = \sum_i c_i \mu(A_i) + \sum_i d_i \mu(A_i) \end{equation*}

by distributivity and rearrangement of a finite unsigned sum.

(v). A finite \(\mathcal{B}_f \subset \mathcal{B}\) making \(f\) measurable also lies in \(\mathcal{B}^{\prime}\), and \(\mu^{\prime}\) agrees with \(\mu\) on it, so both integrals are given by the same formula; independence of the choice of \(\mathcal{B}_f\) is Exercise 1.4.33.

(vi). Pick \(N \in \mathcal{B}\) null with \(f = g\) off \(N\), and atoms \(A_i\) from the lemma. The non-empty sets among \(A_i \cap N\) and \(A_i \setminus N\) partition \(X\) inside \(\mathcal{B}\) and generate a finite \(\sigma\)-algebra (Exercise 1.4.10) refining \(\mathcal{B}^{\prime\prime}\), with atoms \(B_1,\ldots,B_k\) each contained in or disjoint from \(N\), on each of which \(f\) and \(g\) are constant; so their integrals may be computed there (Exercise 1.4.33), say \(f = \sum_j c_j 1_{B_j}\), \(g = \sum_j d_j 1_{B_j}\). If \(B_j \cap N = \emptyset\) then \(c_j = d_j\); if \(B_j \subset N\) then \(\mu(B_j) = 0\) and \(c_j \mu(B_j) = 0 = d_j \mu(B_j)\) even for infinite coefficients. Summing gives the claim.

(vii). With \(f = \sum_i c_i 1_{A_i}\), a finite unsigned sum is finite iff every term is, so the integral is finite iff \(c_i \mu(A_i) < \infty\) for all \(i\); moreover \(\{f = \infty\} = \bigcup_{i : c_i = \infty} A_i\) and \(\{f \ne 0\} = \bigcup_{i : c_i \ne 0} A_i\) lie in \(\mathcal{B}\), with measures the corresponding sums of \(\mu(A_i)\) by finite additivity.

(a) If the integral is finite: \(c_i = \infty\) forces \(\mu(A_i) = 0\), so \(\mu(\{f = \infty\}) = 0\); and \(c_i \ne 0\) forces \(\mu(A_i) < \infty\), so \(\mu(\{f \ne 0\}) < \infty\).

(b) Conversely, if \(f\) is finite a.e. and vanishes off some \(F \in \mathcal{B}\) with \(\mu(F) < \infty\), then \(\mu(\{f = \infty\}) = 0\) and \(\mu(\{f \ne 0\}) \le \mu(F) < \infty\); the term \(c_i \mu(A_i)\) is then \(0\) for \(c_i = 0\), finite for \(0 < c_i < \infty\) (as \(A_i \subset \{f \ne 0\}\)), and \(\infty \cdot 0 = 0\) for \(c_i = \infty\) (as \(A_i \subset \{f = \infty\}\)).

(viii). \(\sum_i c_i \mu(A_i) = 0\) iff each \(i\) has \(c_i = 0\) or \(\mu(A_i) = 0\), which holds iff \(\mu(\{f \ne 0\}) = \sum_{i : c_i \ne 0} \mu(A_i) = 0\), i.e. iff \(f = 0\) almost everywhere.

Write \(A := \bigcup_{i=1}^n A_i\) and \(A_J := \bigcap_{i \in J} A_i\) for \(\emptyset \ne J \subset \{1,\ldots,n\}\). All lie in \(\mathcal{B}\) and have finite measure (\(A_J \subset A_{i_0}\) for any \(i_0 \in J\), and \(\mu(A) \le \sum_i \mu(A_i) < \infty\) by finite subadditivity), which licenses the subtraction at the end.

Pointwise we have the identity of functions

\begin{equation*} 1_A = 1 - \prod_{i=1}^n \big(1 - 1_{A_i}\big) = \sum_{J \ne \emptyset} (-1)^{|J|-1} 1_{A_J} . \end{equation*}

For the first equality, the product of the \(0\)-\(1\) factors equals \(1\) exactly when \(x \notin A_i\) for every \(i\), so it is \(1_{X \setminus A}\). For the second, expanding multilinearly gives \(\prod_i (1 - 1_{A_i}) = \sum_{J} (-1)^{|J|} 1_{A_J}\) with \(A_{\emptyset} := X\) (using \(\prod_{i \in J} 1_{A_i} = 1_{A_J}\)); removing the \(J = \emptyset\) term and negating yields the right-hand side.

To integrate without signed integrals, split off the two signs:

\begin{equation*} P := \sum_{J \ne \emptyset,\ |J| \text{ odd}} 1_{A_J}, \qquad N := \sum_{J \ne \emptyset,\ |J| \text{ even}} 1_{A_J} . \end{equation*}

Both are unsigned simple functions – measurable by Exercise 1.4.29(ii),(viii) and taking values in \(\{0,1,\ldots,2^n\}\) – and the identity rearranges to \(1_A + N = P\). Applying Exercise 1.4.34(iv) repeatedly together with Exercise 1.4.34(ii),

\begin{equation*} \mu(A) + \sum_{J \ne \emptyset,\ |J| \text{ even}} \mu(A_J) = \sum_{J \ne \emptyset,\ |J| \text{ odd}} \mu(A_J). \end{equation*}

Every term is finite and there are finitely many, so subtracting the even-cardinality sum gives

\begin{equation*} \mu\Big( \bigcup_{i=1}^n A_i \Big) = \sum_{J \subset \{1,\ldots,n\} : J \ne \emptyset} (-1)^{|J|-1} \mu\Big( \bigcap_{i \in J} A_i \Big). \end{equation*}

Throughout, \(\int_X f\,d\mu = \sup\{\mathrm{Simp}\int_X h\,d\mu : h \text{ simple},\ 0 \le h \le f\}\) (Definition 1.4.37), and \(h + h^{\prime}\), \(ch\), \(h1_E\) are simple whenever \(h,h^{\prime}\) are. We also record the partition formula: if \(A_1,\ldots,A_k \in \mathcal{B}\) partition \(X\) and \(h \equiv a_j\) on \(A_j\), then

\begin{equation*} \mathrm{Simp}\int_X h\, d\mu = \sum_{j=1}^k a_j \mu(A_j), \end{equation*}

since \(h\) is measurable for the finite algebra \(\langle A_1,\ldots,A_k\rangle\) whose non-empty atoms are the \(A_j\) (Definition 1.4.34, Exercise 1.4.30), and any such algebra computes the same value (Exercise 1.4.33). We prove the parts in the order (ii), (i), (v), (vi), (vii), (viii), (iii), (iv), (ix), (x), (xi).

(ii). Let \(N \in \mathcal{B}\) be null with \(f \le g\) off \(N\), and \(h\) simple with \(0 \le h \le f\). Then \(h^{\prime} := h 1_{X \setminus N}\) is simple with \(h^{\prime} \le g\) pointwise, so \(\mathrm{Simp}\int_X h^{\prime}\,d\mu \le \int_X g\,d\mu\); and \(h = h^{\prime}\) almost everywhere, so the two simple integrals agree by Exercise 1.4.34(vi). Take the supremum over \(h\).

(i). Apply (ii) in both directions.

(v). \(f\) is a competitor for itself, giving \(\ge\); and \(0 \le h \le f\) gives \(\mathrm{Simp}\int_X h\,d\mu \le \mathrm{Simp}\int_X f\,d\mu\) by Exercise 1.4.34(i), giving \(\le\).

(vi). The set \(E := \{f \ge \lambda\} = \bigcap_m \{f > \lambda - \tfrac1m\}\) lies in \(\mathcal{B}\) by Exercise 1.4.29(i), and \(h := \lambda 1_E\) is simple with \(0 \le h \le f\), so by Exercise 1.4.34(ii),(iii),

\begin{equation*} \lambda \mu(E) = \mathrm{Simp}\int_X \lambda 1_E\, d\mu \le \int_X f\, d\mu ; \end{equation*}

divide by \(\lambda \in (0,\infty)\).

(vii). \(\{f = \infty\} \subset \{f \ge \lambda\}\) for every \(\lambda\), so \(\mu(\{f = \infty\}) \le \lambda^{-1} \int_X f\,d\mu \to 0\) as \(\lambda \to \infty\), the integral being finite.

(viii). By (vi), \(\mu(\{f \ge 1/n\}) \le n \int_X f\,d\mu = 0\); since \(\{f > 0\} = \bigcup_n \{f \ge 1/n\}\), countable subadditivity (Exercise 1.4.23(i)) gives \(\mu(\{f > 0\}) = 0\).

(iii). Three cases.

(a) \(0 < c < \infty\): \(h \mapsto ch\) is a bijection from the competitors for \(f\) onto those for \(cf\), with inverse \(k \mapsto c^{-1}k\), and \(\mathrm{Simp}\int_X ch\,d\mu = c\,\mathrm{Simp}\int_X h\,d\mu\) by Exercise 1.4.34(iii); multiplication by \(c\) is an order-isomorphism of \([0,+\infty]\), hence commutes with the supremum.

(b) \(c = 0\): \(cf \equiv 0\) and the only competitor is \(h = 0\), so both sides are \(0\) (using \(0 \cdot \infty = 0\)).

(c) \(c = +\infty\): put \(E := \{f > 0\}\). If \(\mu(E) = 0\) then \(f = 0\) and \(cf = 0\) almost everywhere, so both sides are \(0\) by (i). If \(\mu(E) > 0\) then \(\int_X f\,d\mu > 0\) by (viii), so \(c \int_X f\,d\mu = +\infty\); and \(n 1_E \le cf\) for every \(n\) gives \(\int_X cf\,d\mu \ge n\mu(E) \to \infty\).

(iv). For simple \(0 \le u \le f\) and \(0 \le v \le g\), the function \(u+v\) is simple with \(u+v \le f+g\), so Exercise 1.4.34(iv) gives

\begin{equation*} \int_X (f+g)\, d\mu \ge \mathrm{Simp}\int_X u\, d\mu + \mathrm{Simp}\int_X v\, d\mu . \end{equation*}

Here \(u\) and \(v\) range independently, and addition on \([0,+\infty]\) is increasing and continuous in each variable, so the suprema may be taken one at a time.

(ix). By (ii) the integrals \(\int_X \min(f,n)\,d\mu\) increase to some \(L \le \int_X f\,d\mu\). Conversely let \(h\) be simple with \(0 \le h \le f\), taking the values \(a_j\) on its level sets \(A_1,\ldots,A_k\). Then \(\min(h,n)\) is simple with value \(\min(a_j,n)\) on \(A_j\) and \(\min(h,n) \le \min(f,n)\), so by the partition formula

\begin{equation*} \sum_{j=1}^k \min(a_j,n)\, \mu(A_j) \le \int_X \min(f,n)\, d\mu \le L . \end{equation*}

Each term tends to \(a_j\mu(A_j)\) in \([0,+\infty]\) (Check! – the cases \(a_j = 0\), \(a_j = \infty\), and \(\mu(A_j) \in \{0,\infty\}\) use \(0 \cdot \infty = 0\)), and the sum is finite, so \(\mathrm{Simp}\int_X h\,d\mu \le L\). Take the supremum over \(h\).

(x). Write \(E := \bigcup_n E_n\); by (ii) the integrals \(\int_X f 1_{E_n}\,d\mu\) increase to some \(L \le \int_X f1_E\,d\mu\). Conversely let \(h\) be simple with \(0 \le h \le f1_E\), so that \(h = h1_E\); with values \(a_j\) on level sets \(A_j\), the partition \(A_1 \cap E_n, \ldots, A_k \cap E_n, X \setminus E_n\) and the partition formula give

\begin{equation*} \sum_{j=1}^k a_j\, \mu(A_j \cap E_n) = \mathrm{Simp}\int_X h1_{E_n}\, d\mu \le \int_X f 1_{E_n}\, d\mu \le L . \end{equation*}

As \(A_j \cap E_n\) increases to \(A_j \cap E\), Exercise 1.4.23(ii) gives \(\mu(A_j \cap E_n) \to \mu(A_j \cap E)\), hence \(a_j \mu(A_j \cap E_n) \to a_j \mu(A_j \cap E)\) (for \(a_j = \infty\): both sides vanish if \(\mu(A_j \cap E) = 0\), and are \(+\infty\) for large \(n\) otherwise). So \(\mathrm{Simp}\int_X h\,d\mu = \sum_j a_j \mu(A_j \cap E) \le L\); take the supremum.

(xi). Extension by zero matches simple integrals: if \(w\) is simple on \((Y,\mathcal{B}\!\restriction_Y)\) with values \(b_i\) on level sets \(C_i \subset Y\), its extension \(\tilde w\) by \(0\) is constant on each member of the \(\mathcal{B}\)-partition \(C_1,\ldots,C_r, X \setminus Y\) of \(X\), so the partition formula gives

\begin{equation*} \mathrm{Simp}\int_X \tilde w\, d\mu = \sum_{i=1}^r b_i \mu(C_i) = \mathrm{Simp}\int_Y w\, d\mu\!\restriction_Y ; \end{equation*}

conversely a simple \(h\) on \(X\) with \(h = h1_Y\) restricts to a simple function on \(Y\) (since \((h\!\restriction_Y)^{-1}(U) = h^{-1}(U) \cap Y \in \mathcal{B}\!\restriction_Y\)) of which it is the zero-extension. Now each simple \(0 \le h \le f1_Y\) satisfies \(h = h1_Y\) and \(h\!\restriction_Y \le f\!\restriction_Y\), so \(\mathrm{Simp}\int_X h\,d\mu \le \int_Y f\!\restriction_Y d\mu\!\restriction_Y\); and each simple \(0 \le w \le f\!\restriction_Y\) has \(\tilde w \le f1_Y\), so \(\mathrm{Simp}\int_Y w\,d\mu\!\restriction_Y \le \int_X f1_Y\,d\mu\). Taking suprema gives equality.

By Exercise 1.4.22 both \(c\mu\) and \(\lambda := \sum_{n=1}^{\infty} \mu_n\) are measures on \(\mathcal{B}\), so every integral below is defined; and as all these measures live on the same \(\sigma\)-algebra, the competitor set \(\mathcal{S}_f := \{h \text{ simple} : 0 \le h \le f\}\) of Definition 1.4.37 is the same for all of them.

For simple \(h\) with distinct values \(a_1,\ldots,a_k\) on \(A_j := h^{-1}(\{a_j\})\), the \(A_j\) are exactly the atoms of the finite \(\sigma\)-algebra they generate, so for any measure \(\nu\) on \(\mathcal{B}\),

\begin{equation*} \mathrm{Simp}\int_X h\, d\nu = \sum_{j=1}^k a_j \nu(A_j), \end{equation*}

the sets \(A_j\) depending on \(h\) alone.

(i). Distributivity in \([0,+\infty]\) (with \(0 \cdot \infty = 0\)) gives \(\mathrm{Simp}\int_X h\,d(c\mu) = \sum_j c\, a_j \mu(A_j) = c\,\mathrm{Simp}\int_X h\,d\mu\), whence

\begin{equation*} \int_X f\, d(c\mu) = \sup_{h \in \mathcal{S}_f} c\, \mathrm{Simp}\int_X h\, d\mu = c \int_X f\, d\mu , \end{equation*}

using \(\sup_{a \in A} ca = c \sup A\) for non-empty \(A \subset [0,+\infty]\): for \(0 < c < \infty\) multiplication by \(c\) is an order-isomorphism; for \(c = 0\) both sides are \(0\); for \(c = +\infty\) both sides are \(0\) when \(\sup A = 0\) and \(+\infty\) otherwise.

(ii). For simple \(h\), distributivity followed by interchanging a finite sum with an unsigned infinite one gives

\begin{equation*} \mathrm{Simp}\int_X h\, d\lambda = \sum_{j=1}^k a_j \sum_{n=1}^{\infty} \mu_n(A_j) = \sum_{n=1}^{\infty} \mathrm{Simp}\int_X h\, d\mu_n . \end{equation*}

Upper bound. For \(h \in \mathcal{S}_f\) each \(\mathrm{Simp}\int_X h\,d\mu_n \le \int_X f\,d\mu_n\), so \(\mathrm{Simp}\int_X h\,d\lambda \le \sum_n \int_X f\,d\mu_n\); take the supremum over \(h\).

Lower bound. Fix \(N\) and \(h_1,\ldots,h_N \in \mathcal{S}_f\). Then \(h := \max(h_1,\ldots,h_N)\) is simple – \(\{h > \alpha\} = \bigcup_{n \le N}\{h_n > \alpha\}\) with Exercise 1.4.29(i), and \(h\) takes finitely many values – and lies in \(\mathcal{S}_f\) with \(h \ge h_n\), so Exercise 1.4.34(i) gives

\begin{equation*} \int_X f\, d\lambda \ge \mathrm{Simp}\int_X h\, d\lambda = \sum_{n=1}^{\infty} \mathrm{Simp}\int_X h\, d\mu_n \ge \sum_{n=1}^N \mathrm{Simp}\int_X h_n\, d\mu_n . \end{equation*}

The \(h_n\) range independently, and \(\sup\{a_1 + \cdots + a_N : a_n \in A_n\} = \sum_n \sup A_n\) for non-empty \(A_n \subset [0,+\infty]\), so \(\int_X f\,d\lambda \ge \sum_{n \le N} \int_X f\,d\mu_n\); let \(N \to \infty\).

Write \(\nu := \phi_{\ast}\mu\).

(i). \(\nu\) is well defined since \(\phi^{-1}(E) \in \mathcal{B}\) for every \(E \in \mathcal{C}\), and \(\nu(\emptyset) = \mu(\emptyset) = 0\). Preimages preserve disjointness and countable unions, so for disjoint \(E_1,E_2,\ldots \in \mathcal{C}\),

\begin{equation*} \nu\Big( \bigcup_{n=1}^{\infty} E_n \Big) = \mu\Big( \bigcup_{n=1}^{\infty} \phi^{-1}(E_n) \Big) = \sum_{n=1}^{\infty} \nu(E_n). \end{equation*}

(ii). Here \(f \circ \phi\) is measurable, being a composition of measurable morphisms (Remark 1.4.33). Throughout we use that a simple \(h\) with distinct values \(a_j\) on \(A_j := h^{-1}(\{a_j\})\) satisfies \(\mathrm{Simp}\int h\,d\lambda = \sum_j a_j \lambda(A_j)\) for any measure \(\lambda\) on the ambient \(\sigma\)-algebra, the \(A_j\) being the atoms of the finite algebra they generate (Exercise 1.4.33).

Simple \(h\) on \(Y\). With \(E_j := h^{-1}(\{a_j\})\), the function \(h \circ \phi\) is simple with \((h\circ\phi)^{-1}(\{a_j\}) = \phi^{-1}(E_j)\), so

\begin{equation*} \mathrm{Simp}\int_X (h \circ \phi)\, d\mu = \sum_{j=1}^k a_j\, \mu(\phi^{-1}(E_j)) = \mathrm{Simp}\int_Y h\, d\nu . \end{equation*}

The inequality \(\ge\). For simple \(0 \le h \le f\) on \(Y\), the function \(h \circ \phi\) is simple with \(h \circ \phi \le f \circ \phi\), so \(\mathrm{Simp}\int_Y h\,d\nu \le \int_X (f\circ\phi)\,d\mu\); take the supremum over \(h\).

The inequality \(\le\). Let \(g\) be simple on \(X\) with \(0 \le g \le f \circ \phi\), with distinct values \(a_1 > a_2 > \cdots > a_k \ge 0\) on \(A_j := g^{-1}(\{a_j\})\). Put \(B_j := \{y \in Y : f(y) \ge a_j\} \in \mathcal{C}\), so \(B_1 \subset \cdots \subset B_k\), write \(D_1 := B_1\), \(D_j := B_j \setminus B_{j-1}\), and set

\begin{equation*} h := \sum_{j=1}^k a_j 1_{D_j}, \end{equation*}

a simple function with \(h \le f\) (on \(D_j\), \(h = a_j \le f\); off \(B_k\), \(h = 0\)). With \(a_{k+1} := 0\), Abel summation

\begin{equation*} \sum_{j=1}^k a_j t_j = \sum_{j=1}^k (a_j - a_{j+1})(t_1 + \cdots + t_j), \qquad t_j \in [0,+\infty], \end{equation*}

is valid in \([0,+\infty]\): each \(a_j - a_{j+1} \ge 0\), and the coefficient of \(t_i\) on the right telescopes to \(\sum_{j \ge i}(a_j - a_{j+1}) = a_i\), also when \(a_i = +\infty\) (which forces \(i = 1\)). Applying it with \(t_j := \nu(D_j)\), whose partial sums are \(\nu(B_j)\) by finite additivity, and with \(t_j := \mu(A_j)\),

\begin{equation*} \begin{aligned} \mathrm{Simp}\int_Y h\, d\nu &= \sum_{j=1}^k (a_j - a_{j+1})\, \nu(B_j), \\ \mathrm{Simp}\int_X g\, d\mu &= \sum_{j=1}^k (a_j - a_{j+1})\, \mu(A_1 \cup \cdots \cup A_j). \end{aligned} \end{equation*}

If \(x \in A_i\) with \(i \le j\) then \(f(\phi(x)) \ge g(x) = a_i \ge a_j\), so \(A_1 \cup \cdots \cup A_j \subset \phi^{-1}(B_j)\) and \(\mu(A_1 \cup \cdots \cup A_j) \le \nu(B_j)\) by monotonicity. Comparing termwise, \(\mathrm{Simp}\int_X g\,d\mu \le \mathrm{Simp}\int_Y h\,d\nu \le \int_Y f\,d\nu\); take the supremum over \(g\).

Method (2). With Theorem 1.4.44 available, set \(h_n(y) := \min(2^{-n}\lfloor 2^n f(y)\rfloor, n)\), and \(h_n(y) := n\) where \(f(y) = +\infty\); these are simple and increase pointwise to \(f\), so \(h_n \circ \phi\) increase pointwise to \(f \circ \phi\), and monotone convergence on both spaces together with the simple case gives the identity directly.

Put \(S := T^{-1}\), again an invertible linear map, with \(\det S = 1/\det T\) by multiplicativity of \(\det\). Since \(T\) is a bijection with inverse \(S\),

\begin{equation*} T^{-1}(E) = \{ x : Tx \in E \} = \{ Sy : y \in E \} = S(E), \end{equation*}

which is Lebesgue measurable whenever \(E\) is, by Exercise 1.2.21 applied to \(S\); so \(T\) is a measurable morphism of \((\mathbb{R}^d,\mathcal{L}[\mathbb{R}^d])\) to itself and \(T_{\ast}m\) is a measure (Exercise 1.4.38(i)). The same Exercise 1.2.21 evaluates it: for every Lebesgue measurable \(E\),

\begin{equation*} T_{\ast}m(E) = m(S(E)) = |\det S|\, m(E) = \frac{1}{|\det T|}\, m(E). \end{equation*}

Measurability is automatic: \(\mathcal{B} = 2^X\), so \(f^{-1}(U) \in \mathcal{B}\) for every \(U \subset [0,+\infty]\). Write

\begin{equation*} \sum_{x \in X} f(x) := \sup\Big\{ \sum_{x \in F} f(x) : F \subset X \text{ finite} \Big\}, \end{equation*}

which is monotone in \(f\).

The identity for simple \(h\). With distinct values \(a_1,\ldots,a_k\) on \(A_j := h^{-1}(\{a_j\})\), the \(A_j\) are the atoms of the finite \(\sigma\)-algebra they generate, so \(\mathrm{Simp}\int_X h\,d\# = \sum_j a_j \#(A_j)\) by Definitions 1.4.34 and 1.4.35. Indices with \(a_j = 0\) contribute \(0\) to both sides (using \(0 \cdot \infty = 0\)); let \(J := \{ j : a_j > 0\}\), so \(\{h > 0\} = \bigcup_{j \in J} A_j\).

(a) If every \(j \in J\) has \(a_j < \infty\) and \(A_j\) finite, then \(F_0 := \bigcup_{j \in J} A_j\) is finite with

\begin{equation*} \sum_{j=1}^k a_j \#(A_j) = \sum_{j \in J} a_j |A_j| = \sum_{x \in F_0} h(x), \end{equation*}

and since \(h\) vanishes off \(F_0\), every finite \(F\) has \(\sum_{x \in F} h(x) \le \sum_{x \in F_0} h(x)\); so the supremum is attained at \(F_0\) and the two sides agree.

(b) Otherwise some \(j \in J\) has \(a_j = \infty\) (whence \(A_j \ne \emptyset\) and \(\#(A_j) \ge 1\)) or has \(A_j\) infinite (whence \(\#(A_j) = \infty\) and \(a_j > 0\)); either way \(a_j \#(A_j) = \infty\), so the left side is \(+\infty\). Correspondingly \(\sum_{x \in X} h(x) \ge h(x_0) = \infty\) for any \(x_0 \in A_j\) in the first case, and \(\sum_{x \in X} h(x) \ge N a_j \to \infty\) on choosing \(F \subset A_j\) with \(|F| = N\) in the second.

The inequality \(\le\). Every simple \(0 \le h \le f\) has \(\mathrm{Simp}\int_X h\,d\# = \sum_{x \in X} h(x) \le \sum_{x \in X} f(x)\); take the supremum over \(h\).

The inequality \(\ge\). For finite \(F \subset X\) the function \(f1_F\) takes at most \(|F|+1\) values, so it is simple with \(0 \le f1_F \le f\), and

\begin{equation*} \int_X f\, d\# \ge \mathrm{Simp}\int_X f1_F\, d\# = \sum_{x \in F} f(x); \end{equation*}

take the supremum over finite \(F\).

Throughout we use Definition 1.4.39, Theorem 1.4.38 (unsigned additivity, extended to finitely many summands by induction), Exercise 1.4.36, and measurability of sums, products, absolute values, real and imaginary parts and \(\pm\)-parts (Exercise 1.4.29(iv),(v),(vii),(viii)).

(i). \(L^1\) is a subset of the complex vector space of all functions \(X \to \mathbb{C}\) containing \(0\), so only closure matters: from \(|f+g| \le |f| + |g|\) with monotonicity (Exercise 1.4.36(ii)) and Theorem 1.4.38,

\begin{equation*} \int_X |f+g|\, d\mu \le \int_X |f|\, d\mu + \int_X |g|\, d\mu < \infty , \end{equation*}

and from \(|cf| = |c|\,|f|\) with Exercise 1.4.36(iii), \(\int_X |cf|\,d\mu = |c| \int_X |f|\,d\mu < \infty\).

(ii). (a) Real additivity. For real absolutely integrable \(f,g\) and \(h := f+g\), all finite-valued, \(h_+ - h_- = (f_+ - f_-) + (g_+ - g_-)\) rearranges into the unsigned identity \(h_+ + f_- + g_- = h_- + f_+ + g_+\). Integrating with Theorem 1.4.38 and noting all six integrals are finite (\(f_{\pm} \le |f|\), \(g_{\pm} \le |g|\), \(h_{\pm} \le |h|\)), we may rearrange to

\begin{equation*} \int_X h_+ d\mu - \int_X h_- d\mu = \Big( \int_X f_+ d\mu - \int_X f_- d\mu \Big) + \Big( \int_X g_+ d\mu - \int_X g_- d\mu \Big), \end{equation*}

which by Definition 1.4.39 is \(\int_X (f+g)\,d\mu = \int_X f\,d\mu + \int_X g\,d\mu\).

(b) Real homogeneity. For \(c \ge 0\), \((cf)_{\pm} = c f_{\pm}\), so Exercise 1.4.36(iii) gives \(\int_X cf\,d\mu = c \int_X f\,d\mu\); for \(c = -1\), \((-f)_{\pm} = f_{\mp}\) gives \(\int_X (-f)\,d\mu = -\int_X f\,d\mu\); a general real \(c\) is \(\pm|c|\).

(c) Complex case. The parts \(\mathrm{Re}\,f, \mathrm{Im}\,f\) are absolutely integrable (both are \(\le |f|\) in modulus), and \(\mathrm{Re}, \mathrm{Im}\) are additive, so Definition 1.4.39 and (a) give additivity. Real scalars commute with \(\mathrm{Re}, \mathrm{Im}\); and \(\mathrm{Re}(if) = -\mathrm{Im}\,f\), \(\mathrm{Im}(if) = \mathrm{Re}\,f\) give \(\int_X if\,d\mu = i \int_X f\,d\mu\), so for \(c = a+ib\) write \(cf = af + b(if)\). The value lies in \(\mathbb{C}\) because all four unsigned integrals are finite.

(iii). The two displays in (i) are exactly \(\|f+g\|_{L^1(\mu)} \le \|f\|_{L^1(\mu)} + \|g\|_{L^1(\mu)}\) and \(\|cf\|_{L^1(\mu)} = |c|\,\|f\|_{L^1(\mu)}\), the latter an equality because Exercise 1.4.36(iii) is one.

(iv). If \(f = g\) almost everywhere then \((\mathrm{Re}\,f)_{\pm} = (\mathrm{Re}\,g)_{\pm}\) and \((\mathrm{Im}\,f)_{\pm} = (\mathrm{Im}\,g)_{\pm}\) outside the same null set, so the four pairs of unsigned integrals agree by Exercise 1.4.36(i); assemble by Definition 1.4.39.

(v). Claim: if \(F : X \to [0,+\infty]\) is \(\mathcal{B}\)-measurable with \(\int_X F\,d\mu < \infty\), then \(\int_X F\,d\mu^{\prime} = \int_X F\,d\mu\). (Note every \(\mathcal{B}\)-measurable function is \(\mathcal{B}^{\prime}\)-measurable, since \(\mathcal{B} \subset \mathcal{B}^{\prime}\).)

(a) One inequality. Each \(\mathcal{B}\)-simple \(0 \le h \le F\) is \(\mathcal{B}^{\prime}\)-simple with \(\mathrm{Simp}\int_X h\,d\mu = \mathrm{Simp}\int_X h\,d\mu^{\prime} \le \int_X F\,d\mu^{\prime}\) by Exercise 1.4.34(v); take the supremum to get \(\int_X F\,d\mu \le \int_X F\,d\mu^{\prime}\).

(b) The reverse, truncated case. Assume also \(0 \le F \le M < \infty\) and \(F = 0\) off some \(E \in \mathcal{B}\) with \(\mu(E) < \infty\). Set \(A_j := \{x \in E : jM/n \le F(x) < (j+1)M/n\}\) for \(0 \le j < n-1\) and \(A_{n-1} := \{x \in E : (n-1)M/n \le F(x) \le M\}\); these \(\mathcal{B}\)-measurable sets partition \(E\), and the \(\mathcal{B}\)-simple functions \(u_n := \sum_{j} \tfrac{jM}{n} 1_{A_j}\), \(v_n := u_n + \tfrac{M}{n} 1_E\) satisfy \(u_n \le F \le v_n\) everywhere. Hence any \(\mathcal{B}^{\prime}\)-simple \(0 \le g \le F\) has \(g \le v_n\) and, by Exercise 1.4.34(i)-(v),

\begin{equation*} \begin{aligned} \mathrm{Simp}\int_X g\, d\mu^{\prime} &\le \mathrm{Simp}\int_X v_n\, d\mu^{\prime} = \mathrm{Simp}\int_X v_n\, d\mu \\ &= \mathrm{Simp}\int_X u_n\, d\mu + \tfrac{M}{n}\mu(E) \le \int_X F\, d\mu + \tfrac{M}{n}\mu(E) . \end{aligned} \end{equation*}

Let \(n \to \infty\) (both \(M\) and \(\mu(E)\) are finite) and take the supremum over \(g\).

(c) General \(F\). By Markov’s inequality (Exercise 1.4.36(vi)) the sets \(E_k := \{F > 1/k\} \in \mathcal{B}\) have finite measure and increase to \(\{F > 0\}\). Each \(\min(F,n)1_{E_k}\) is bounded by \(n\) and vanishes off \(E_k\), so (b) gives \(\int_X \min(F,n)1_{E_k}\,d\mu^{\prime} = \int_X \min(F,n)1_{E_k}\,d\mu\). Letting \(k \to \infty\) with horizontal truncation (Exercise 1.4.36(x)) in each of the two spaces, and using \(\min(F,n)1_{\{F>0\}} = \min(F,n)\), gives \(\int_X \min(F,n)\,d\mu^{\prime} = \int_X \min(F,n)\,d\mu\); now let \(n \to \infty\) with vertical truncation (Exercise 1.4.36(ix)).

Applying the Claim to \(|f|\) puts \(f\) in \(L^1(X,\mathcal{B}^{\prime},\mu^{\prime})\) with the same norm, and applying it to each of \((\mathrm{Re}\,f)_{\pm}\), \((\mathrm{Im}\,f)_{\pm}\) – all dominated by \(|f|\), hence of finite \(\mu\)-integral – gives \(\int_X f\,d\mu^{\prime} = \int_X f\,d\mu\) by Definition 1.4.39.

(vi). \(\int_X |f|\,d\mu = 0\) forces \(|f| = 0\) almost everywhere by Exercise 1.4.36(viii), and conversely \(f = 0\) almost everywhere gives \(\int_X |f|\,d\mu = \int_X 0\,d\mu = 0\) by Exercise 1.4.36(i).

(vii). The restriction is measurable since \((f\!\restriction_Y)^{-1}(U) = f^{-1}(U) \cap Y \in \mathcal{B}\!\restriction_Y\) (Exercise 1.4.29(iii)), and Exercise 1.4.36(xi) applied to \(|f|\) gives

\begin{equation*} \int_Y |f\!\restriction_Y|\, d\mu\!\restriction_Y = \int_X |f| 1_Y\, d\mu \le \int_X |f|\, d\mu < \infty , \end{equation*}

so \(f\!\restriction_Y \in L^1(Y,\mathcal{B}\!\restriction_Y,\mu\!\restriction_Y)\) and \(f1_Y \in L^1(X,\mathcal{B},\mu)\). Real parts, imaginary parts and \(\pm\)-parts commute both with restriction to \(Y\) and with multiplication by \(1_Y\) (e.g. \(\max(a1_Y,0) = \max(a,0)1_Y\)), so Exercise 1.4.36(xi) applied to each of \((\mathrm{Re}\,f)_{\pm}\), \((\mathrm{Im}\,f)_{\pm}\) – all eight quantities finite – assembles by Definition 1.4.39 into \(\int_X f1_Y\,d\mu = \int_Y f\!\restriction_Y d\mu\!\restriction_Y\).

On a finite measure space constants are integrable: for \(\varepsilon \in (0,\infty)\), Exercises 1.4.36(v) and 1.4.34(ii),(iii) give

\begin{equation*} \int_X \varepsilon\, d\mu = \mathrm{Simp}\int_X \varepsilon 1_X\, d\mu = \varepsilon \mu(X) < \infty . \end{equation*}

Uniform convergence is pointwise, so \(f\) is measurable by Exercise 1.4.29(vi).

Unsigned case. Given \(\varepsilon > 0\), fix \(N\) with \(f - \varepsilon \le f_n \le f + \varepsilon\) pointwise for all \(n \ge N\) (conventions \(\infty \pm \varepsilon = \infty\)). Monotonicity (Exercise 1.4.36(ii)) and Theorem 1.4.38 give, for \(n \ge N\),

\begin{equation*} \int_X f_n\, d\mu \le \int_X f\, d\mu + \varepsilon\mu(X), \qquad \int_X f\, d\mu \le \int_X f_n\, d\mu + \varepsilon\mu(X). \end{equation*}

(i) If \(\int_X f\,d\mu = \infty\), the second bound together with \(\varepsilon\mu(X) < \infty\) forces \(\int_X f_n\,d\mu = \infty\) for every \(n \ge N\), so the integrals are eventually equal to \(\int_X f\,d\mu\).

(ii) Otherwise the first bound makes \(\int_X f_n\,d\mu\) finite for \(n \ge N\), and the two bounds combine to \(|\int_X f_n\,d\mu - \int_X f\,d\mu| \le \varepsilon\mu(X)\); let \(\varepsilon \downarrow 0\).

Absolutely integrable case. Given \(\varepsilon > 0\), fix \(N\) with \(|f_n - f| \le \varepsilon\) everywhere for \(n \ge N\). Taking \(\varepsilon = 1\) first, the bound \(|f| \le |f_N| + 1\) gives \(\int_X |f|\,d\mu \le \int_X |f_N|\,d\mu + \mu(X) < \infty\), so \(f \in L^1\). For \(n \ge N\) the function \(g := f_n - f\) lies in \(L^1\) (Exercise 1.4.41(i)) with \(|g| \le \varepsilon\), so each of \((\mathrm{Re}\,g)_{\pm}\), \((\mathrm{Im}\,g)_{\pm}\) is bounded by \(\varepsilon\) and has integral in \([0,\varepsilon\mu(X)]\); hence \(|\int_X \mathrm{Re}\,g\,d\mu| \le \varepsilon\mu(X)\) and likewise for \(\mathrm{Im}\,g\), and by linearity (Exercise 1.4.41(ii)) and the triangle inequality in \(\mathbb{C}\),

\begin{equation*} \Big| \int_X f_n\, d\mu - \int_X f\, d\mu \Big| = \Big| \int_X g\, d\mu \Big| \le 2\varepsilon\mu(X) . \end{equation*}

Take \(X = \mathbb{R}\) with Lebesgue measure \(m\) and \(f_n := 1_{[n,\,n+1)} - 1_{[n+1,\,n+2)}\).

Each \(f_n\) is simple with \(\int_{\mathbb{R}} |f_n|\,dm = m([n,n+2)) = 2 < \infty\), and at each \(x\) at most two of the \(f_n\) are non-zero (the intervals \([n,n+1)\) are disjoint), so \(\sum_n |f_n(x)| \le 2\) and the series converges absolutely at every point. The partial sums telescope,

\begin{equation*} \sum_{n=1}^{N} f_n = 1_{[1,2)} - 1_{[N+1,\,N+2)} \longrightarrow 1_{[1,2)} \end{equation*}

pointwise, the subtracted term vanishing at \(x\) once \(N + 1 > x\). Hence

\begin{equation*} \int_{\mathbb{R}} \sum_{n=1}^{\infty} f_n\, dm = m([1,2)) = 1 , \end{equation*}

whereas linearity gives \(\int_{\mathbb{R}} f_n\,dm = m([n,n+1)) - m([n+1,n+2)) = 0\) for every \(n\), so \(\sum_n \int_{\mathbb{R}} f_n\,dm = 0 \ne 1\). Corollary 1.4.46 therefore fails for absolutely integrable \(f_n\), the mass \(-1\) escaping to horizontal infinity (Example 1.4.40).

Let \(F := \sum_{n=1}^{\infty} 1_{E_n}\), the increasing limit of the measurable partial sums, hence unsigned measurable by Exercise 1.4.29(ii),(vi),(viii); at each \(x\),

\begin{equation*} F(x) = \sum_{n=1}^{\infty} 1_{E_n}(x) = \#\{ n \in \mathbb{N} : x \in E_n \} \in [0,+\infty], \end{equation*}

the partial sums counting \(\{n \le N : x \in E_n\}\). By Corollary 1.4.46 (Tonelli for sums and integrals) and \(\int_X 1_{E_n}\,d\mu = \mu(E_n)\) (Exercises 1.4.36(v), 1.4.34(ii)),

\begin{equation*} \int_X F\, d\mu = \sum_{n=1}^{\infty} \int_X 1_{E_n}\, d\mu = \sum_{n=1}^{\infty} \mu(E_n) < \infty . \end{equation*}

So \(F\) is finite almost everywhere by Exercise 1.4.36(vii), i.e. almost every \(x\) lies in only finitely many \(E_n\).

(i). The set of points lying in infinitely many \(E_n\) is

\begin{equation*} E := \bigcap_{N=1}^{\infty} F_N \in \mathcal{B}, \qquad F_N := \bigcup_{n \ge N} E_n \in \mathcal{B}, \end{equation*}

since \(x \in \bigcap_N F_N\) says exactly that \(\{n : x \in E_n\}\) is unbounded, hence infinite. By countable subadditivity (Exercise 1.4.23(i)) and monotonicity (Exercise 1.4.20(i)),

\begin{equation*} \mu(E) \le \mu(F_N) \le \sum_{n=N}^{\infty} \mu(E_n) =: t_N , \end{equation*}

and \(t_N \to 0\) as the tail of a convergent series, so \(\mu(E) = 0\).

(ii). On \(X = [0,1]\) with Lebesgue measure take the typewriter sequence: writing \(n = 2^k + j\) with \(k \ge 0\) and \(0 \le j < 2^k\), set \(E_n := [j2^{-k}, (j+1)2^{-k}]\). Then \(m(E_n) = 2^{-k} < 2/n \to 0\), since \(n < 2^{k+1}\). Yet every \(x \in [0,1]\) lies in infinitely many \(E_n\): for each \(k\) the \(2^k\) intervals of generation \(k\) cover \([0,1]\), so some \(n \in [2^k, 2^{k+1})\) has \(x \in E_n\), and these indices are distinct for distinct \(k\). The exceptional set is all of \([0,1]\), of measure \(1\).

Apply Theorem 1.4.49 to \(h_n := |f_n - f|\) with dominating function \(2G\), where \(G\) is the dominator in the hypotheses.

Discarding the countable union of null sets off which \(f_n \to f\) and \(|f_n| \le G\) hold (null by Exercise 1.4.23(i), and integrals are unchanged), we may assume both hold everywhere. Then letting \(n \to \infty\) in \(|f_n| \le G\) gives \(|f| \le G\) pointwise, so

\begin{equation*} h_n = |f_n - f| \le |f_n| + |f| \le 2G, \qquad \int_X 2G \, d\mu < \infty, \end{equation*}

while \(h_n \to 0\) pointwise. The \(h_n\) are measurable, finite-valued and dominated by the absolutely integrable \(2G\), so Theorem 1.4.49 gives

\begin{equation*} \| f_n - f \|_{L^1(\mu)} = \int_X h_n \, d\mu \longrightarrow \int_X 0 \, d\mu = 0 . \end{equation*}

Split \(h_n := |f_n - f|\) at the truncation level \(2G\): the truncated part dies by dominated convergence, the overshoot is at most \(g_n\).

Discarding the countable union of null sets off which \(f_n \to f\) and \(|f_n| \le G + g_n\) hold (Exercise 1.4.23(i)), we may assume both hold everywhere; each \(f_n\) is then absolutely integrable, as \(\int_X(G+g_n)\,d\mu < \infty\).

The limit is dominated by \(G\) alone. Pick \(n_1 < n_2 < \cdots\) with \(\int_X g_{n_k}\,d\mu \le 2^{-k}\). Then \(\int_X \sum_k g_{n_k} \, d\mu = \sum_k \int_X g_{n_k}\,d\mu < \infty\) by Corollary 1.4.46, so \(\sum_k g_{n_k}(x) < \infty\) for a.e. \(x\) (Exercise 1.4.36(vii)) and hence \(g_{n_k}(x) \to 0\) there. Letting \(k \to \infty\) in \(|f_{n_k}| \le G + g_{n_k}\) gives \(|f| \le G\) a.e.; discard that null set too, so \(|f| \le G\) everywhere and \(f\) is absolutely integrable.

The splitting. From \(h_n \le |f_n| + |f| \le 2G + g_n\) and the identity \(a = \min(a,b) + (a-b)^+\) (valid for \(a \in [0,\infty)\), \(b \in [0,+\infty]\), with \((a-b)^+ := 0\) when \(b = +\infty\)),

\begin{equation*} h_n = \min(h_n, 2G) + (h_n - 2G)^+, \qquad (h_n - 2G)^+ \le g_n , \end{equation*}

the second bound being trivial where \(G = +\infty\) and obtained by subtracting \(2G\) from \(h_n \le 2G + g_n\) where \(G < \infty\). Both terms are unsigned and measurable, so additivity (Theorem 1.4.38) and monotonicity (Exercise 1.4.36(ii)) give

\begin{equation*} \int_X h_n \, d\mu \le \int_X \min(h_n, 2G) \, d\mu + \int_X g_n \, d\mu . \end{equation*}

The functions \(\min(h_n,2G)\) are measurable, finite, tend to \(0\) pointwise (as \(h_n \to 0\)) and are dominated by the absolutely integrable \(2G\), so Theorem 1.4.49 sends the first integral to \(0\); the second tends to \(0\) by hypothesis. Hence

\begin{equation*} \begin{aligned} \Big| \int_X f_n \, d\mu - \int_X f \, d\mu \Big| &\le \int_X |f_n - f| \, d\mu = \|f_n - f\|_{L^1(\mu)} \longrightarrow 0 . \end{aligned} \end{equation*}

The quantity equals \(2(\int_X h_n \, d\mu - \int_X f \, d\mu)\) for \(h_n := \min(f_n, f)\), and dominated convergence kills it.

All of \(f_1, f_2, \ldots, f\) are finite off a null set (Exercise 1.4.36(vii)), whose union is null (Exercise 1.4.23(i)); multiplying by the indicator of its complement alters no integral, so assume all take values in \([0,\infty)\).

For \(a, b \ge 0\) one has \(|a-b| = (a - \min(a,b)) + (b - \min(a,b))\) (check the cases \(a \le b\), \(a > b\)), so pointwise \(|f_n - f| = (f_n - h_n) + (f - h_n)\), all terms unsigned and measurable. Since \(0 \le h_n \le f\) forces \(\int_X h_n \,d\mu \le \int_X f \, d\mu < \infty\), additivity of the unsigned integral (Theorem 1.4.38) may be rearranged to give

\begin{equation*} \| f - f_n \|_{L^1(\mu)} = \int_X f_n \, d\mu + \int_X f \, d\mu - 2 \int_X h_n \, d\mu , \end{equation*}

whence

\begin{equation*} \int_X f_n \, d\mu - \int_X f \, d\mu - \| f - f_n \|_{L^1(\mu)} = 2 \Big( \int_X h_n \, d\mu - \int_X f \, d\mu \Big) . \end{equation*}

The \(h_n\) are measurable, finite, satisfy \(0 \le h_n \le f\) with \(f\) absolutely integrable, and converge pointwise to \(\min(f,f) = f\); so Theorem 1.4.49 gives \(\int_X h_n \, d\mu \to \int_X f \, d\mu\) and the right-hand side tends to \(0\).

Countable additivity is Corollary 1.4.46 (Tonelli for sums and integrals) applied to the pointwise identity \(1_E g = \sum_n 1_{E_n} g\).

Each \(1_E g\) is unsigned and measurable (Exercise 1.4.29(ii), (viii)), so \(\mu_g(E) \in [0,+\infty]\) is defined and equals \(\int_E g \, d\mu\) by the restriction property (Exercise 1.4.36(xi)). Taking \(E = \emptyset\) gives \(1_\emptyset g \equiv 0\) and so \(\mu_g(\emptyset) = 0\).

Let \(E_1, E_2, \ldots \in \mathcal{B}\) be pairwise disjoint with union \(E \in \mathcal{B}\). Fixing \(x\): if \(x \notin E\) both sides of \(1_E g = \sum_{n} 1_{E_n} g\) vanish, and if \(x \in E\) then disjointness leaves exactly one nonzero summand \(g(x)\) (valid also when \(g(x) = +\infty\)). Since each \(1_{E_n} g\) is unsigned measurable, Corollary 1.4.46 gives

\begin{equation*} \mu_g(E) = \int_X \sum_{n=1}^\infty 1_{E_n} g \, d\mu = \sum_{n=1}^\infty \int_X 1_{E_n} g \, d\mu = \sum_{n=1}^\infty \mu_g(E_n) . \end{equation*}

The measure is forced to be \(\mu(E) := I(1_E)\), and the three axioms match it to the integral on simple functions, then on all of \(\mathcal{U}(X,\mathcal{B})\) by monotone approximation. Throughout, \(0 \cdot \infty = 0\).

Uniqueness. If \(I(f) = \int_X f \, d\mu\) for all \(f\), then for \(E \in \mathcal{B}\) the indicator \(1_E\) is unsigned simple, so Exercises 1.4.36(v) and 1.4.34(ii) give \(I(1_E) = \int_X 1_E \, d\mu = \mu(E)\). So at most one \(\mu\) can work, and it must be this one.

The formula defines a measure. Homogeneity (i) with \(f = 0\), \(c = 0\) gives \(I(0) = 0 \cdot I(0) = 0\) (valid even if \(I(0) = +\infty\), by the convention), so \(\mu(\emptyset) = 0\). For pairwise disjoint \(E_1, E_2, \ldots\) with union \(E\), disjointness gives \(1_{E_1 \cup \cdots \cup E_N} = \sum_{n \le N} 1_{E_n}\), so (ii) and induction give \(\mu(\bigcup_{n \le N} E_n) = \sum_{n \le N} \mu(E_n)\); the indicators \(1_{\bigcup_{n \le N} E_n}\) are unsigned measurable, non-decreasing in \(N\), with pointwise limit \(1_E\), so (iii) gives

\begin{equation*} \mu(E) = \lim_{N \to \infty} \sum_{n=1}^N \mu(E_n) = \sum_{n=1}^\infty \mu(E_n). \end{equation*}

Simple functions. An unsigned simple \(f\) with distinct values \(a_1, \ldots, a_k\) on \(A_i := f^{-1}(\{a_i\}) \in \mathcal{B}\) satisfies \(f = \sum_i a_i 1_{A_i}\), so iterating (ii) and applying (i),

\begin{equation*} I(f) = \sum_{i=1}^k a_i I(1_{A_i}) = \sum_{i=1}^k a_i \mu(A_i) = \mathrm{Simp}\!\int_X f\,d\mu = \int_X f\,d\mu, \end{equation*}

the last two equalities by Exercise 1.4.34(ii)–(iv) and Exercise 1.4.36(v).

General \(f\). Put \(f_n := \min(2^{-n}\lfloor 2^n f\rfloor, n)\), read as \(n\) where \(f = +\infty\); equivalently \(f_n = \sum_{k=1}^{n2^n} 2^{-n} 1_{\{f \ge k2^{-n}\}}\), a simple function since each level set is measurable (Exercise 1.4.29(i)). From \(\lfloor 2^{n+1}t\rfloor \ge 2\lfloor 2^n t\rfloor\) the \(f_n\) are non-decreasing, and \(f_n \to f\) pointwise: \(f_n = n \to \infty\) where \(f = +\infty\), while \(f_n \in [f - 2^{-n}, f]\) once \(n > f(x)\). Axiom (iii) gives \(I(f) = \lim_n I(f_n)\) and Theorem 1.4.44 gives \(\int_X f \, d\mu = \lim_n \int_X f_n \, d\mu\); the terms agree by the previous paragraph, so \(I(f) = \int_X f \, d\mu\).

Squeeze \(f\) between a decreasing sequence of simple majorants and an increasing sequence of simple minorants; their limits \(G \ge f \ge H\) are measurable and agree off a null set \(N\), and completeness makes \(\{f > \lambda\}\) measurable by correcting \(\{G > \lambda\}\) within \(N\).

Here “simple” means real-valued simple (Definition 1.4.35 with values in \(\mathbb{R}\)), the only reading for which minorants of a sign-changing \(f\) exist; such a \(g\) is bounded, hence absolutely integrable since \(\mu(X) < \infty\). If \(|f| \le M\) the constants \(\pm M\) are competitors, so both classes are non-empty. Monotonicity of the integral on absolutely integrable functions, used repeatedly, is Exercise 1.4.41(ii) applied to \(\int_X v \, d\mu = \int_X u \, d\mu + \int_X (v-u)\,d\mu\).

Reduction to the unsigned case. Put \(F := f + M\), so \(0 \le F \le C := 2M\). Adding \(M\) is a bijection between the competitors for \(f\) and those for \(F\), shifting every integral by the finite constant \(M\mu(X)\), so the upper and lower integrals of \(F\) agree too, at a common finite value \(S \in [0, C\mu(X)]\). If \(F\) is measurable then so are \(\{F > c\}\), \(\{F \ge c\} = \bigcap_m \{F > c - \tfrac1m\}\) and \(\{F < c\}\); since for \(\lambda \ge 0\)

\begin{equation*} \begin{aligned} \{\max(f,0) > \lambda\} &= \{F > \lambda + M\}, \\ \{\max(-f,0) > \lambda\} &= \{F < M - \lambda\} \end{aligned} \end{equation*}

(and both level sets are \(X\) for \(\lambda < 0\)), Exercise 1.4.29(i),(v) then make \(f\) measurable. So it suffices to prove \(F\) measurable.

Competitors in \([0,C]\). Replacing a majorant \(g \ge F\) by \(\min(g,C)\) and a minorant \(h \le F\) by \(\max(h,0)\) keeps them simple competitors and does not increase, resp. decrease, their integrals. So choose simple \(g_n, h_n : X \to [0,C]\) with

\begin{equation*} \begin{aligned} h_n \le F \le g_n, \qquad \int_X g_n \, d\mu \le S + \tfrac1n, \\ \int_X h_n\, d\mu \ge S - \tfrac1n . \end{aligned} \end{equation*}

Put \(G_n := \min(g_1,\dots,g_n)\) and \(H_n := \max(h_1,\dots,h_n)\), again simple with values in \([0,C]\) (finitely many values, and measurable since \(\{\min(u,v)>\lambda\} = \{u>\lambda\}\cap\{v>\lambda\}\), \(\{\max(u,v)>\lambda\} = \{u>\lambda\}\cup\{v>\lambda\}\)). Then \(G_n\) decreases, \(H_n\) increases, \(H_n \le F \le G_n\), and \(G_n \le g_n\), \(H_n \ge h_n\) give, via additivity of the simple integral (Exercise 1.4.34(iv)) and \(\int_X H_n \, d\mu \le C\mu(X) < \infty\),

\begin{equation*} \int_X (G_n - H_n)\,d\mu = \int_X G_n\,d\mu - \int_X H_n\,d\mu \le \tfrac2n . \end{equation*}

The limits. The monotone bounded limits \(G := \inf_n G_n\) and \(H := \sup_n H_n\) take values in \([0,C]\) and are measurable (Exercise 1.4.29(vi)), with \(H \le F \le G\) pointwise. Their difference is measurable (Exercise 1.4.29(iv),(viii)) and satisfies \(0 \le G - H \le G_n - H_n\), so \(\int_X (G-H)\,d\mu \le 2/n\) for every \(n\), i.e. \(\int_X (G-H)\,d\mu = 0\). Hence

\begin{equation*} N := \{G > H\} = \bigcup_{m=1}^\infty \{G - H \ge \tfrac1m\} \end{equation*}

is measurable with \(\mu(N) = 0\), by Markov’s inequality (Exercise 1.4.36(vi)) and countable subadditivity. Since \(H \le F \le G\), we get \(F = G\) on \(X \setminus N\).

Completeness. Fix \(\lambda\) and set \(P := \{G > \lambda\} \in \mathcal{B}\), \(Q := \{F > \lambda\}\). As \(F = G\) off \(N\), both \(P \setminus Q\) and \(Q \setminus P\) lie in \(N\), hence are null and in particular measurable by completeness (Definition 1.4.31). Then

\begin{equation*} Q = \big(P \setminus (P \setminus Q)\big) \cup (Q \setminus P) \in \mathcal{B} . \end{equation*}

As \(\lambda\) was arbitrary (the level set being \(X\) for \(\lambda < 0\)), \(F\) is measurable by Exercise 1.4.29(i), and hence so is \(f = F - M\).

(i) Each of the seven definitions is a statement about the numbers \(|f_n(x)-f(x)|\) alone, and \(\bigl||f_n(x)-f(x)| - 0\bigr| = |f_n(x)-f(x)|\), so substituting the pair \((|f_n - f|, 0)\) for \((f_n, f)\) changes nothing in any of them. Hence \(f_n \to f\) in a mode iff \(|f_n - f| \to 0\) in that mode.

(ii) Sums: run the pointwise triangle inequality \(|(f_n+g_n)-(f+g)| \le |f_n - f| + |g_n - g|\) through each definition with tolerance \(\varepsilon/2\).

• Pointwise, uniform, pointwise a.e., \(L^\infty\). Take \(N := \max(N_1,N_2)\) for the two \(\varepsilon/2\)-thresholds; the exceptional set is the union of the two exceptional null sets, again null (Exercise 1.4.23(i)).
• Almost uniform. Take \(E \cup F\) with \(\mu(E), \mu(F) \le \varepsilon/2\); subadditivity gives \(\mu(E \cup F) \le \varepsilon\), and the uniform case applies on the complement.
• \(L^1\). \(\int_X |(f_n+g_n)-(f+g)| \, d\mu \le \int_X |f_n-f|\,d\mu + \int_X |g_n-g|\,d\mu \to 0\), by Exercise 1.4.36(ii) and Theorem 1.4.38.
• In measure. \(\{|(f_n+g_n)-(f+g)| \ge \varepsilon\} \subseteq \{|f_n-f| \ge \varepsilon/2\} \cup \{|g_n-g| \ge \varepsilon/2\}\), then finite subadditivity.

Scalars: \(c = 0\) is trivial; for \(c \ne 0\) use \(|cf_n - cf| = |c||f_n - f|\), replacing \(\varepsilon\) by \(\varepsilon/|c|\) in the first five modes, homogeneity (Exercise 1.4.36(iii)) for \(L^1\), and the set identity \(\{|cf_n - cf| \ge \varepsilon\} = \{|f_n - f| \ge \varepsilon/|c|\}\) for measure.

(iii) The hypothesis forces \(f_n \ge 0\), so \(|g_n - 0| \le |f_n - 0|\) pointwise, and every mode is monotone: if \(0 \le h_n \le k_n\) and \(k_n \to 0\) in a mode then \(h_n \to 0\) in it. For the first five modes, \(k_n \le \varepsilon\) at a point forces \(h_n \le \varepsilon\) there, so the same thresholds and exceptional sets work; for \(L^1\), \(\int_X h_n \, d\mu \le \int_X k_n \, d\mu \to 0\) (Exercise 1.4.36(ii)); in measure, \(\{h_n \ge \varepsilon\} \subseteq \{k_n \ge \varepsilon\}\). Apply this with \(h_n := |g_n|\), \(k_n := f_n\), then convert back by part (i).

Each implication is the definition with the same threshold, an exceptional set built from a countable union or intersection, or Markov’s inequality.

One remark is used throughout, and dispenses with any completeness assumption: a \(\mathcal{B}\)-measurable subset of a sub-null set is null, by monotonicity. So whenever a bound \(|f_n - f| \le \varepsilon\) holds almost everywhere, the measurable set \(\{|f_n - f| > \varepsilon\}\) is genuinely null.

• (i) The uniform threshold \(N\) for \(\varepsilon\) works at each fixed \(x\).
• (ii) Forward: the uniform bound holds for every \(x\), hence a.e., with empty exceptional set. Converse: for each \(k\) take \(N_k\) with \(|f_n - f| \le 1/k\) a.e. for \(n \ge N_k\), put \(E_{n,k} := \{|f_n - f| > 1/k\}\) for \(n \ge N_k\) and \(\emptyset\) otherwise (null, by the remark), and set \(E := \bigcup_{k,n} E_{n,k}\), null by countable subadditivity (Exercise 1.4.23(i)). Given \(\eta > 0\) pick \(k\) with \(1/k \le \eta\); for \(x \notin E\) and \(n \ge N_k\) we get \(|f_n(x) - f(x)| \le 1/k \le \eta\), with \(N_k\) independent of \(x\). That is uniform convergence on \(X \setminus E\).
• (iii) The set \(E\) produced in (ii) lies in \(\mathcal{B}\) and has \(\mu(E) = 0 \le \varepsilon\), so it serves as the exceptional set for every \(\varepsilon\).
• (iv) Take \(E_k \in \mathcal{B}\) with \(\mu(E_k) \le 1/k\) and uniform convergence on \(X \setminus E_k\); then \(E := \bigcap_k E_k\) has \(\mu(E) \le \inf_k 1/k = 0\), and each \(x \notin E\) misses some \(E_k\), so \(f_n(x) \to f(x)\) by (i).
• (v) Pointwise convergence is the a.e. statement with exceptional set \(\emptyset\).
• (vi) Markov’s inequality (Exercise 1.4.36(vi)) for the unsigned measurable \(|f_n - f|\) at level \(\varepsilon\) gives

\begin{equation*} \mu(\{|f_n-f| \ge \varepsilon\}) \le \tfrac{1}{\varepsilon}\|f_n-f\|_{L^1(\mu)} \to 0 . \end{equation*}

• (vii) Fix \(\varepsilon > 0\) and let \(\delta > 0\). Choose \(E\) with \(\mu(E) \le \delta\) and uniform convergence on \(X \setminus E\), so \(\{|f_n - f| \ge \varepsilon\} \subseteq E\) for all large \(n\) and \(\mu(\{|f_n-f|\ge\varepsilon\}) \le \delta\) there. As \(\delta\) was arbitrary, the limsup is \(0\).

Three standing facts. Since \(\mu(E_n) > 0\), \(\sup_x f_n(x) = A_n\) and the bound “\(f_n \le \varepsilon\) a.e.” holds iff \(A_n \le \varepsilon\). By the standing dichotomy, \(A_n \not\to 0\) gives \(c > 0\) with \(A_n \ge c\) for all \(n\); this drives every “only if” direction. And the \(E_N^*\) decrease, with \(x \in \bigcap_N E_N^*\) iff \(x\) lies in infinitely many \(E_n\).

(i), (ii) Uniform convergence to \(0\) means \(\sup_x f_n(x) = A_n \le \varepsilon\) eventually, for each \(\varepsilon > 0\); \(L^\infty\) convergence means \(f_n \le \varepsilon\) a.e. eventually, which by the first standing fact is again \(A_n \le \varepsilon\) eventually. Both say \(A_n \to 0\).

(iii) If \(A_n \to 0\), (i) gives uniform hence almost uniform convergence (take \(E = \emptyset\)). If \(\mu(E_N^*) \to 0\), given \(\varepsilon > 0\) pick \(N\) with \(\mu(E_N^*) \le \varepsilon\) and take \(E := E_N^*\): every \(f_n\) with \(n \ge N\) vanishes on \(X \setminus E\). Conversely, if \(f_n \to 0\) almost uniformly with \(A_n \ge c > 0\), take \(E\) with \(\mu(E) \le \varepsilon\) and uniform convergence off \(E\); then \(f_n < c\) on \(X \setminus E\) for \(n \ge N\), while \(f_n = A_n \ge c\) on \(E_n\), so \(E_n \subseteq E\) for \(n \ge N\) and \(\mu(E_M^*) \le \mu(E) \le \varepsilon\) for all \(M \ge N\). As \(\varepsilon\) was arbitrary, \(\mu(E_N^*) \to 0\).

(iv) If \(A_n \to 0\), (i) and Exercise 1.5.2(i) give pointwise convergence. If \(\bigcap_N E_N^* = \emptyset\), each \(x\) misses some \(E_N^*\), so \(f_n(x) = 0\) for all \(n \ge N\). Conversely, with \(A_n \ge c > 0\), any \(x \in \bigcap_N E_N^*\) lies in infinitely many \(E_n\), so \(f_n(x) = A_n \ge c\) infinitely often and \(f_n(x) \not\to 0\); hence the intersection is empty.

(v) Same argument off \(Z := \bigcap_N E_N^*\): if \(A_n \to 0\) or \(\mu(Z) = 0\) then \(f_n \to 0\) pointwise outside \(Z\). Conversely, with \(A_n \ge c > 0\), convergence fails at every point of \(Z\), so \(Z\) is contained in a null set \(F\); \(Z\) is measurable (a countable intersection of countable unions of the \(E_n\)), so \(\mu(Z) \le \mu(F) = 0\) with no completeness needed.

(vi) For \(\varepsilon > 0\),

\begin{equation*} \{f_n \ge \varepsilon\} = \begin{cases} E_n & \text{if } A_n \ge \varepsilon,\\ \emptyset & \text{if } A_n < \varepsilon.\end{cases} \end{equation*}

If \(A_n \to 0\) the set is eventually empty; if \(\mu(E_n) \to 0\) its measure is at most \(\mu(E_n) \to 0\). Conversely, with \(A_n \ge c > 0\), taking \(\varepsilon := c\) makes the set equal \(E_n\) for every \(n\), so convergence in measure says \(\mu(E_n) \to 0\).

(vii) \(f_n\) is unsigned simple, so by Exercise 1.4.36(v)

\begin{equation*} \|f_n - 0\|_{L^1(\mu)} = \int_X A_n 1_{E_n} \, d\mu = A_n \mu(E_n) \end{equation*}

(with \(A_n \cdot (+\infty) = +\infty\), legitimate as \(A_n > 0\)), and the criterion is \(A_n\mu(E_n) \to 0\).

An a.e. bound integrates against the finite total measure: \(\|f_n - f\|_{L^1(\mu)} \le \varepsilon\mu(X)\).

If \(\mu(X) = 0\) every integral vanishes, so assume \(0 < \mu(X) < \infty\). Given \(\eta > 0\), set \(\varepsilon := \eta/\mu(X)\) and take \(N\) with \(|f_n - f| \le \varepsilon\) a.e. for every \(n \ge N\). Then by monotonicity up to null sets and compatibility with the simple integral (Exercise 1.4.36(ii),(v)),

\begin{equation*} \|f_n - f\|_{L^1(\mu)} \le \int_X \varepsilon 1_X \, d\mu = \varepsilon \mu(X) = \eta \qquad (n \ge N) . \end{equation*}

As \(\eta > 0\) was arbitrary, \(\|f_n - f\|_{L^1(\mu)} \to 0\).

Work with the tail sums \(G_N := \sum_{n \ge N} |f_n - f| : X \to [0,+\infty]\), measurable as increasing limits of the partial sums. Writing \(\delta_n := \|f_n - f\|_{L^1(\mu)}\), Corollary 1.4.46 (Tonelli for sums and integrals) gives

\begin{equation*} \int_X G_N \, d\mu = \sum_{n \ge N} \delta_n =: T_N, \qquad T_N \to 0, \quad T_1 < \infty . \end{equation*}

(i) Since \(\int_X G_1 \, d\mu = T_1 < \infty\), Exercise 1.4.36(vii) gives \(G_1(x) < \infty\) for a.e. \(x\); at such \(x\) the convergent series \(\sum_n |f_n(x)-f(x)|\) has terms tending to \(0\), i.e. \(f_n(x) \to f(x)\).

(ii) Given \(\varepsilon > 0\), pick for each \(k \ge 1\) an index \(N_k\) with \(T_{N_k} \le \varepsilon/(k2^k)\), and put \(B_k := \{G_{N_k} \ge 1/k\}\). Markov’s inequality (Exercise 1.4.36(vi)) at level \(1/k\) and countable subadditivity (Exercise 1.4.23(i)) give

\begin{equation*} \mu(B_k) \le k\,T_{N_k} \le \frac{\varepsilon}{2^k}, \qquad \mu(E) \le \varepsilon \quad \text{for } E := \bigcup_{k \ge 1} B_k \in \mathcal{B} . \end{equation*}

For \(\eta > 0\) choose \(k\) with \(1/k < \eta\); every \(x \in X \setminus E\) misses \(B_k\), so since each term of the series is at most the whole sum,

\begin{equation*} |f_n(x) - f(x)| \le G_{N_k}(x) < \tfrac1k < \eta \qquad (n \ge N_k), \end{equation*}

with \(N_k\) independent of \(x\). So \(f_n \to f\) uniformly on \(X \setminus E\), and \(\varepsilon\) was arbitrary.

Choose \(n_1 < n_2 < \cdots\) recursively so that the measurable sets

\begin{equation*} A_j := \{x \in X : |f_{n_j}(x)-f(x)| \ge 1/j\} \quad \text{satisfy} \quad \mu(A_j) \le 2^{-j}, \end{equation*}

possible since convergence in measure at \(\varepsilon = 1/j\) sends \(\mu(\{|f_n - f| \ge 1/j\}) \to 0\).

Given \(\varepsilon > 0\), pick \(J\) with \(2^{-J+1} \le \varepsilon\) and set \(E := \bigcup_{j \ge J} A_j \in \mathcal{B}\), so \(\mu(E) \le \sum_{j \ge J} 2^{-j} \le \varepsilon\) by countable subadditivity (Exercise 1.4.23(i)). Every \(x \in X \setminus E\) misses each \(A_j\) with \(j \ge J\), so \(|f_{n_j}(x)-f(x)| < 1/j\) there; given \(\eta > 0\), any \(J’ \ge J\) with \(1/J’ < \eta\) then bounds \(|f_{n_j} - f| < \eta\) on \(X \setminus E\) for all \(j \ge J’\), uniformly in \(x\). Hence \(f_{n_j} \to f\) almost uniformly, and pointwise almost everywhere by Exercise 1.5.2(iv).

Pick \(m_n\) making \(\mu(\{|f_{n,m_n} - f_n| \ge 1/n\}) \le 2^{-n}\), then run Borel–Cantelli.

Borel–Cantelli. If \(A_1, A_2, \dots \in \mathcal{B}\) have \(\sum_n \mu(A_n) < \infty\), then \(A := \bigcap_N \bigcup_{n \ge N} A_n\) is measurable with \(\mu(A) \le \sum_{n \ge N}\mu(A_n) \to 0\) by monotonicity and countable subadditivity (Exercise 1.4.23(i)), so \(\mu(A) = 0\).

(i) Fix \(n\). Since \(\mu(X) < \infty\), Egorov’s theorem (Theorem 1.5.9) upgrades \(f_{n,m} \to f_n\) pointwise a.e. to almost uniform convergence, and Exercise 1.5.2(vii) turns that into convergence in measure; at \(\varepsilon := 1/n\) this lets us choose \(m_n\) with

\begin{equation*} \mu(A_n) \le 2^{-n}, \qquad A_n := \{x \in X : |f_{n,m_n}(x) - f_n(x)| \ge 1/n\} . \end{equation*}

Then \(\sum_n \mu(A_n) \le 1 < \infty\), so \(A := \bigcap_N \bigcup_{n \ge N} A_n\) is null; let \(Z\) be a null set off which \(f_n \to f\) pointwise and put \(Z’ := A \cup Z\). For \(x \notin Z’\) there is \(N\) with \(x \notin A_n\) for all \(n \ge N\), so \(|f_{n,m_n}(x) - f_n(x)| < 1/n \to 0\), and

\begin{equation*} |f_{n,m_n}(x) - f(x)| \le |f_{n,m_n}(x)-f_n(x)| + |f_n(x)-f(x)| \longrightarrow 0 . \end{equation*}

(ii) Write \(X = \bigcup_k Y_k\) with \(\mu(Y_k) < \infty\) and replace \(Y_k\) by \(X_k := Y_1 \cup \cdots \cup Y_k\), so the \(X_k\) increase to \(X\) with finite measure. For each \(n\), the restriction of the space to \(X_n\) is a finite measure space (Example 1.4.30) on which \(f_{n,m}|_{X_n} \to f_n|_{X_n}\) pointwise a.e. (intersect the exceptional null set with \(X_n\)), so as in (i) choose \(m_n\) with

\begin{equation*} \mu(A_n) \le 2^{-n}, \qquad A_n := \{x \in X_n : |f_{n,m_n}(x)-f_n(x)| \ge 1/n\} , \end{equation*}

the truncation to \(X_n\) being what keeps the measure controllable when \(\mu(X) = \infty\). Borel–Cantelli again makes \(A := \bigcap_N \bigcup_{n \ge N} A_n\) null; set \(Z’ := A \cup Z\) as before. Any \(x \notin Z’\) lies in some \(X_k\), hence in \(X_n\) for all \(n \ge k\), and misses \(A_n\) for all \(n \ge N\); so for \(n \ge \max(N,k)\) the definition of \(A_n\) forces \(|f_{n,m_n}(x)-f_n(x)| < 1/n\), and the triangle inequality finishes as in (i).

Write \(a_n(\varepsilon) := \mu(\{|f_n - f| \ge \varepsilon\})\); the whole exercise is the sub-subsequence principle applied to these numbers.

(i) implies (ii): for each \(\varepsilon > 0\) the subsequence \((a_{n_j}(\varepsilon))_j\) of a null sequence is null, so \(f_{n_j} \to f\) in measure, and Exercise 1.5.6 extracts a further subsequence converging almost uniformly to \(f\).

(ii) implies (i): if \(a_n(\varepsilon) \not\to 0\) for some \(\varepsilon > 0\), then (failure of convergence to \(0\) in \([0,+\infty]\)) there are \(\delta > 0\) and \(n_1 < n_2 < \cdots\) with \(a_{n_j}(\varepsilon) \ge \delta\) for all \(j\). Hypothesis (ii) gives a further subsequence converging almost uniformly, hence in measure by Exercise 1.5.2(vii), so \(a_{n_{j_i}}(\varepsilon) \to 0\) – contradicting \(a_{n_{j_i}}(\varepsilon) \ge \delta > 0\).

“Only if” is Exercise 1.5.2(vi) and needs no domination; “if” is the sub-subsequence principle applied to \(a_n := \|f_n - f\|_{L^1(\mu)}\), with Exercise 1.5.6 and Theorem 1.4.49 supplying the further subsequences.

Let \(g\) be the dominator. As \(|f_n| \ge 0\), the bound forces \(g \ge 0\) a.e.; discarding a null set (no integral or mode of convergence changes), assume \(g : X \to [0,+\infty)\) with \(\int_X g \,d\mu < \infty\) and \(|f_n| \le g\) everywhere.

Suppose \(f_n \to f\) in measure. Exercise 1.5.6 gives a subsequence converging to \(f\) pointwise a.e., so letting \(k \to \infty\) in \(|f_{n_k}| \le g\) gives \(|f| \le g\) a.e.; thus \(f\) and every \(f_n - f\) is absolutely integrable, with

\begin{equation*} |f_n - f| \le |f_n| + |f| \le 2g \qquad \text{a.e., for every } n . \end{equation*}

It suffices that every subsequence of \((a_n)\) have a further subsequence tending to \(0\): for if \(a_n \not\to 0\), some subsequence is bounded below by \(\varepsilon > 0\) and no further subsequence of it can be null. Given \(f_{n_j}\), it too converges in measure to \(f\) (Exercise 1.5.8), so Exercise 1.5.6 gives a further subsequence with \(h_i := |f_{n_{j_i}} - f| \to 0\) pointwise a.e.; these are measurable, nonnegative and dominated a.e. by the absolutely integrable \(2g\), so Theorem 1.4.49 gives \(a_{n_{j_i}} = \int_X h_i \, d\mu \to 0\).

(iii) is \(f_n := 1_{[n,n+1]}\) on \(\mathbb{R}\); (i) and (ii) are monotone and dominated convergence applied to the truncations. Definition 1.5.11 asks for (a) \(\sup_n \|f_n\|_{L^1(\mu)} < \infty\), (b) \(\sup_n \int_{|f_n| \ge M}|f_n|\,d\mu \to 0\) as \(M \to +\infty\), (c) \(\sup_n \int_{|f_n| \le \delta}|f_n|\,d\mu \to 0\) as \(\delta \to 0\).

(i) (a) is \(\|f\|_{L^1(\mu)} < \infty\). For (b), the functions \(|f|1_{\{|f|<m\}}\) increase pointwise to \(|f|\) (as \(f\) is \(\mathbb{C}\)-valued, so finite), so Theorem 1.4.44 and subtraction of the finite \(\int_X |f| \, d\mu\) give

\begin{equation*} \int_{|f| \ge m} |f| \, d\mu = \int_X |f| \, d\mu - \int_X |f| 1_{\{|f|<m\}} \, d\mu \to 0 ; \end{equation*}

since \(M \mapsto \int_{|f| \ge M}|f|\,d\mu\) is nonincreasing, convergence along integers suffices. For (c), the same argument with \(|f|1_{\{|f| > 1/m\}} \uparrow |f|\) (the terms equal \(0 = |f(x)|\) where \(f(x) = 0\), and \(|f(x)|\) once \(1/m < |f(x)|\)) gives \(\int_{|f| \le 1/m}|f|\,d\mu \to 0\), and \(\delta \mapsto \int_{|f| \le \delta}|f|\,d\mu\) is nondecreasing.

(ii) As in Exercise 1.5.9, modify \(g\) on a null set so that \(g : X \to [0,+\infty)\) is absolutely integrable with \(|f_n| \le g\) everywhere; then each \(f_n\) is absolutely integrable and (a) holds with bound \(\|g\|_{L^1(\mu)}\). For (b), \(\{|f_n| \ge M\} \subseteq \{g \ge M\}\) and \(|f_n| \le g\) give

\begin{equation*} \int_{|f_n| \ge M} |f_n| \, d\mu \le \int_{g \ge M} g \, d\mu , \end{equation*}

independent of \(n\) and tending to \(0\) by (i) applied to \(g\). For (c), \(|f_n| \le \min(\delta, g)\) on \(\{|f_n| \le \delta\}\), so the integral is at most \(\int_X \min(\delta, g)\,d\mu\); along \(\delta = 1/m\) the functions \(\min(1/m, g)\) tend to \(0\) pointwise (\(g\) is finite) under the dominating \(g\), so Theorem 1.4.49 sends this to \(0\), and \(\delta \mapsto \int_X\min(\delta,g)\,d\mu\) is nondecreasing.

(iii) \(f_n := 1_{[n,n+1]}\) on \(\mathbb{R}\) (Example 1.5.2). Here \(\|f_n\|_{L^1(m)} = 1\) for all \(n\), giving (a); \(\{|f_n| \ge M\} = \emptyset\) for \(M > 1\), giving (b); and \(|f_n|\) vanishes on \(\{|f_n| \le \delta\}\) for \(\delta < 1\), giving (c). But if \(g \ge |f_n|\) a.e. for every \(n\), then \(N_n := \{x \in [n,n+1] : g(x) < 1\}\) is null for each \(n\), so \(N := \bigcup_n N_n\) is null and \(g \ge 1\) on \([1,\infty) \setminus N\), forcing \(\int_{\mathbb{R}} g \, dm = +\infty\): not absolutely integrable.

The stated condition \((\ast)\) is literally clause (b) of Definition 1.5.11, so “only if” is immediate (and does not use \(\mu(X) < \infty\)); conversely \((\ast)\) plus finite measure forces clauses (a) and (c).

Clause (a) and absolute integrability. By \((\ast)\) fix \(M_0 > 0\) with \(\sup_n \int_{|f_n| \ge M_0}|f_n|\,d\mu \le 1\). Splitting \(X\) at the level \(M_0\), where the integrand is bounded by \(M_0\) on a set of measure at most \(\mu(X)\),

\begin{equation*} \int_X |f_n| \, d\mu \le 1 + M_0 \mu(X) < \infty , \end{equation*}

uniformly in \(n\); so each \(f_n\) is absolutely integrable and (a) holds.

Clause (c). On \(\{|f_n| \le \delta\}\) the integrand is at most \(\delta\), so

\begin{equation*} \sup_n \int_{|f_n| \le \delta} |f_n| \, d\mu \le \delta \, \mu(X) \to 0 \qquad (\delta \to 0) . \end{equation*}

On \(\{|f_n| \ge M\}\) one has \(|f_n| \le M^{1-p}|f_n|^p\), which gives the tail bound \(A M^{1-p}\) with \(A := \sup_n \int_X |f_n|^p\,d\mu < \infty\).

Since \(\mu(X) < \infty\), Exercise 1.5.11 reduces uniform integrability to the single condition \(\sup_n \int_{|f_n| \ge M}|f_n|\,d\mu \to 0\). Where \(|f_n(x)| \ge M > 0\), the map \(t \mapsto t^{1-p}\) is decreasing (as \(p > 1\)), so \(|f_n(x)|^{1-p} \le M^{1-p}\) and hence \(|f_n(x)| = |f_n(x)|^p|f_n(x)|^{1-p} \le M^{1-p}|f_n(x)|^p\). Integrating and enlarging the domain to \(X\) (the integrand \(|f_n|^p\) being unsigned),

\begin{equation*} \sup_n \int_{|f_n| \ge M} |f_n| \, d\mu \le M^{1-p} \sup_n \int_X |f_n|^p \, d\mu = \frac{A}{M^{p-1}} \to 0 \end{equation*}

as \(M \to +\infty\), since \(p - 1 > 0\).

Take \(\delta := \varepsilon/(2M)\), where clause (ii) of Definition 1.5.11 supplies \(M > 0\) with

\begin{equation*} \int_{|f_n| \ge M} |f_n| \, d\mu \le \frac{\varepsilon}{2} \qquad \text{for every } n \ge 1 . \end{equation*}

Given \(n\) and \(E \in \mathcal{B}\) with \(\mu(E) \le \delta\), split \(E\) at the level \(M\) and bound the high piece by enlarging its domain to \(\{|f_n| \ge M\}\) (the integrand being unsigned), the low piece by \(|f_n| < M\) on a set of measure at most \(\delta\):

\begin{equation*} \int_E |f_n| \, d\mu \le \int_{|f_n| \ge M} |f_n| \, d\mu + M\mu(E) \le \frac{\varepsilon}{2} + M\delta = \varepsilon . \end{equation*}

Markov’s inequality makes \(\{|f_n| \ge M\}\) a legitimate test set for the hypothesis once \(M\) is large. Put \(A := \sup_n \|f_n\|_{L^1(\mu)} < \infty\) and \(\mu(X) = 1\).

Clause (i) of Definition 1.5.11 is the hypothesis itself. Clause (iii) is automatic on a finite measure space, exactly as in Exercise 1.5.11: \(\sup_n \int_{|f_n| \le \delta}|f_n|\,d\mu \le \delta\mu(X) = \delta \to 0\).

For clause (ii), let \(\varepsilon > 0\) and take the \(\delta > 0\) the hypothesis supplies. Markov’s inequality (Exercise 1.4.36(vi)) for the unsigned absolutely integrable \(|f_n|\) gives \(\mu(\{|f_n| \ge M\}) \le A/M\), so with \(M_0 := (A+1)/\delta\),

\begin{equation*} \mu(\{|f_n| \ge M\}) \le \frac{A\delta}{A+1} < \delta \qquad (M \ge M_0,\ n \ge 1) . \end{equation*}

Applying the hypothesis with \(E := \{|f_n| \ge M\}\) – legitimate, as it is asserted for all \(n\) and all measurable \(E\) of measure at most \(\delta\), so \(E\) may depend on \(n\) – gives \(\int_{|f_n| \ge M}|f_n|\,d\mu \le \varepsilon\) for every \(n\) and every \(M \ge M_0\). As \(\varepsilon > 0\) was arbitrary, \(\sup_n \int_{|f_n| \ge M}|f_n|\,d\mu \to 0\).

Take \(f_n := 1_{[n,n+1]}\) on \(X = \mathbb{R}\) with Lebesgue measure \(m\), the escape to horizontal infinity of Example 1.5.2.

Uniformly integrable. \(\|f_n\|_{L^1(m)} = m([n,n+1]) = 1\) for all \(n\), giving clause (i) of Definition 1.5.11; \(f_n\) takes only the values \(0\) and \(1\), so \(\{|f_n| \ge M\} = \emptyset\) for \(M > 1\), giving (ii); and \(\{|f_n| \le \delta\} = \{f_n = 0\}\) for \(0 < \delta < 1\), on which the integrand vanishes, giving (iii).

Pointwise to zero. \(f_n(x) = 0\) for every \(n > |x|\).

Failures. \(\|f_n - 0\|_{L^1(m)} = 1 \not\to 0\). For \(\varepsilon = 1/2\), \(\{|f_n| \ge 1/2\} = [n,n+1]\) has measure \(1\), so convergence in measure fails; hence almost uniform convergence fails too, by Exercise 1.5.2(vii). By Proposition 1.5.7 no other limit is available: since \(f_n \to 0\) pointwise a.e., any limit in these modes would equal \(0\) a.e.

Monotonicity turns the \(L^1\) distance into a difference of integrals: \(\|f - f_n\|_{L^1(\mu)} = A - \int_X f_n \, d\mu\), where \(f := \sup_n f_n\) and \(A := \sup_n \int_X f_n \, d\mu < \infty\).

By monotonicity, \(f(x) = \lim_n f_n(x)\) in \([0,+\infty]\) for every \(x\), so \(f\) is measurable (Exercise 1.4.29(vi)), and Theorem 1.4.44 gives \(\int_X f \, d\mu = \lim_n \int_X f_n \, d\mu = A < \infty\) (the limit is the supremum since \(n \mapsto \int_X f_n \, d\mu\) is non-decreasing, Exercise 1.4.36(ii)). Markov’s inequality (Exercise 1.4.36(vi)) then gives \(\mu(\{f \ge \lambda\}) \le A/\lambda\) for every \(\lambda > 0\), and \(\{f = +\infty\} \subseteq \{f \ge \lambda\}\), so \(\mu(\{f = +\infty\}) = 0\): the limit is finite a.e., and \(\|f - f_n\|_{L^1(\mu)}\) is read as the unsigned integral \(\int_X |f - f_n|\,d\mu\), which is the modification the statement alludes to.

Fix \(n\). Since \(0 \le f_n < \infty\) and \(f_n \le f\), the difference \(f - f_n : X \to [0,+\infty]\) is unsigned and measurable (it is the increasing limit of the measurable \(f_m - f_n\), \(m \ge n\)), and \(f_n + (f - f_n) = f\) everywhere. Additivity (Theorem 1.4.38) and the finiteness of \(\int_X f_n \, d\mu \le A\) permit subtraction, so

\begin{equation*} \|f - f_n\|_{L^1(\mu)} = \int_X (f - f_n)\, d\mu = A - \int_X f_n\, d\mu \to 0 , \end{equation*}

using \(|f - f_n| = f - f_n\) and \(\int_X f_n \, d\mu \to A\).

Both directions come from the identity \(\|f_n - f\|_{L^1(\mu)} = \int_X f_n \, d\mu + \int_X f \, d\mu - 2\int_X g_n \, d\mu\) with \(g_n := \min(f_n, f)\).

First, \(f\) is absolutely integrable: Fatou’s lemma (Corollary 1.4.47) gives \(\int_X f \, d\mu \le \liminf_n \int_X f_n \, d\mu \le A := \sup_n \int_X f_n\,d\mu < \infty\), applied to \(\liminf_n f_n\), which equals \(f\) a.e. and so has the same integral (Exercise 1.4.36(i)). So every integral below is finite.

Only if. From \(f_n \le f + |f_n - f|\) and \(f \le f_n + |f_n - f|\), monotonicity and additivity (Exercise 1.4.36(ii), Theorem 1.4.38) give, all terms being finite,

\begin{equation*} \Big| \int_X f_n\, d\mu - \int_X f\, d\mu \Big| \le \|f_n - f\|_{L^1(\mu)} \to 0 . \end{equation*}

If. Each \(g_n\) is unsigned, finite and measurable (\(\{g_n > \lambda\} = \{f_n > \lambda\} \cap \{f > \lambda\}\), Exercise 1.4.29(i)), and for reals \(a, b \ge 0\) one has \(|a-b| = a + b - 2\min(a,b)\) (from \(\max + \min = a+b\) and \(|a-b| = \max - \min\)). So \(f_n + f = |f_n - f| + 2g_n\) pointwise, and additivity (Theorem 1.4.38) with all terms finite rearranges to

\begin{equation*} \|f_n - f\|_{L^1(\mu)} = \int_X f_n\, d\mu + \int_X f\, d\mu - 2 \int_X g_n\, d\mu . \end{equation*}

The \(g_n\) satisfy \(0 \le g_n \le f\) with \(f\) absolutely integrable, and \(g_n \to \min(f,f) = f\) pointwise a.e. by continuity of \(\min\), so Theorem 1.4.49 gives \(\int_X g_n \, d\mu \to \int_X f \, d\mu\). With the hypothesis \(\int_X f_n \, d\mu \to \int_X f \, d\mu\), the right-hand side tends to \(0\).

One direction is Exercise 1.5.2(iv); the other is Egorov’s theorem with domination replacing finiteness of \(\mu(X)\), the substitution happening at the downward monotone convergence step (Exercise 1.4.23(iii)).

Replacing the dominator \(g\) by \(|g|\), assume \(g : X \to [0,+\infty)\) with \(\int_X g \, d\mu < \infty\).

Let \(Z\) be the union of the measurable null sets \(\{|f_n| > g\}\) (\(n \ge 1\)) and \(\{\limsup_n |f_n - f| > 0\}\), a measurable null set, so no completeness is needed. Off \(Z\) we get \(|f| = \lim_n |f_n| \le g\), hence the bound

\begin{equation*} |f_n(x) - f(x)| \le |f_n(x)| + |f(x)| \le 2g(x) \qquad (n \ge 1,\ x \notin Z) . \end{equation*}

For \(\eta > 0\) put \(S_N^\eta := \bigcup_{n \ge N}\{x \in X \setminus Z : |f_n(x)-f(x)| > \eta\}\), measurable and non-increasing in \(N\). By that bound, \(S_1^\eta \subseteq \{g > \eta/2\}\), so Markov’s inequality (Exercise 1.4.36(vi)) gives \(\mu(S_1^\eta) \le \frac{2}{\eta}\int_X g\,d\mu < \infty\); and \(\bigcap_N S_N^\eta = \emptyset\), since a point of the intersection would have \(|f_n - f| > \eta\) infinitely often off \(Z\). Downward monotone convergence for sets (Exercise 1.4.23(iii)), whose finite-measure hypothesis is exactly what the displayed bound supplied, gives \(\mu(S_N^\eta) \to 0\).

Given \(\varepsilon > 0\), choose for each \(k\) an \(N_k\) with \(\mu(S_{N_k}^{1/k}) \le \varepsilon 2^{-k}\) and set \(E := Z \cup \bigcup_k S_{N_k}^{1/k}\), so \(\mu(E) \le \varepsilon\) by countable subadditivity. For \(\eta > 0\) pick \(k\) with \(1/k \le \eta\); any \(x \in X \setminus E\) misses \(Z\) and \(S_{N_k}^{1/k}\), whose defining union runs over all \(n \ge N_k\), so \(|f_n(x)-f(x)| \le 1/k \le \eta\) for all \(n \ge N_k\), uniformly in \(x\). Hence \(f_n \to f\) almost uniformly.

All three counterexamples run on \(X = [0,1]\) with Lebesgue measure and the pair \(1_{[0,1/2)}\), \(1_{[1/2,1]}\), which share a cumulative distribution function but agree nowhere.

(i) Every mode implies convergence in measure. By Exercise 1.5.2, uniform implies \(L^\infty\) (ii), \(L^\infty\) implies almost uniform (iii), almost uniform implies in measure (vii); \(L^1\) implies in measure (vi); pointwise implies pointwise a.e. (v); and since \(\mu(X) = 1 < \infty\), Egorov’s theorem (Theorem 1.5.9) sends pointwise a.e. to almost uniform.

In measure implies in distribution. Let \(\lambda\) be a continuity point of \(F\) and \(\eta > 0\). If \(f_n(x) \le \lambda\) and \(|f_n(x)-f(x)| < \eta\) then \(f(x) \le \lambda + \eta\); symmetrically, \(f(x) \le \lambda - \eta\) with the same bound gives \(f_n(x) \le \lambda\). So

\begin{equation*} \{f_n \le \lambda\} \subseteq \{f \le \lambda+\eta\} \cup \{|f_n - f| \ge \eta\}, \end{equation*}

\begin{equation*} \{f \le \lambda-\eta\} \subseteq \{f_n \le \lambda\} \cup \{|f_n - f| \ge \eta\}, \end{equation*}

and monotonicity with finite subadditivity, together with \(\mu(\{|f_n-f|\ge\eta\}) \to 0\), give

\begin{equation*} \begin{aligned} F(\lambda - \eta) \le \liminf_{n \to \infty} F_n(\lambda) \le \limsup_{n \to \infty} F_n(\lambda) \le F(\lambda + \eta) . \end{aligned} \end{equation*}

Letting \(\eta \to 0^+\) and using continuity of \(F\) at \(\lambda\) gives \(F_n(\lambda) \to F(\lambda)\).

(ii) Take \(f := 1_{[0,1/2)}\) and \(f_n := 1_{[1/2,1]}\) for every \(n\). Each takes the value \(1\) on a set of measure \(1/2\), so

\begin{equation*} F(\lambda) = F_n(\lambda) = \begin{cases} 0, & \lambda < 0 \\ 1/2, & 0 \le \lambda < 1 \\ 1, & \lambda \ge 1 , \end{cases} \end{equation*}

and \(f_n \to f\) in distribution. But exactly one of the two indicators is \(1\) at each \(x\), so \(|f_n - f| \equiv 1\) and \(\mu(\{|f_n - f| \ge 1/2\}) = 1 \not\to 0\): convergence in measure fails, hence by (i) so do all seven modes.

(iii) With \(f := 1_{[0,1/2)}\), \(g := 1_{[1/2,1]}\) and \(f_n := g_n := g\), part (ii) gives \(f_n \to f\) in distribution and \(g_n \to g\) trivially. But \(f_n + g_n = 2g\) while \(f + g \equiv 1\), so

\begin{equation*} F_{f+g}(\lambda) = \begin{cases} 0, & \lambda < 1 \\ 1, & \lambda \ge 1, \end{cases} \qquad F_{2g}(\lambda) = \begin{cases} 0, & \lambda < 0 \\ 1/2, & 0 \le \lambda < 2 \\ 1, & \lambda \ge 2 . \end{cases} \end{equation*}

At \(\lambda = 3/2\), a continuity point of \(F_{f+g}\), we get \(F_{f_n+g_n}(3/2) = 1/2 \ne 1 = F_{f+g}(3/2)\).

(iv) Take \(f_n := f := 1_{[0,1/2)}\) and \(g := 1_{[1/2,1]}\). Then \(F_n = F\) identically, so \(f_n \to f\) in distribution, and \(F = F_g\) by (ii), so \(f_n \to g\) in distribution too. Yet \(f \ne g\) at every point of \([0,1]\), a set of measure \(1\).

Both assertions come from the difference quotient itself, and \(F^{\prime}\) is measurable as a pointwise limit of difference quotients at scale \(1/n\).

Continuity. If \(F\) is differentiable at \(x\), then as \(y \to x\) in \([a,b]\setminus\{x\}\),

\begin{equation*} F(y)-F(x) = \frac{F(y)-F(x)}{y-x}\,(y-x) \longrightarrow F^{\prime}(x)\cdot 0 = 0, \end{equation*}

so \(F\) is continuous at \(x\); in the everywhere differentiable case, on all of \([a,b]\).

Measurability, everywhere differentiable case. Extend \(F\) to \(\hat F : \mathbb{R}\to\mathbb{R}\) by the constants \(F(a)\) on \((-\infty,a)\) and \(F(b)\) on \((b,\infty)\); by the previous paragraph \(\hat F\) is continuous, hence measurable (Exercise 1.3.8(i)), and so is each

\begin{equation*} g_n(x) := n\left(\hat F\left(x+\tfrac1n\right)-\hat F(x)\right). \end{equation*}

For \(x \in [a,b)\) and \(n\) large enough that \(\tfrac1n < b-x\) the point \(x+\tfrac1n\) lies in \([a,b]\), so \(g_n(x)\) is a difference quotient of \(F\) at \(x\) and \(g_n(x)\to F^{\prime}(x)\) by (1.19). Hence \(g_n 1_{[a,b)} \to F^{\prime}1_{[a,b)}\) pointwise on \(\mathbb{R}\), so \(F^{\prime}1_{[a,b)}\) is measurable (Exercise 1.3.8(iv),(vi)), and altering it at the single point \(b\) – a null set – leaves it measurable (Exercise 1.3.8(iii)).

Measurability, almost everywhere differentiable case. Let \(E\) be the set of differentiability and \(N := [a,b]\setminus E\), which is null and hence measurable, so \(E\) is measurable. Only the measurability of \(\hat F := F 1_{[a,b]}\) needs a new argument, \(F\) no longer being continuous. For \(\lambda \in \mathbb{R}\) put

\begin{equation*} U_\lambda := \{x \in \mathbb{R} : F > \lambda \text{ on } [a,b]\cap(x-\delta,x+\delta) \text{ for some } \delta>0\}, \end{equation*}

which is open (halve the witness \(\delta\)). If \(x \in E\) and \(F(x)>\lambda\), continuity of \(F\) at \(x\) supplies such a \(\delta\); conversely \(y:=x\) in the definition gives \(F(x)>\lambda\). So \(\{x\in E: F(x)>\lambda\} = E \cap U_\lambda\) is measurable, whence \(\tilde F := F 1_E\) has measurable superlevel sets (for \(\lambda<0\) adjoin the measurable set \(\mathbb{R}\setminus E\)) and is measurable by Exercise 1.3.7(iv), since \(\tilde F^{-1}((c,d)) = \{\tilde F>c\}\setminus\bigcap_n \{\tilde F > d-\tfrac1n\}\) and Lemma 1.2.13. As \(\hat F\) agrees with \(\tilde F\) off the null set \(N\), \(\hat F\) is measurable (Exercise 1.3.8(iii)), and so is each \(g_n\) above, the translate \(\hat F(\cdot+\tfrac1n)\) being measurable by translation invariance (Exercise 1.2.20). For \(x \in E \cap [a,b)\) and \(n\) large, \(g_n(x)\to F^{\prime}(x)\) exactly as before; hence \(F^{\prime}1_{E\cap[a,b)} = \lim_n g_n 1_{E\cap[a,b)}\) is measurable, and it agrees with \(F^{\prime}\) outside the null set \(N \cup \{b\}\).

Discontinuous example. On \([0,1]\) take \(F := 1_{\{1/2\}}\). For \(x \neq \tfrac12\), \(F\) vanishes identically on \([0,1]\cap(x-\delta,x+\delta)\) with \(\delta := |x-\tfrac12|\), so \(F^{\prime}(x)=0\); thus \(F\) is differentiable off the null set \(\{\tfrac12\}\), yet \(F(y)\to 0 \neq 1 = F(\tfrac12)\).

Take \([a,b] := [-1,1]\) and

\begin{equation*} F(x) := \begin{cases} x^2 \sin(1/x), & x \neq 0, \\ 0, & x = 0. \end{cases} \end{equation*}

For \(x \neq 0\) the product and chain rules give

\begin{equation*} F^{\prime}(x) = 2x\sin(1/x) - \cos(1/x), \end{equation*}

and at the origin

\begin{equation*} \left|\frac{F(h)-F(0)}{h}\right| = |h|\,|\sin(1/h)| \leq |h| \longrightarrow 0, \end{equation*}

so \(F^{\prime}(0) = 0\) and \(F\) is everywhere differentiable on \([-1,1]\). But along \(x_n := \frac{1}{2\pi n} \to 0\) we have \(\sin(1/x_n)=0\) and \(\cos(1/x_n)=1\), so \(F^{\prime}(x_n) = -1\) for every \(n\), whereas \(F^{\prime}(0)=0\). Hence \(F^{\prime}\) is discontinuous at the origin.

Take \(f(x) := \left|x - \tfrac12\right|\) on \([0,1]\), which is continuous with \(f(0) = \tfrac12 = f(1)\), as Theorem 1.6.2 requires.

Near any \(x \neq \tfrac12\) the function \(f\) is affine: for \(|y-x| < |x-\tfrac12|\) one has \(f(y) = \tfrac12 - y\) when \(x < \tfrac12\) and \(f(y) = y - \tfrac12\) when \(x > \tfrac12\), so

\begin{equation*} f^{\prime}(x) = -1 \ \ (x < \tfrac12), \qquad f^{\prime}(x) = +1 \ \ (x > \tfrac12), \end{equation*}

and \(f\) is differentiable off the null set \(\{\tfrac12\}\). At \(\tfrac12\) the difference quotient \(\frac{|y-1/2|}{y-1/2}\) is \(+1\) for \(y > \tfrac12\) and \(-1\) for \(y < \tfrac12\), so \(f^{\prime}(\tfrac12)\) does not exist. Hence no \(x \in (0,1)\) has \(f^{\prime}(x) = 0\), and the conclusion of Theorem 1.6.2 fails.

Both directions are statements about \(H := F - G\).

If \(F = G + C\), the difference quotients of \(F\) and \(G\) at any \(x\) coincide identically in \(y\), so the limit (1.19) for \(F\) exists and equals \(G^{\prime}(x)\).

Conversely, suppose \(F^{\prime} = G^{\prime}\) on \([a,b]\). By the limit laws applied to

\begin{equation*} \frac{H(y)-H(x)}{y-x} = \frac{F(y)-F(x)}{y-x} - \frac{G(y)-G(x)}{y-x}, \end{equation*}

\(H\) is differentiable with \(H^{\prime} = F^{\prime}-G^{\prime} = 0\) throughout \([a,b]\). Fix \(x \in (a,b]\); the restriction \(H|_{[a,x]}\) is differentiable at each \(z \in [a,x]\), its difference quotients being a restriction of those on \([a,b]\setminus\{z\}\), so the mean value theorem (Corollary 1.6.5) supplies \(\xi \in (a,x)\) with

\begin{equation*} \frac{H(x)-H(a)}{x-a} = H^{\prime}(\xi) = 0. \end{equation*}

Thus \(H(x) = H(a) =: C\) for every \(x \in [a,b]\), i.e. \(F = G + C\).

Continuity is dominated convergence applied to \(h_n := f\left(1_{[-\infty,x_n]} - 1_{[-\infty,x]}\right)\).

Fix \(x \in \mathbb{R}\) and \(x_n \to x\); sequential continuity suffices on a metric space. By linearity of the integral (Exercise 1.3.19),

\begin{equation*} F(x_n) - F(x) = \int_{\mathbb{R}} h_n(t)\ dt. \end{equation*}

The bracket takes values in \(\{-1,0,1\}\), so \(|h_n| \leq |f|\) pointwise with \(|f|\) absolutely integrable. For \(t \neq x\): if \(t < x\) then \(x_n > t\) for all large \(n\), and if \(t > x\) then \(x_n < t\) for all large \(n\); either way both indicators agree at \(t\) eventually and \(h_n(t) = 0\). Thus \(h_n \to 0\) outside the null set \(\{x\}\), and the dominated convergence theorem (Theorem 1.4.49), with dominating function \(|f|\), gives \(F(x_n) - F(x) \to 0\). Hence \(F\) is continuous.

Continuity of \(F\) is Exercise 1.6.5; differentiability is (1.20) and (1.21) read as the two one-sided difference quotients of \(F\).

For \(h > 0\) one has \(1_{[-\infty,x+h]} - 1_{[-\infty,x]} = 1_{(x,x+h]}\) and \(1_{[-\infty,x]} - 1_{[-\infty,x-h]} = 1_{(x-h,x]}\), so by linearity of the integral, and since altering the integrand on the null sets \(\{x\}\), \(\{x-h\}\) does not change its integral,

\begin{align*} \frac{F(x+h)-F(x)}{h} &= \frac1h \int_{[x,x+h]} f(t)\ dt, \\ \frac{F(x-h)-F(x)}{(x-h)-x} &= \frac1h \int_{[x-h,x]} f(t)\ dt. \end{align*}

Let \(A\) and \(B\) be the sets where (1.20) and (1.21) respectively hold; by Theorem 1.6.12 their complements are null, hence so is \(\mathbb{R}\setminus(A\cap B)\), and \(A \cap B\) has full measure.

Fix \(x \in A \cap B\) and \(\varepsilon > 0\), and let \(\delta > 0\) be smaller than the two thresholds furnished by (1.20) and (1.21) for this \(\varepsilon\). For \(y \neq x\) with \(|y-x| < \delta\), putting \(h := |y-x| \in (0,\delta)\) and using the first display when \(y > x\), the second when \(y < x\),

\begin{equation*} \left|\frac{F(y)-F(x)}{y-x} - f(x)\right| \leq \varepsilon. \end{equation*}

As \(\varepsilon\) was arbitrary, the limit (1.19) exists at \(x\) with \(F^{\prime}(x) = f(x)\). So \(F\) is differentiable with derivative \(f\) on a set of full measure, which with the continuity above is Theorem 1.6.11.

Fix \(M \geq 0\) and a null set \(N\) with \(|g| \leq M\) off \(N\); then \(|f*g| \leq M\|f\|_{L^1}\) everywhere, and continuity follows from \(L^1\)-continuity of translation (Proposition 1.6.13).

The reflection-translation \(T_x(y) := x-y\) is its own inverse and preserves Lebesgue measurability and measure, by Exercise 1.2.20 (translation) and Exercise 1.2.21 (linear maps, here \(|\det| = 1\)). Hence \(F \circ T_x\) is measurable whenever \(F\) is, via \((F\circ T_x)^{-1}(U) = T_x(F^{-1}(U))\) and Exercise 1.3.7(iv), and

\begin{equation*} \int_{\mathbb{R}^d} F(T_x(y))\ dy = \int_{\mathbb{R}^d} F(z)\ dz \tag{\dagger} \end{equation*}

for \(F\) unsigned measurable or absolutely integrable (Exercises 1.3.15, 1.3.16, 1.3.20).

Well-definedness and boundedness. Fix \(x\). Then \(y \mapsto g(x-y) = g(T_x(y))\) is measurable, hence so is \(y \mapsto f(y)g(x-y)\) (Exercise 1.3.8(vi)); and \(x - N\) is null, so \(|g(x-y)| \leq M\) for almost every \(y\) and

\begin{equation*} \int_{\mathbb{R}^d} |f(y)g(x-y)|\ dy \leq M\int_{\mathbb{R}^d} |f(y)|\ dy = M\|f\|_{L^1} < \infty \end{equation*}

by monotonicity and homogeneity of the unsigned integral (Exercise 1.3.10(ii),(iii)). So \(f*g(x)\) is well defined, and \(|f*g(x)| \leq M\|f\|_{L^1}\) by the triangle inequality (Lemma 1.3.19).

Continuity. Applying \((\dagger)\) to \(F(z) := f(x-z)g(z)\) gives \(f*g(x) = \int f(x-z)g(z)\ dz\), so for \(x, x^{\prime} \in \mathbb{R}^d\) with \(h := x - x^{\prime}\), linearity, the triangle inequality and \(|g| \leq M\) almost everywhere yield

\begin{align*} |f*g(x) - f*g(x^{\prime})| &\leq M \int_{\mathbb{R}^d} |f(x-z)-f(x^{\prime}-z)|\ dz \\ &= M \int_{\mathbb{R}^d} |f(w) - f(w-h)|\ dw, \end{align*}

the last step by \((\dagger)\) with \(w := x-z\), since then \(x^{\prime}-z = w-h\). By Proposition 1.6.13 the right-hand side tends to \(0\) as \(h \to 0\), so \(f*g\) is (uniformly) continuous.

The convolution \(1_E * 1_{-E}(x) = m\big(E \cap (E+x)\big)\) is continuous and strictly positive at the origin, and its positivity at \(x\) forces \(x \in E-E\).

Reduction to bounded \(E\). As \(E = \bigcup_N (E \cap B(0,N))\) increases to \(E\), upward monotone convergence of Lebesgue measure gives \(m(E \cap B(0,N)) > 0\) for some \(N\); replacing \(E\) by \(E \cap B(0,N)\) only shrinks \(E-E\), so we may assume \(0 < m(E) < \infty\).

The convolution. Here \(f := 1_E\) is absolutely integrable with \(\int |f| = m(E)\), and \(g := 1_{-E}\) is measurable (reflection preserves Lebesgue measurability, Exercise 1.2.21) and bounded by \(1\), so Exercise 1.6.7 makes \(f*g\) bounded and continuous. Since \(x - y \in -E\) exactly when \(y \in E+x\), the integrand is \(1_{E \cap (E+x)}(y)\) and

\begin{equation*} f*g(x) = m\big(E \cap (E+x)\big). \end{equation*}

Conclusion. Thus \(f*g(0) = m(E) > 0\), so by continuity there is \(\delta > 0\) with \(f*g > 0\) on \(B(0,\delta)\). For such an \(x\) the set \(E \cap (E+x)\) is non-empty, say \(z = w + x\) with \(z, w \in E\), whence \(x = z - w \in E-E\). Therefore \(B(0,\delta) \subseteq E - E\).

Every homomorphism is rationally homogeneous: \(x=y=0\) gives \(f(0)=0\), then \(f(-x)=-f(x)\), induction gives \(f(nx)=nf(x)\) for \(n \in \mathbb{Z}\), and \(n f((m/n)x) = f(mx) = mf(x)\), so

\begin{equation*} f(qx) = q f(x) \qquad (q \in \mathbb{Q},\ x \in \mathbb{R}^d). \end{equation*}

(i) It suffices to prove continuity at the origin, since \(f(x)-f(x_0) = f(x-x_0)\). Fix \(\varepsilon > 0\), put \(D := \{|w| < \varepsilon\}\) and \(S := \{\tfrac{\varepsilon}{2}(j+ik) : j,k \in \mathbb{Z}\}\); every \(w \in \mathbb{C}\) is within \(\tfrac{\varepsilon}{2}\sqrt2 < \varepsilon\) of \(S\), so \(\mathbb{C} = \bigcup_{z\in S}(z+D)\) and

\begin{equation*} \mathbb{R}^d = \bigcup_{z \in S} f^{-1}(z+D) \end{equation*}

is a countable union of measurable sets. Were all of them null, countable subadditivity would give \(m(\mathbb{R}^d)=0\); so \(A := f^{-1}(z+D)\) has positive measure for some \(z\). By the Steinhaus theorem (Exercise 1.6.8), \(A - A \supseteq B(0,\delta)\) for some \(\delta>0\), and writing \(x = a-a^{\prime}\) with \(a,a^{\prime}\in A\) gives

\begin{equation*} f(x) = f(a)-f(a^{\prime}) \in (z+D)-(z+D) = D-D, \qquad |x| < \delta, \end{equation*}

so \(|f(x)| < 2\varepsilon\). Running this with \(\varepsilon/2\) yields continuity at \(0\), hence everywhere.

(ii) Maps \(x \mapsto \sum_j x_j z_j\) are additive and continuous, hence measurable. Conversely let \(f\) be a measurable homomorphism and set \(z_j := f(e_j)\). Additivity and rational homogeneity give

\begin{equation*} f(q_1,\dots,q_d) = \sum_{j=1}^d f(q_j e_j) = \sum_{j=1}^d q_j z_j \qquad (q_j \in \mathbb{Q}), \end{equation*}

so \(f\) agrees with the continuous map \(g(x) := \sum_j x_j z_j\) on the dense set \(\mathbb{Q}^d\); by (i) \(f\) is continuous too, so \(f=g\).

(iii) Give \(\mathbb{R}^d\) a basis \(B\) over \(\mathbb{Q}\) (Zorn), so each \(x\) has a unique expansion \(x = \sum_{b\in B} q_b(x)\,b\) with finitely many non-zero rational coefficients; fix \(b_0 \in B\) and put \(f(x) := q_{b_0}(x)\). Uniqueness makes each \(q_b\) additive, so \(f\) is a homomorphism. Its range lies in \(\mathbb{Q}\), hence is countable, whereas \(x \mapsto \sum_j x_j z_j\) with some \(z_j \neq 0\) already takes uncountably many values along \(t \mapsto te_j\), and with all \(z_j=0\) it vanishes identically, contradicting \(f(b_0)=1\). So \(f\) has no such form, and by (ii) \(f\) is not measurable: some \(f^{-1}(U)\), \(U \subseteq \mathbb{C}\) open, is non-measurable.

Take for each \(x \in U\) the maximal interval \(I_x := (a_x,b_x)\) about \(x\), where

\begin{equation*} a_x := \inf\{a < x : (a,x] \subseteq U\}, \qquad b_x := \sup\{b > x : [x,b) \subseteq U\}; \end{equation*}

both sets are non-empty because \(U\) is open, so \(a_x < x < b_x\).

\(I_x \subseteq U\). For \(y \in (a_x,x]\) there is \(a < y\) with \((a,x]\subseteq U\), so \(y \in U\); symmetrically on \([x,b_x)\).

Maximality. If \(x \in J = (\alpha,\beta) \subseteq U\) then \((a,x] \subseteq U\) for every \(a \in (\alpha,x)\), so \(a_x \leq \alpha\), and symmetrically \(b_x \geq \beta\); hence \(J \subseteq I_x\).

Endpoints outside \(U\). If \(a_x\) were finite and lay in \(U\), openness would give \((a_x-\varepsilon,a_x+\varepsilon)\subseteq U\), hence \((a_x-\varepsilon,x]\subseteq U\) and \(a_x \leq a_x-\varepsilon\), absurd; symmetrically for \(b_x\), the infinite case being vacuous.

Disjointness. If \(I_x \cap I_y \neq \emptyset\) then \(I_x \cup I_y\) is an open interval inside \(U\) containing both \(x\) and \(y\), so maximality gives \(I_x \cup I_y \subseteq I_x\) and \(\subseteq I_y\), i.e. \(I_x = I_y\). Distinct members of \(\mathcal{I} := \{I_x : x \in U\}\) are therefore disjoint.

Countability. Each \(I \in \mathcal{I}\) contains a rational \(q(I)\), and disjointness makes \(I \mapsto q(I)\) injective into \(\mathbb{Q}\), so \(\mathcal{I}\) is at most countable. Finally \(x \in I_x \subseteq U\) gives \(U = \bigcup_{I \in \mathcal{I}} I\).

Every interval through \(x\) splits at \(x\) into two one-sided ones, so \(Mf\) is controlled by the one-sided maximal functions

\begin{equation*} M^+f(x) := \sup_{h>0}\frac1h\int_{[x,x+h]}|f|, \qquad M^-f(x) := \sup_{h>0}\frac1h\int_{[x-h,x]}|f|, \end{equation*}

where \(Mf(x)\) denotes the two-sided maximal function of the statement.

Splitting. Let \(\mu > 0\) and \(Mf(x) > \mu\), witnessed by an interval \(I \ni x\) with \(\int_I |f| > \mu|I|\); then \(|I| < \infty\), since \(\int_I|f| \leq \int_{\mathbb{R}}|f| < \infty\). Writing \(c \leq x \leq d\) for its endpoints and using that \(\{x\}\) is null,

\begin{equation*} \int_{[c,x]}|f| + \int_{[x,d]}|f| \geq \int_I |f| > \mu(x-c) + \mu(d-x), \end{equation*}

so at least one summand beats its counterpart, and a summand of zero length cannot:

• (i) \(\int_{[c,x]}|f| > \mu(x-c)\), forcing \(x-c>0\) and \(M^-f(x) > \mu\);
• (ii) \(\int_{[x,d]}|f| > \mu(d-x)\), forcing \(d-x>0\) and \(M^+f(x) > \mu\).

Hence \(\{Mf > \mu\} \subseteq \{M^+f > \mu\} \cup \{M^-f > \mu\}\).

Measurability. With \(H(x) := \int_{[-\infty,x]}|f|\), continuous by Exercise 1.6.5, we have \(\frac1h\int_{[x,x+h]}|f| = \frac{H(x+h)-H(x)}{h}\) and \(\frac1h\int_{[x-h,x]}|f| = \frac{H(x)-H(x-h)}{h}\), each continuous in \(x\) for fixed \(h\); so \(M^{\pm}f\) are lower semicontinuous and \(\{M^{\pm}f>\mu\}\) are open. So is \(\{Mf>\mu\}\): enlarging the witness to \(I_\varepsilon := (c-\varepsilon,d+\varepsilon)\) preserves \(\int_{I_\varepsilon}|f| \geq \int_I|f| > \mu|I|\), and for \(\varepsilon\) small the strict inequality gives \(\int_{I_\varepsilon}|f| > \mu|I_\varepsilon|\), so \(Mf > \mu\) on the whole neighbourhood \(I_\varepsilon\) of \(x\). Consequently \(\{Mf \geq \lambda\} = \bigcap_{n>1/\lambda}\{Mf > \lambda-\tfrac1n\}\) is measurable.

One-sided bounds. Lemma 1.6.16 gives \(m(\{M^+f \geq \mu\}) \leq \frac1\mu\int_{\mathbb{R}}|f|\) for all \(\mu>0\). Applying it to \(\tilde f(x) := f(-x)\), which is absolutely integrable with the same integral of \(|\tilde f|\), and substituting \(t=-s\) to get \(M^-f(x) = M^+\tilde f(-x)\), reflection invariance of Lebesgue measure yields the same bound for \(m(\{M^-f\geq\mu\})\).

Conclusion. For \(0<\mu<\lambda\) we have \(\{Mf\geq\lambda\}\subseteq\{Mf>\mu\}\), so the splitting and subadditivity give

\begin{equation*} m(\{Mf \geq \lambda\}) \leq m(\{M^+f \geq \mu\}) + m(\{M^-f\geq\mu\}) \leq \frac{2}{\mu}\int_{\mathbb{R}}|f(t)|\,dt, \end{equation*}

and letting \(\mu \uparrow \lambda\) gives the stated inequality.

Set \(F(x) := \int_{[-\infty,x]} f\), finite and continuous by Exercise 1.6.5, with \(F(-\infty)=0\) and \(F(+\infty) = L := \int_{\mathbb{R}} f\) by dominated convergence, and put \(G(x) := F(x) - \lambda x\). Since a point is null, \(\frac1h\int_{[x,x+h]}f > \lambda\) precisely when \(G(x+h) > G(x)\), so

\begin{equation*} E_\lambda := \{f^* > \lambda\} = \{x \in \mathbb{R} : G(y) > G(x) \text{ for some } y > x\}, \end{equation*}

the rising sun shadow of \(G\) on \(\mathbb{R}\).

Decomposition. \(E_\lambda\) is open: given \(x \in E_\lambda\) with witness \(y\), continuity of \(G\) at \(x\) supplies a neighbourhood \(V \subseteq (-\infty,y)\) of \(x\) on which \(G < G(y)\), and \(V \subseteq E_\lambda\). By Exercise 1.6.10, \(E_\lambda = \bigsqcup_n I_n\) with \(I_n = (a_n,b_n)\) disjoint non-empty open intervals whose finite endpoints lie outside \(E_\lambda\).

Rising sun estimate. (a) If \(b := b_n < \infty\) then \(G \leq G(b)\) on \(I_n\). Else \(G(x) > G(b)\) for some \(x \in I_n\); since \(b \notin E_\lambda\) we have \(G \leq G(b)\) on \((b,\infty)\), and \(A := \{s \in [x,b] : G(s) \geq G(x)\}\) is compact, contains \(x\) and misses \(b\), so \(s^* := \sup A \in A \cap [x,b)\). But \(s^* \in E_\lambda\) yields \(y > s^*\) with \(G(y) > G(s^*) \geq G(x) > G(b)\), excluding \(y \geq b\); so \(y \in (s^*,b)\) lies in \(A\), contradicting \(s^* = \sup A\).

(b) If \(I_n = (a_n,\infty)\) and \(G(y) \to \ell \in [-\infty,\infty)\) as \(y \to +\infty\), then \(G \leq \ell\) on \(I_n\): else pick \(\ell < c < G(x)\), so that \(A := \{y \geq x : G(y) \geq c\}\) is closed and bounded above and has a maximum \(s^*\), while \(s^* \in E_\lambda\) supplies \(y > s^*\) with \(G(y) > G(s^*) \geq c\), i.e. \(y \in A\).

On a bounded component, (a) and continuity (let \(x \downarrow a_n\)) give \(G(a_n) \leq G(b_n)\), i.e.

\begin{equation*} \int_{I_n} f = F(b_n)-F(a_n) \geq \lambda(b_n-a_n) = \lambda\, m(I_n). \end{equation*}

Disjointness with \(\sum_n \int_{I_n}|f| \leq \int_{\mathbb{R}}|f| < \infty\) makes every series below absolutely convergent, with \(\int_{E_\lambda} f = \sum_n \int_{I_n} f\) by countable additivity.

• (i) \(\lambda > 0\): \(G(y) = F(y)-\lambda y \to -\infty\), so by (b) with \(\ell = -\infty\) no component has right endpoint \(+\infty\); and a component \((-\infty,b)\) is impossible, since (a) would force \(G \leq G(b)\) on it while \(G(x) \to +\infty\) as \(x \to -\infty\). Every component is thus bounded, and summing gives \(\int_{E_\lambda} f \geq \lambda\, m(E_\lambda)\).
• (ii) \(\lambda = 0\), so \(G = F\) with finite limits \(0\) and \(L\). Bounded components are as above; for \((a_n,\infty)\), (b) with \(\ell = L\) gives \(F(a_n) \leq L\), so \(\int_{I_n} f = L - F(a_n) \geq 0\); for \((-\infty,b_n)\), (a) gives \(0 = F(-\infty) \leq F(b_n) = \int_{I_n} f\); for \(\mathbb{R}\), (b) gives \(0 \leq L\). Summing, \(\int_{E_0} f \geq 0 = 0\cdot m(E_0)\).
• (iii) \(\lambda < 0\): if \(m(E_\lambda) = \infty\) the left side is \(-\infty\) and the right side is real (bounded by \(\int_{\mathbb{R}}|f|\)), so the inequality is trivial; otherwise every component has finite length, hence finite endpoints, and summation proceeds as in (i).

Lemma 1.6.16. Apply case (i) to the real-valued absolutely integrable \(g := |f|\): for \(0 < \mu < \lambda\),

\begin{equation*} \mu\, m(\{g^* > \mu\}) \leq \int_{\{g^*>\mu\}} |f| \leq \int_{\mathbb{R}}|f(t)|\,dt, \end{equation*}

and \(\{g^* \geq \lambda\} \subseteq \{g^* > \mu\}\), so \(m(\{g^*\geq\lambda\}) \leq \frac1\mu\int_{\mathbb{R}}|f|\); letting \(\mu \uparrow \lambda\) gives Lemma 1.6.16.

Equality holds because both endpoints of each component of \(E_\lambda\) lie outside it. (Following the published errata, “Lemma 1.6.16” in the printed statement reads “Exercise 1.6.12” and \(\lambda > 0\) is assumed; as printed it is false, e.g. \(f = 1_{[0,1]}\), \(\lambda = 2\) gives \(0 \neq \tfrac12\).)

Keep \(F\), \(G(x) = F(x)-\lambda x\) and \(E_\lambda = \{f^*>\lambda\} = \bigsqcup_n I_n\), \(I_n = (a_n,b_n)\), as in Exercise 1.6.12; since \(\lambda > 0\), every component there is bounded with both endpoints finite and outside \(E_\lambda\), and \(m(E_\lambda) \leq \frac1\lambda\int_{\mathbb{R}}|f| < \infty\).

Fix \(n\). The rising sun estimate of Exercise 1.6.12 gives \(G \leq G(b_n)\) on \(I_n\), hence \(G(a_n) \leq G(b_n)\) on letting \(x \downarrow a_n\). Conversely \(a_n \notin E_\lambda\) means \(G(y) \leq G(a_n)\) for every \(y > a_n\), and \(y := b_n\) gives \(G(b_n) \leq G(a_n)\). So \(G(a_n) = G(b_n)\), i.e.

\begin{equation*} \int_{I_n} f = F(b_n)-F(a_n) = \lambda(b_n-a_n) = \lambda\, m(I_n). \end{equation*}

Summing over \(n\), absolutely convergent since \(\sum_n \int_{I_n}|f| \leq \int_{\mathbb{R}}|f| < \infty\) and \(\sum_n m(I_n) = m(E_\lambda) < \infty\),

\begin{equation*} \int_{x : f^*(x)>\lambda} f(x)\,dx = \sum_n \int_{I_n} f = \lambda\sum_n m(I_n) = \lambda\, m(\{f^*(x)>\lambda\}). \end{equation*}

(i) If \(\int_{B(0,r)}|f| < \infty\) for every \(r>0\), then \(B(0,|x|+1)\) is an open neighbourhood of \(x\) on which \(f\) is absolutely integrable. Conversely, let \(f\) be locally integrable and \(r>0\). For each \(x \in \overline{B}(0,r)\) pick an open \(U_x \ni x\) with \(\int_{U_x}|f| < \infty\) and a ball \(B(x,\rho_x) \subseteq U_x\); the set \(\overline{B}(0,r)\) is compact (Heine-Borel), so finitely many of these balls cover it and

\begin{equation*} B(0,r) \subseteq \bigcup_{j=1}^k U_{x_j}, \qquad \int_{B(0,r)}|f| \leq \sum_{j=1}^k \int_{U_{x_j}}|f| < \infty \end{equation*}

by monotonicity and finite subadditivity of the unsigned integral (\(f\) being measurable on the finite union).

(ii) The generalisation asserts that a locally integrable \(f\) has almost every \(x\) as a Lebesgue point, so that (1.26) and the averaging conclusion \(\frac{1}{m(B(x,r))}\int_{B(x,r)} f \to f(x)\) hold almost everywhere. Put \(f_n := f 1_{B(0,n)}\), which by (i) is measurable with \(\int_{\mathbb{R}^d}|f_n| = \int_{B(0,n)}|f| < \infty\), and let \(N_n\) be the null set off which Theorem 1.6.19 gives (1.26) for \(f_n\). Then \(N := \bigcup_n N_n\) is null. For \(x \notin N\) choose \(n > |x|\): then \(f_n(x)=f(x)\), and for \(0<r<n-|x|\) we have \(B(x,r) \subseteq B(0,n)\), so \(f_n = f\) on \(B(x,r)\) and

\begin{equation*} \frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dy = \frac{1}{m(B(x,r))}\int_{B(x,r)}|f_n(y)-f_n(x)|\,dy \longrightarrow 0 \end{equation*}

as \(r \to 0\), since \(x \notin N_n\). The averaging conclusion follows from the triangle inequality,

\begin{equation*} \left|\frac{1}{m(B(x,r))}\int_{B(x,r)} f(y)\,dy - f(x)\right| \leq \frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dy, \end{equation*}

all integrals being finite because \(B(x,r) \subseteq B(0,n)\) for small \(r\).

The whole matter is the inclusion \(x + E_h \subseteq B(x,h)\) together with \(m(E_h) \geq c\, m(B(x,h))\).

The \(E_h\) are understood measurable. Translation invariance gives \(m(x+E_h) = m(E_h) \geq c\,m(B(0,h)) = c\,\omega_d h^d > 0\), where \(\omega_d := m(B(0,1)) \in (0,\infty)\), and \(\int_{B(x,h)}|f| < \infty\) by Exercise 1.6.14(i); so every quantity below is finite with non-zero denominator. Writing \(f(x) = \frac{1}{m(E_h)}\int_{x+E_h} f(x)\,dy\), the triangle inequality, monotonicity of the unsigned integral over \(x+E_h \subseteq x + B(0,h) = B(x,h)\), and the measure hypothesis give

\begin{align*} \left|\frac{1}{m(E_h)}\int_{x+E_h} f(y)\,dy - f(x)\right| &\leq \frac{1}{m(E_h)}\int_{B(x,h)}|f(y)-f(x)|\,dy \\ &\leq \frac1c \cdot \frac{1}{m(B(x,h))}\int_{B(x,h)}|f(y)-f(x)|\,dy, \end{align*}

which tends to \(0\) as \(h \to 0\) since \(c\) is independent of \(h\) and \(x\) is a Lebesgue point of \(f\).

Theorem 1.6.12 from Theorem 1.6.19. Take \(d=1\) and \(f\) absolutely integrable, hence locally integrable, so that almost every \(x\) is a Lebesgue point by Theorem 1.6.19. With \(E_h := [0,h) \subseteq B(0,h)\) we have \(m(E_h) = h = \tfrac12 m(B(0,h))\), so \(c = \tfrac12\) works, and \(x+E_h = [x,x+h)\) differs from \([x,x+h]\) by a null set, giving (1.20). With \(E_h := (-h,0]\), likewise \(c = \tfrac12\) and \(x + E_h = (x-h,x]\) differs from \([x-h,x]\) by a null set, giving (1.21).

When \(f\) is continuous, every point is a Lebesgue point, so the exceptional null set is empty.

Fix \(x\) and \(\varepsilon > 0\), and take \(\delta > 0\) with \(|f(y)-f(x)| \leq \varepsilon\) for \(|y-x| < \delta\). For \(0 < r < \delta\) the continuous integrand \(y \mapsto |f(y)-f(x)|\) is measurable and bounded by \(\varepsilon\) on \(B(x,r)\), and \(m(B(x,r)) = \omega_d r^d \in (0,\infty)\), so monotonicity of the unsigned integral gives

\begin{equation*} \frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dy \leq \varepsilon. \end{equation*}

As \(\varepsilon\) was arbitrary, (1.26) holds at \(x\). The averaging conclusion follows by the triangle inequality,

\begin{equation*} \left|\frac{1}{m(B(x,r))}\int_{B(x,r)} f(y)\,dy - f(x)\right| \leq \frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dy \longrightarrow 0. \end{equation*}

Write \(M\) for the Hardy-Littlewood maximal function of Theorem 1.6.20 and

\begin{equation*} \Lambda F(x) := \limsup_{r\to0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|F(y)-F(x)|\,dy \in [0,+\infty]; \end{equation*}

conclusion (1.26) says exactly that \(\Lambda f = 0\) almost everywhere, the second conclusion of Theorem 1.6.19 following from it by the triangle inequality as in the text.

Sublinearity. For absolutely integrable \(g\) and \(h := f-g\), the pointwise bound \(|f(y)-f(x)| \leq |g(y)-g(x)| + |h(y)| + |h(x)|\), averaged over \(B(x,r)\) and passed to the \(\limsup\), gives

\begin{equation*} \Lambda f(x) \leq \Lambda g(x) + Mh(x) + |h(x)| \qquad (x \in \mathbb{R}^d). \end{equation*}

Density. Given \(\varepsilon > 0\), Littlewood’s second principle (Theorem 1.3.20(iii)) supplies a continuous, compactly supported \(g\) with \(\|f-g\|_{L^1} \leq \varepsilon\); by Exercise 1.6.16 every point is a Lebesgue point of \(g\), so \(\Lambda g \equiv 0\) and \(\Lambda f \leq Mh + |h|\) with \(\|h\|_{L^1} \leq \varepsilon\).

Estimate. Hence for \(\lambda > 0\),

\begin{equation*} \{\Lambda f > \lambda\} \subseteq \{Mh \geq \lambda/2\} \cup \{|h| \geq \lambda/2\}, \end{equation*}

and the two pieces have measure at most \(\frac{2C_d}{\lambda}\varepsilon\) by Theorem 1.6.20 and \(\frac{2}{\lambda}\varepsilon\) by Markov’s inequality (Lemma 1.3.15) applied to the unsigned \(|h|\). Monotonicity and finite subadditivity of outer measure – used because the left set is not yet known to be measurable – give

\begin{equation*} m^*(\{\Lambda f > \lambda\}) \leq \frac{2(C_d+1)}{\lambda}\varepsilon, \end{equation*}

whose left side is independent of \(\varepsilon\), so \(\{\Lambda f > \lambda\}\) is null. Therefore \(\{\Lambda f > 0\} = \bigcup_n \{\Lambda f > 1/n\}\) is null by countable subadditivity, i.e. \(\Lambda f = 0\) almost everywhere; a vanishing \(\limsup\) of a non-negative quantity is a limit, so this is (1.26).

Record at each time only the single index chosen then; the algorithm’s family after \(t\) steps is the initial segment \((j_1,\dots,j_t)\) of one fixed sequence, so nothing is ever redefined.

Let \(B_i = B(c_i,r_i)\) for \(i \in [n] := \{1,\dots,n\}\), with \(r_i > 0\) (radius-zero balls are empty and may be discarded, affecting neither conclusion). Recursively, given distinct \(j_1,\dots,j_t \in [n]\) – for \(t=0\) the empty list – set

\begin{equation*} A_t := \{i \in [n] : B_i \cap B_{j_s} = \emptyset \text{ for all } 1 \leq s \leq t\}, \end{equation*}

so \(A_0 = [n]\). If \(A_t = \emptyset\), put \(t_* := t\) and stop; otherwise \(\rho_t := \max_{i \in A_t} r_i\) is attained on the finite non-empty \(A_t\) and we set

\begin{equation*} j_{t+1} := \min\{i \in A_t : r_i = \rho_t\}, \end{equation*}

the least index of maximal radius – a definite prescription replacing the arbitrary tie-break, and needing no choice. Each of \(A_t\), \(\rho_t\), \(j_{t+1}\), \(t_*\) is defined exactly once, from previously defined objects only.

Termination. If \(j_{t+1}\) is defined then \(j_{t+1} \in A_t\), so \(B_{j_{t+1}}\) is disjoint from \(B_{j_1},\dots,B_{j_t}\); a non-empty ball is not disjoint from itself, so \(j_1,\dots,j_t\) are distinct and \(t \leq n\). Hence \(A_t = \emptyset\) for some \(t \leq n\). Put \(m := t_*\) and \(B^{\prime}_s := B_{j_s}\) for \(1 \leq s \leq m\). (In the hint’s notation, \(m_t := \min(t,t_*)\) and \(B^{\prime}_{s,t} := B_{j_s}\).)

Disjointness. For \(s < s^{\prime} \leq m\) the \(A_t\) decrease, so \(j_{s^{\prime}} \in A_{s^{\prime}-1} \subseteq A_s\) and \(B_{j_{s^{\prime}}} \cap B_{j_s} = \emptyset\).

Covering. Fix \(i \in [n]\). As \(A_m = \emptyset\), \(B_i\) meets some \(B^{\prime}_s\); take the least such \(s\), so \(B_i\) misses \(B_{j_1},\dots,B_{j_{s-1}}\), i.e. \(i \in A_{s-1}\), and maximality of \(\rho_{s-1}\) gives \(r_i \leq r_{j_s}\). Picking \(z \in B_i \cap B^{\prime}_s\), the triangle inequality gives \(|c_i-c_{j_s}| < r_i + r_{j_s} \leq 2r_{j_s}\), so \(c_i \in 2B^{\prime}_s\), and for \(y \in B_i\),

\begin{equation*} |y-c_{j_s}| \leq |y-c_i| + |c_i-z| + |z-c_{j_s}| < 2r_i + r_{j_s} \leq 3r_{j_s}, \end{equation*}

so \(B_i \subseteq 3B^{\prime}_s\). Hence \(\bigcup_{i=1}^n B_i \subseteq \bigcup_{j=1}^m 3B^{\prime}_j\), which is (1.27), and finite subadditivity with \(m(3B^{\prime}_j) = 3^d m(B^{\prime}_j)\) gives \(m(\bigcup_{i=1}^n B_i) \leq 3^d\sum_{j=1}^m m(B^{\prime}_j)\).

The greedy selection of Lemma 1.6.22 in fact places each centre \(c_i\) in \(2B^{\prime}_{s(i)}\), and shrinking every ball about its own centre turns this into a covering, giving \(C_d = 2^d\).

Refined covering. Let \(B_i = B(c_i,r_i)\), \(r_i>0\), and let \(B^{\prime}_s = B(c^{\prime}_s,r^{\prime}_s)\) be the disjoint subcollection selected greedily. As shown in Exercise 1.6.18, if \(B^{\prime}_s\) is the first selected ball met by \(B_i\) then \(B_i\) was still available when \(B^{\prime}_s\) was chosen, so \(r_i \leq r^{\prime}_s\) by maximality, and any \(z \in B_i \cap B^{\prime}_s\) gives \(|c_i-c^{\prime}_s| < r_i + r^{\prime}_s \leq 2r^{\prime}_s\). Hence for every \(\varepsilon>0\),

\begin{equation*} B(c_i,\varepsilon r_i) \subseteq B(c^{\prime}_s, 2r^{\prime}_s + \varepsilon r_i) \subseteq (2+\varepsilon)B^{\prime}_s, \end{equation*}

again by \(r_i \leq r^{\prime}_s\). The shrinking is the epsilon of room: the centres alone form a finite set.

The inequality. It suffices to bound \(m(\{Mf>\lambda\})\), the non-strict form following by applying that bound with \(\lambda^{\prime}<\lambda\) and letting \(\lambda^{\prime}\uparrow\lambda\). This set is open: if \(\int_{B(x,r)}|f| > \lambda\omega_d r^d\), choose \(\rho>r\) with \(\lambda\omega_d\rho^d\) still below that integral; then \(|y-x| < \rho-r\) gives \(B(x,r) \subseteq B(y,\rho)\) and so \(Mf(y)>\lambda\). By inner regularity it is enough to bound \(m(K)\) for compact \(K \subseteq \{Mf>\lambda\}\).

Fix \(\varepsilon>0\). Each \(x \in K\) admits \(r_x>0\) with \(\int_{B(x,r_x)}|f| > \lambda\,m(B(x,r_x))\), and the shrunken balls \(B(x,\varepsilon r_x)\) cover \(K\), so finitely many do, with centres \(x_1,\dots,x_n\) and radii \(r_i := r_{x_i}\). Greedy selection applied to \(B_i := B(x_i,r_i)\) yields disjoint \(B^{\prime}_1,\dots,B^{\prime}_m\), and the refined covering gives

\begin{equation*} K \subseteq \bigcup_{i=1}^n B(x_i,\varepsilon r_i) \subseteq \bigcup_{j=1}^m (2+\varepsilon)B^{\prime}_j. \end{equation*}

Each \(B^{\prime}_j\) is one of the \(B_i\), so \(\lambda\,m(B^{\prime}_j) < \int_{B^{\prime}_j}|f|\), and finite subadditivity, \(m(cB) = c^d m(B)\) and disjointness give

\begin{align*} m(K) \leq (2+\varepsilon)^d\sum_{j=1}^m m(B^{\prime}_j) &\leq \frac{(2+\varepsilon)^d}{\lambda}\sum_{j=1}^m \int_{B^{\prime}_j}|f(y)|\,dy \\ &\leq \frac{(2+\varepsilon)^d}{\lambda}\|f\|_{L^1}. \end{align*}

Letting \(\varepsilon \to 0^+\) gives \(m(K) \leq \frac{2^d}{\lambda}\|f\|_{L^1}\), hence Theorem 1.6.20 with \(C_d = 2^d\).

Nesting makes the maximal dyadic cubes on which the average exceeds \(\mu\) pairwise disjoint, which replaces the covering lemma and costs no constant.

Write \(\mathcal{D}\) for the countable family of dyadic cubes and

\begin{equation*} M_{\mathcal{D}}f(x) := \sup_{Q \in \mathcal{D},\, x \in Q} \frac{1}{|Q|}\int_Q |f(y)|\,dy, \end{equation*}

a supremum over a non-empty set, since for each \(n\) the cubes of sidelength \(2^{-n}\) partition \(\mathbb{R}^d\).

Nesting. If \(Q,R \in \mathcal{D}\) meet, one contains the other. Say \(Q\) has sidelength \(2^{-n}\) and \(R\) sidelength \(2^{-n^{\prime}}\) with \(k := n-n^{\prime} \geq 0\). Writing \(i = q2^k + s\) with \(0 \leq s < 2^k\), in each coordinate

\begin{equation*} \frac{q}{2^{n^{\prime}}} \leq \frac{i}{2^n}, \qquad \frac{i+1}{2^n} = \frac{q2^k+s+1}{2^n} \leq \frac{q+1}{2^{n^{\prime}}}, \end{equation*}

so \(Q\) lies in a dyadic cube of sidelength \(2^{-n^{\prime}}\); that cube meets \(R\), and dyadic cubes of equal sidelength are equal or disjoint, so it is \(R\).

Strict estimate. Fix \(\mu > 0\) and set \(\mathcal{Q}_\mu := \{Q \in \mathcal{D} : \frac{1}{|Q|}\int_Q|f| > \mu\}\) and \(\Omega_\mu := \bigcup_{Q \in \mathcal{Q}_\mu} Q\), measurable as a countable union of boxes. Every \(Q \in \mathcal{Q}_\mu\) obeys \(|Q| < \frac1\mu\|f\|_{L^1} < \infty\), while by nesting the dyadic cubes containing a given \(Q\) form a chain \(Q = Q^{(0)} \subset Q^{(1)} \subset \cdots\) with \(|Q^{(k)}| = 2^{dk}|Q| \to \infty\); so only finitely many of them lie in \(\mathcal{Q}_\mu\) and \(Q\) is contained in a maximal element of \(\mathcal{Q}_\mu\). Hence \(\Omega_\mu\) is the union of the set \(\mathcal{M}_\mu\) of maximal elements, which are pairwise disjoint by nesting and maximality, and countable additivity of measure and of the integral gives

\begin{equation*} m(\Omega_\mu) = \sum_{Q \in \mathcal{M}_\mu} |Q| \leq \frac1\mu \sum_{Q \in \mathcal{M}_\mu} \int_Q |f| = \frac1\mu \int_{\Omega_\mu}|f| \leq \frac1\mu \|f\|_{L^1}. \end{equation*}

Non-strict form. For \(\lambda > 0\), \(M_{\mathcal{D}}f(x) \geq \lambda\) holds exactly when \(x \in \Omega_\mu\) for every \(0 < \mu < \lambda\), so

\begin{equation*} E_\lambda := \{M_{\mathcal{D}}f \geq \lambda\} = \bigcap_{n \geq 2} \Omega_{\lambda(1-1/n)} \end{equation*}

is measurable with \(m(E_\lambda) \leq \frac{1}{\lambda(1-1/n)}\|f\|_{L^1}\) for every \(n\); letting \(n \to \infty\) gives the stated inequality, with constant \(1\).

Prune the family until no member is covered by the others; a point can then not lie in three of them.

Refinement. Repeatedly delete from the current list a member \(I\) with \(I \subseteq \bigcup\{J : J \text{ in the list},\ J \neq I\}\), members being compared as list entries so that a repeated interval counts twice. Deletion preserves the union, and each step shortens the list, so after at most \(n\) steps we reach a subfamily \(I^{\prime}_1,\dots,I^{\prime}_m\) with \(\bigcup_j I^{\prime}_j = \bigcup_i I_i\), which is (i), and with

\begin{equation*} (\ast)\qquad I^{\prime}_j \not\subseteq \bigcup_{k \neq j} I^{\prime}_k \quad \text{for every } j. \end{equation*}

In particular every \(I^{\prime}_j\) is non-empty, and the \(I^{\prime}_j\) are pairwise distinct as sets.

No point lies in three. Suppose \(x \in I^{\prime}_{j_1} \cap I^{\prime}_{j_2} \cap I^{\prime}_{j_3}\) with \(j_1,j_2,j_3\) distinct, and write \(I^{\prime}_{j_l} = (a_l,b_l)\), so \(a_l < x < b_l\) for \(l=1,2,3\) (endpoints possibly infinite). Choose \(u\) with \(a_u = \min_l a_l\) and \(v\) with \(b_v = \max_l b_l\).

• (i) \(u = v\): then \(a_u \leq a_l\) and \(b_l \leq b_u\) give \(I^{\prime}_{j_l} \subseteq I^{\prime}_{j_u}\) for the two \(l \neq u\), contradicting \((\ast)\).
• (ii) \(u \neq v\): both \((a_u,b_u)\) and \((a_v,b_v)\) contain \(x\), so their union is the interval \((a_u,b_v)\), and the remaining index \(r\) obeys \(a_u \leq a_r\), \(b_r \leq b_v\), whence \(I^{\prime}_{j_r} \subseteq (a_u,b_v) \subseteq \bigcup_{k \neq j_r} I^{\prime}_k\), again contradicting \((\ast)\).

So every \(x \in \mathbb{R}\) lies in at most two of the \(I^{\prime}_j\), which is (ii).

Run the maximal-function argument with Exercise 1.6.21 in place of Lemma 1.6.22: bounded overlap costs the factor \(2\) for any measure, whereas the Vitali-type lemma would need a doubling bound \(\mu(3I) \leq C\mu(I)\), which a general Borel \(\mu\) lacks – for \(d\mu = e^{-y^2}dy\) and \(I = (a,a+1)\) one has \(\mu(3I)/\mu(I) \to \infty\) as \(a \to +\infty\).

Write \(A_I := \frac{1}{\mu(I)}\int_I |f|\,d\mu\), well defined since a non-empty open interval has positive length and hence \(0 < \mu(I) < \infty\), and put

\begin{equation*} U_\nu := \bigcup\Big\{I \text{ open interval} : \int_I |f|\,d\mu > \nu\,\mu(I)\Big\}, \end{equation*}

an open set with \(x \in U_\nu\) exactly when \(\sup_{I \ni x} A_I > \nu\). Hence

\begin{equation*} E_\lambda := \Big\{x : \sup_{I \ni x} A_I \geq \lambda\Big\} = \bigcap_{n \geq 1} U_{\lambda - 1/n} \end{equation*}

is Borel, so \(\mu(E_\lambda)\) makes sense.

Estimate on \(U_\nu\). Let \(\nu > 0\); by inner regularity it suffices to bound \(\mu(K)\) for compact \(K \subseteq U_\nu\). Each \(x \in K\) lies in an open interval \(I_x\) with \(\int_{I_x}|f|\,d\mu > \nu\,\mu(I_x)\), and Heine-Borel extracts finitely many \(I_1,\dots,I_n\) covering \(K\); Exercise 1.6.21 then yields a subfamily \(I^{\prime}_1,\dots,I^{\prime}_m\) with the same union and \(\sum_j 1_{I^{\prime}_j} \leq 2\) pointwise. Finite subadditivity, the defining property of each \(I^{\prime}_j\) and linearity of the integral give

\begin{align*} \mu(K) \leq \sum_{j=1}^m \mu(I^{\prime}_j) &< \frac1\nu \sum_{j=1}^m \int_{I^{\prime}_j}|f(y)|\,d\mu(y) \\ &= \frac1\nu \int_{\mathbb{R}} \Big(\sum_{j=1}^m 1_{I^{\prime}_j}(y)\Big) |f(y)|\,d\mu(y) \leq \frac2\nu \int_{\mathbb{R}}|f(y)|\,d\mu(y), \end{align*}

so \(\mu(U_\nu) \leq \frac2\nu\int_{\mathbb{R}}|f|\,d\mu\).

Conclusion. For \(\lambda > 0\) and \(n > 1/\lambda\), \(E_\lambda \subseteq U_{\lambda-1/n}\) gives \(\mu(E_\lambda) \leq \frac{2}{\lambda-1/n}\int_{\mathbb{R}}|f|\,d\mu\); letting \(n \to \infty\) gives the stated inequality.

Call \([c,d] \subseteq [a,b]\) with \(c<d\) good if it admits a tagged partition of the stated kind. Concatenating tagged partitions of \([c,e]\) and \([e,d]\) (listing \(e\) once, keeping the tags) shows that \([c,d]\) is good whenever both halves are; contrapositively, if \([c,d]\) is not good then neither is at least one of \([c,\tfrac{c+d}{2}]\), \([\tfrac{c+d}{2},d]\).

Suppose \([a,b]\) is not good. Bisecting repeatedly by that principle produces nested non-good intervals

\begin{equation*} [a,b] = [c_0,d_0] \supset [c_1,d_1] \supset \cdots, \qquad d_i - c_i = (b-a)2^{-i}, \end{equation*}

so by the nested interval property there is \(x^* \in \bigcap_i [c_i,d_i] \subseteq [a,b]\), and \(\delta(x^*)>0\). Choose \(i\) with \((b-a)2^{-i} \leq \delta(x^*)\). Then the one-interval partition \(t_0 := c_i < t_1 := d_i\) with tag \(t^*_1 := x^* \in [t_0,t_1]\) obeys

\begin{equation*} t_1 - t_0 = (b-a)2^{-i} \leq \delta(x^*) = \delta(t^*_1), \end{equation*}

so \([c_i,d_i]\) is good, contradicting its construction. Hence \([a,b]\) is good.

Both assertions are the Lebesgue differentiation theorem applied to \(f := 1_E\).

This \(f\) is locally integrable, since \(\int_{B(0,r)} 1_E \leq m(B(0,r)) < \infty\) for every \(r>0\), so Theorem 1.6.19 in the locally integrable form of Exercise 1.6.14(ii) gives a null set \(N\) such that for \(x \notin N\)

\begin{equation*} \frac{m(E \cap B(x,r))}{m(B(x,r))} = \frac{1}{m(B(x,r))}\int_{B(x,r)} 1_E(y)\,dy \longrightarrow 1_E(x) \end{equation*}

as \(r \to 0\), the denominators \(m(B(x,r)) = \omega_d r^d\) being positive. For \(x \in E\setminus N\) the limit is \(1\), so \(x\) is a point of density for \(E\); for \(x \in E^c\setminus N\) the limit exists and equals \(0 \neq 1\), so \(x\) is not.

We may assume \(0 < \varepsilon < 1\), the claim for a smaller \(\varepsilon\) being stronger.

(i) By Exercise 1.6.24 almost every point of \(E\) is a point of density, and \(m(E)>0\), so \(E\) contains such a point \(x\); for \(x \in E\) one has \(|1_E(y)-1_E(x)| = 1_{E^c}(y)\), so the Lebesgue average of \(1_E\) at \(x\) is \(1 - \frac{m(E\cap B(x,r))}{m(B(x,r))}\) and \(x\) is a Lebesgue point of \(1_E\). Take the inscribed cubes \(R_h := (-h/\sqrt d,\ h/\sqrt d)^d\): for \(y \in R_h\), \(|y|^2 < d(h/\sqrt d)^2 = h^2\), so \(R_h \subseteq B(0,h)\), and

\begin{equation*} m(R_h) = \Big(\frac{2h}{\sqrt d}\Big)^d = \frac{(2/\sqrt d)^d}{\omega_d}\, m(B(0,h)), \end{equation*}

so Exercise 1.6.15 applies with \(c := (2/\sqrt d)^d/\omega_d > 0\), independent of \(h\), and gives

\begin{equation*} \frac{m(E \cap (x+R_h))}{m(x+R_h)} = \frac{1}{m(R_h)}\int_{x+R_h} 1_E(y)\,dy \longrightarrow 1_E(x) = 1 \end{equation*}

as \(h \to 0\), using \(m(x+R_h) = m(R_h)\). Choose \(h\) with this ratio above \(1-\varepsilon\) and take \(Q := x + R_h\), a cube of sidelength \(2h/\sqrt d > 0\).

(ii) First, \(m(E \cap B(0,n)) > 0\) for some \(n\) by upward monotone convergence, and replacing \(E\) by \(E \cap B(0,n)\) only decreases \(m(E \cap Q)\); so assume \(0 < m(E) < \infty\). Pick \(\eta > 0\) with \(m(E) + \eta < \frac{m(E)}{1-\varepsilon}\), possible as \(\varepsilon \in (0,1)\), and by outer regularity an open \(U \supseteq E\) with \(m(U \setminus E) \leq \eta\), so \(m(U) < \frac{m(E)}{1-\varepsilon} < \infty\). Write \(U = \bigcup_n Q_n\) as an almost disjoint union of closed cubes of positive sidelength (Lemma 1.2.11), so that \(m(U) = \sum_n m(Q_n)\) (Lemma 1.2.9), while \(E = \bigcup_n (E \cap Q_n)\) gives \(m(E) \leq \sum_n m(E \cap Q_n)\). Were \(m(E \cap Q_n) \leq (1-\varepsilon)m(Q_n)\) for every \(n\), then

\begin{equation*} m(E) \leq (1-\varepsilon)\sum_n m(Q_n) = (1-\varepsilon)m(U) < m(E), \end{equation*}

absurd, all quantities being finite. So some \(Q_n\) has \(m(E \cap Q_n) > (1-\varepsilon)m(Q_n)\), and no differentiation theorem was used.

(iii) Apply (i) with \(\varepsilon := \tfrac14\): there is a cube \(Q\) of sidelength \(\ell > 0\) with \(F := E \cap Q\) of measure \(m(F) > \tfrac34\ell^d\). Put \(\eta := 1-2^{-1/d} \in (0,1)\), so \((1-\eta)^d = \tfrac12\), and let \(|v_i| < \eta\ell\) for all \(i\). If \(F \cap (F+v) = \emptyset\), then disjointness and translation invariance give

\begin{equation*} 2m(F) = m\big(F \cup (F+v)\big) \leq m\big(Q \cup (Q+v)\big) = 2m(Q) - m\big(Q\cap(Q+v)\big), \end{equation*}

while \(Q \cap (Q+v)\) is a box of sidelengths \(\ell - |v_i| > 0\), so

\begin{equation*} m\big(Q\cap(Q+v)\big) = \prod_{i=1}^d (\ell-|v_i|) \geq (1-\eta)^d\ell^d = \tfrac12\ell^d, \end{equation*}

whence \(\tfrac32\ell^d < 2m(F) \leq 2\ell^d - \tfrac12\ell^d = \tfrac32\ell^d\), absurd. So there are \(y,z \in F \subseteq E\) with \(y = z+v\), i.e. \(v \in E-E\), and the open box \(\{v : |v_i| < \eta\ell\ \forall i\}\) lies in \(E-E\).

(i). Take \(K := [0,1]\setminus U\), where \((q_n)\) enumerates \(\mathbb{Q}\cap[0,1]\) and

\begin{equation*} U := \bigcup_{n=1}^\infty \big(q_n - 2^{-n-3},\ q_n + 2^{-n-3}\big). \end{equation*}

Then \(K\) is closed and bounded, hence compact, and countable subadditivity gives \(m(U) \leq \sum_n 2^{-n-2} = \tfrac14\), whence

\begin{equation*} m(K) \geq m([0,1]) - m(U) \geq 1 - \tfrac14 = \tfrac34 > 0 . \end{equation*}

Let \(I\) be an interval of positive length. (i) If \(|I| = \infty\) then \(m(K\cap I) \leq 1 < \infty = |I|\). (ii) If \(I \cap (0,1) = \emptyset\) then \(K \cap I \subset \{0,1\}\), so \(m(K\cap I) = 0 < |I|\). (iii) Otherwise \(I^\circ \cap (0,1)\) is a nonempty open set (if \(x \in I\cap(0,1)\) is an endpoint of \(I\), nearby interior points still lie in \((0,1)\)), so it contains some \(q_n\) together with a neighbourhood of \(q_n\), forcing \(m(I \cap U) > 0\); since \(K \cap I \subset I \setminus U\),

\begin{equation*} m(K \cap I) \leq |I| - m(I \cap U) < |I| . \end{equation*}

(ii). Take \(E := \bigcup_{n\geq1} A_n\) for the sets \(A_n\) built below. For \(\alpha < \beta\) let \(K_{[\alpha,\beta]}\) be the image of \(K\) under \(t \mapsto \alpha + (\beta-\alpha)t\); this is compact, nowhere dense (\(K\) is closed and contains no rational, so \(K^\circ = \emptyset\)), and by the scaling law for Lebesgue measure

\begin{equation*} m\big(K_{[\alpha,\beta]}\big) = (\beta-\alpha)\, m(K) \in (0,\ \beta-\alpha] . \end{equation*}

Let \(J_1, J_2, \dots\) enumerate the open intervals with rational endpoints. Recursively, with \(S := \bigcup_{i<n}(A_i \cup B_i)\) a finite union of closed sets of empty interior, hence closed with empty interior, the set \(J_n \setminus S\) is open and nonempty, so it contains two disjoint closed intervals \([\alpha_1,\beta_1], [\alpha_2,\beta_2]\) with \(\beta_1 - \alpha_1 \leq 2^{-n}\); put

\begin{equation*} A_n := K_{[\alpha_1,\beta_1]}, \qquad B_n := K_{[\alpha_2,\beta_2]} \subset J_n . \end{equation*}

Then \(\{A_n\}\cup\{B_n\}\) is pairwise disjoint, so \(E \cap B_n = \emptyset\) for every \(n\), while \(m(A_n) > 0\) and \(m(E) \leq \sum_n 2^{-n} = 1\).

Given \(I\) of positive length, choose rationals \(p<q\) with \(J_n := (p,q) \subset I^\circ\), so \(A_n, B_n \subset I\). Then \(m(E \cap I) \geq m(A_n) > 0\); and \(m(E\cap I) \leq 1 < |I|\) if \(|I| = \infty\), while if \(|I| < \infty\) then \(B_n \subset I \setminus E\) gives

\begin{equation*} m(E \cap I) = |I| - m(I \setminus E) \leq |I| - m(B_n) < |I| . \qquad \square \end{equation*}

Write \(\omega_d := m(B(0,1))\), so \(m(B(x,r)) = \omega_d r^d\), and put

\begin{equation*} A_n := \{x : 2^{n-1} < |x| \leq 2^n\}, \qquad m(A_n) = \omega_d(1-2^{-d})\,2^{nd} \qquad (n \in \mathbb{Z}), \end{equation*}

a disjoint decomposition of \(\mathbb{R}^d\setminus\{0\}\). Each \(P_t\) is itself a good kernel: it is radial with non-increasing profile \(\widetilde{P_t}( r) = t^{-d}\tilde P(r/t)\), and \(\int P_t(x)\,dx = \int P(u)\,du = 1\) by the scaling law for Lebesgue measure.

(i). For the heat kernel take \(P(x) := (4\pi)^{-d/2}e^{-|x|^2/4}\), so that \(t^{-d}P(x/t) = (4\pi t^2)^{-d/2}e^{-|x|^2/4t^2} = P_t(x)\); it is continuous, positive, radial with \(\tilde P( r) = (4\pi)^{-d/2}e^{-r^2/4}\) non-increasing, and substituting \(x = 2u\),

\begin{equation*} \int_{\mathbb{R}^d}(4\pi)^{-d/2}e^{-|x|^2/4}dx = (4\pi)^{-d/2}2^d\int_{\mathbb{R}^d}e^{-|u|^2}du = (4\pi)^{-d/2}2^d\pi^{d/2} = 1, = (4\pi)^{-d/2}2^d\pi^{d/2} = 1, \end{equation*}

using \(\int e^{-|u|^2}du = \pi^{d/2}\) (Tonelli from the one-dimensional case). For the Poisson kernel take \(P(x) := c_d(1+|x|^2)^{-(d+1)/2}\), for which

\begin{equation*} t^{-d}P(x/t) = \frac{c_d\,t^{-d}t^{d+1}}{(t^2+|x|^2)^{(d+1)/2}} = c_d\frac{t}{(t^2+|x|^2)^{(d+1)/2}}, \end{equation*}

again continuous, positive and radially non-increasing; the choice \(c_d := 1/M\) with \(M := \int(1+|x|^2)^{-(d+1)/2}dx\) normalises the mass, and \(M \in (0,\infty)\) since the integrand is \(\leq 1\) on \(B(0,1)\) and \(\leq |x|^{-(d+1)} \leq 2^{-(n-1)(d+1)}\) on \(A_n\), \(n \geq 1\):

\begin{equation*} M \leq \omega_d + \sum_{n\geq1}2^{-(n-1)(d+1)}\omega_d 2^{nd} = \omega_d + \omega_d 2^{d+1}\sum_{n\geq1}2^{-n} < \infty . \end{equation*}

(ii). Monotonicity of \(\tilde P\) gives \(\tilde P(2^n) \leq P \leq \tilde P(2^{n-1})\) on \(A_n\), and \(1 = \int P = \sum_n \int_{A_n}P\) since \(\{0\}\) is null. The lower bound yields

\begin{equation*} 1 \geq \sum_n \tilde P(2^n)m(A_n) = \omega_d(1-2^{-d})\sum_n 2^{dn}\tilde P(2^n), \end{equation*}

and the upper bound, reindexed by \(m := n-1\), yields

\begin{equation*} 1 \leq \sum_n \tilde P(2^{n-1})m(A_n) = \omega_d(2^d-1)\sum_m 2^{md}\tilde P(2^m). \end{equation*}

So the stated bounds hold with \(c_d := \tfrac{1}{2\omega_d(2^d-1)}\) and \(C_d := \tfrac{2^d}{\omega_d(2^d-1)}\).

(iii). Write \(Mf(x) := \sup_{r>0} m(B(x,r))^{-1}\int_{B(x,r)}|f|\) for the Hardy-Littlewood maximal function. For any good kernel \(Q\), decomposing over the annuli \(2^{n-1} < |x-y| \leq 2^n\), bounding \(Q(x-y) \leq \tilde Q(2^{n-1})\) there and enlarging to \(B(x,2^n)\) (spheres are null),

\begin{equation*} \begin{aligned} \int_{\mathbb{R}^d}|f(y)|Q(x-y)\,dy &\leq \sum_n \tilde Q(2^{n-1})\int_{B(x,2^n)}|f(y)|\,dy \\ &\leq \sum_n \tilde Q(2^{n-1})\,\omega_d 2^{nd}\,Mf(x) \\ &= \omega_d 2^d\,Mf(x)\sum_n 2^{(n-1)d}\tilde Q(2^{n-1}) \ \leq\ \omega_d 2^d C_d\, Mf(x), \end{aligned} \end{equation*}

by (ii). Taking \(Q := P_t\) (a good kernel, with \(C_d\) depending only on \(d\)) and \(C^{\prime}_d := \omega_d 2^d C_d = 4^d/(2^d-1)\) gives the claim.

(iv). First, \(s^d\tilde P(s) \to 0\) as \(s \to +\infty\): the series in (ii) converges, so \(2^{nd}\tilde P(2^n) \to 0\), and for \(2^{n-1} < s \leq 2^n\),

\begin{equation*} s^d\tilde P(s) \leq 2^{nd}\tilde P(2^{n-1}) = 2^d\cdot 2^{(n-1)d}\tilde P(2^{n-1}) \to 0 . \end{equation*}

Now let \(x\) be a Lebesgue point. Since \(\int P_t = 1\),

\begin{equation*} |f*P_t(x) - f(x)| \leq \int_{\mathbb{R}^d}|f(y)-f(x)|\,P_t(x-y)\,dy . \end{equation*}

Fix \(\varepsilon>0\) and choose \(r_0>0\) with \(m(B(x,r))^{-1}\int_{B(x,r)}|f-f(x)| \leq \varepsilon\) for all \(0<r\leq r_0\); let \(r_1 := 2^{n_0} \leq r_0\) with \(n_0\) largest such.

(i) Near part: the chain of (iii) applied to \(|f-f(x)|\), summed only over \(n \leq n_0\) (these annuli cover \(0<|x-y|\leq r_1\), and each average is \(\leq\varepsilon\)), gives

\begin{equation*} \int_{|x-y|\leq r_1}|f(y)-f(x)|P_t(x-y)\,dy \leq \omega_d 2^d C_d\,\varepsilon = C^{\prime}_d\,\varepsilon, \end{equation*}

uniformly in \(t\).

(ii) Far part: \(P_t(x-y) \leq \widetilde{P_t}(r_1)\) there, so

\begin{equation*} \int_{|x-y|>r_1}|f(y)-f(x)|P_t(x-y)\,dy \leq \widetilde{P_t}(r_1)\int|f| + |f(x)|\int_{|z|>r_1}P_t(z)\,dz . \end{equation*}

With \(s := r_1/t \to \infty\) as \(t \to 0\) we have \(\widetilde{P_t}(r_1) = r_1^{-d}s^d\tilde P(s) \to 0\) and \(\int|f|<\infty\); and \(\int_{|z|>r_1}P_t(z)\,dz = \int_{|u|>s}P(u)\,du \to 0\) by downward monotone convergence for the finite measure \(A \mapsto \int_A P\). So the far part tends to \(0\).

Hence \(\limsup_{t\to0}|f*P_t(x)-f(x)| \leq C^{\prime}_d\varepsilon\) for every \(\varepsilon>0\), i.e. \(f*P_t(x) \to f(x)\). By Theorem 1.6.19 almost every \(x\) is a Lebesgue point of \(f\), so \(f*P_t \to f\) pointwise almost everywhere. \(\square\)

(i). \(\sup_x|4^{-n}\sin(8^n\pi x)| \leq 4^{-n}\) with \(\sum_n 4^{-n} = \tfrac13 < \infty\), so the series converges absolutely at every \(x\), \(|F| \leq \tfrac13\), and by the Weierstrass \(M\)-test the partial sums converge uniformly; a uniform limit of continuous functions is continuous, so \(F\) is continuous.

(ii). The printed inequality is false at \(n=1\), and we prove in its place a second-difference version, which is what (iii) needs. Indeed \(8^m(j/8^n) \in \mathbb{Z}\) for \(m \geq n\), so those terms vanish and

\begin{equation*} F\Big(\frac{j}{8^n}\Big) = \sum_{m=1}^{n-1}4^{-m}\sin\Big(\frac{\pi j}{8^{n-m}}\Big); \end{equation*}

for \(n=1\) the sum is empty, so \(F \equiv 0\) on \(\tfrac18\mathbb{Z}\) and the asserted bound reads \(0 \geq c/4\).

Lemma. For \(x \in \mathbb{R}\), \(n \geq 1\) and \(\theta \in \{1,2\}\) put \(h := \theta 8^{-n}\) and

\begin{equation*} D_\theta(x,n) := F(x+h) - 2F\big(x+\tfrac h2\big) + F(x). \end{equation*}

Choosing \(\theta := 1\) if \(|\cos(8^n\pi x)| \geq 2^{-1/2}\) and \(\theta := 2\) otherwise, \(|D_\theta(x,n)| \geq \tfrac65 4^{-n}\).

Proof. Write \(\alpha_m := 8^m\pi x\) and \(\phi_m := 8^m\pi h = \theta 8^{m-n}\pi\). Absolute convergence lets us combine the three series termwise, and \(\sin(u+\phi) - 2\sin(u+\tfrac\phi2) + \sin u = 2\sin(u+\tfrac\phi2)(\cos\tfrac\phi2 - 1)\) gives

\begin{equation*} D_\theta(x,n) = \sum_{m=1}^\infty 4^{-m}\cdot 2\sin\big(\alpha_m + \tfrac{\phi_m}{2}\big)\Big(\cos\tfrac{\phi_m}{2} - 1\Big). \Big(\cos\tfrac{\phi_m}{2} - 1\Big). \end{equation*}

(i) For \(m>n\), \(\tfrac{\phi_m}{2} = \tfrac\theta2 8^{m-n}\pi\) is an even multiple of \(\pi\) (for \(\theta=1\), \(\tfrac{8^{m-n}}2 = 4\cdot8^{m-n-1}\) is even), so \(\cos\tfrac{\phi_m}2 = 1\) and the term vanishes. (ii) For \(m=n\): if \(\theta=1\) then \(\tfrac{\phi_n}2 = \tfrac\pi2\) and the term is \(-2\cdot4^{-n}\cos(8^n\pi x)\); if \(\theta=2\) then \(\tfrac{\phi_n}2 = \pi\) and the term is \(4\cdot4^{-n}\sin(8^n\pi x)\). (iii) For \(m<n\), using \(|\sin|\leq1\) and \(|\cos u - 1| \leq \tfrac{u^2}2\),

\begin{equation*} \sum_{m<n}4^{-m}\Big(\tfrac\theta2 8^{m-n}\pi\Big)^2 = \frac{\theta^2\pi^2}{4}8^{-2n}\sum_{m<n}16^m \leq \frac{\theta^2\pi^2}{60}4^{-n}, \end{equation*}

since \(\sum_{m<n}16^m < 16^n/15\) and \(16^n 8^{-2n} = 4^{-n}\). Hence by the triangle inequality

\begin{equation*} |D_1(x,n)| \geq \Big(\tfrac{2}{\sqrt2} - \tfrac{\pi^2}{60}\Big)4^{-n} > 1.24\cdot4^{-n}, \end{equation*}

in the case \(\theta=1\), and in the case \(\theta=2\) (where \(|\sin(8^n\pi x)| > 2^{-1/2}\)),

\begin{equation*} |D_2(x,n)| \geq \Big(\tfrac{4}{\sqrt2} - \tfrac{\pi^2}{15}\Big)4^{-n} > 2.17\cdot4^{-n}. \qquad \square \end{equation*}

(iii). If \(F\) were differentiable at \(x\), then with \(\rho(h) := F(x+h)-F(x)-hF^{\prime}(x) = o(h)\),

\begin{equation*} F(x+h) - 2F\big(x+\tfrac h2\big) + F(x) = \rho(h) - 2\rho\big(\tfrac h2\big) = o(h) \qquad (h \to 0). \end{equation*}

But the Lemma supplies \(h_n = \theta_n 8^{-n} \in (0, 2\cdot8^{-n}]\) with \(h_n \to 0\) and

\begin{equation*} \frac{\big|F(x+h_n) - 2F(x+\tfrac{h_n}2) + F(x)\big|}{h_n} \geq \frac{\tfrac65 4^{-n}}{2\cdot 8^{-n}} = \tfrac35\cdot 2^n \to \infty, \end{equation*}

a contradiction. So \(F\) is nowhere differentiable. \(\square\)

Termwise differentiation proves nothing because differentiation does not commute with infinite summation absent uniform convergence of the differentiated series: \(\sum_{n\geq1}\big(\tfrac{\sin(n!x)}{n} - \tfrac{\sin((n+1)!x)}{n+1}\big)\) telescopes to \(\sin x\), which is everywhere differentiable, yet its termwise derivative series has partial sums \(1 - N! \to -\infty\) at \(x=0\).

Every monotone \(F\) is in fact Borel measurable, because each superlevel set \(\{F>\lambda\}\) is an interval. We may assume \(F\) non-decreasing, since otherwise \(-F\) is and \(F = -(-F)\) is then measurable by Exercise 1.3.8(v). By Exercise 1.3.7(iv) it is enough that \(F^{-1}(U)\) be measurable for every open \(U \subseteq \mathbb{R}\) (\(F\) is real-valued, and every open \(U \subseteq \mathbb{R}\) is \(V \cap \mathbb{R}\) for the open set \(V := \{z : \mathrm{Re}(z) \in U, |\mathrm{Im}(z)|<1\}\)); by Exercise 1.6.10 such a \(U\) is a countable union of open intervals, so it suffices to treat

\begin{equation*} F^{-1}((\alpha,\beta)) = S_\alpha \setminus S_{\beta^-}, \qquad S_\lambda := \{F > \lambda\}, \quad S_{\beta^-} := \bigcap_{k\geq1} S_{\beta - 1/k}, \end{equation*}

with \(S_{-\infty} := \mathbb{R}\) and \(S_{+\infty^-} := \emptyset\). Fix \(\lambda \in \mathbb{R}\). Then \(S_\lambda\) is an up-set: \(x \in S_\lambda\) and \(y > x\) give \(F(y) \geq F(x) > \lambda\). So if \(S_\lambda \neq \emptyset\), put \(c := \inf S_\lambda\); if \(c = -\infty\) the up-set property forces \(S_\lambda = \mathbb{R}\), and if \(c \in \mathbb{R}\) then every \(y>c\) exceeds some element of \(S_\lambda\) while no \(y<c\) lies in it, so

\begin{equation*} (c,+\infty) \subseteq S_\lambda \subseteq [c,+\infty), \end{equation*}

i.e. \(S_\lambda\) is \((c,\infty)\) or \([c,\infty)\) according as \(F( c) \leq \lambda\) or \(F( c) > \lambda\). In every case \(S_\lambda\) is an interval, hence Borel. \(\square\)

The point is that for monotone \(F\) the suprema and infima defining the Dini derivatives may be taken over rational \(h\) alone. We may assume \(F\) non-decreasing: if \(F\) is non-increasing then \(-F\) is non-decreasing, \(D^+F = -D_+(-F)\) and \(D_+F = -D^+(-F)\) (likewise on the left), and \(-G\) is measurable whenever the unsigned \(G\) is, since \(\{-G > \lambda\} = \{G < -\lambda\}\) is measurable by Lemma 1.3.9(vii). Moreover \(\tilde F(x) := -F(-x)\) is again non-decreasing and

\begin{equation*} D^-F(x) = D^+\tilde F(-x), \qquad D_-F(x) = D_+\tilde F(-x), \end{equation*}

while \(y \mapsto -y\) preserves Lebesgue measurability, so only \(D^+F\) and \(D_+F\) need treating. For \(h>0\) put \(\Delta(x,h) := (F(x+h)-F(x))/h \in [0,+\infty)\) and, for \(\delta>0\),

\begin{equation*} S_\delta(x) := \sup_{0<h<\delta}\Delta(x,h), \qquad I_\delta(x) := \inf_{0<h<\delta}\Delta(x,h), \end{equation*}

so that, by monotonicity in \(\delta\), \(D^+F = \inf_{n\geq1}S_{1/n}\) and \(D_+F = \sup_{n\geq1}I_{1/n}\).

Claim: \(S_\delta = S_\delta^{\mathbb{Q}}\) and \(I_\delta = I_\delta^{\mathbb{Q}}\), the superscript meaning that \(h\) is restricted to \(\mathbb{Q}\cap(0,\delta)\). One inequality each is trivial. For the others fix \(0<h<\delta\) and \(\varepsilon>0\). (i) Choose a rational \(h^{\prime} \in (h, \min(\delta,(1+\varepsilon)h))\); then \(F(x+h^{\prime}) \geq F(x+h) \geq F(x)\), so

\begin{equation*} \Delta(x,h^{\prime}) \geq \frac{F(x+h)-F(x)}{(1+\varepsilon)h} = \frac{\Delta(x,h)}{1+\varepsilon}, \end{equation*}

whence \(S^{\mathbb{Q}}_\delta \geq S_\delta/(1+\varepsilon)\) and, as \(\varepsilon \to 0\), \(S^{\mathbb{Q}}_\delta \geq S_\delta\) (also when \(S_\delta = +\infty\)). (ii) Choose a rational \(h^{\prime} \in (h/(1+\varepsilon), h)\); then \(F(x+h^{\prime}) \leq F(x+h)\), so

\begin{equation*} \Delta(x,h^{\prime}) \leq \frac{F(x+h)-F(x)}{h^{\prime}} \leq (1+\varepsilon)\Delta(x,h), \end{equation*}

whence \(I^{\mathbb{Q}}_\delta \leq (1+\varepsilon)I_\delta\) for all \(\varepsilon>0\), i.e. \(I^{\mathbb{Q}}_\delta \leq I_\delta\).

By Exercise 1.6.29 \(F\) is measurable, hence so is each translate \(x \mapsto F(x+h)\) and, by Exercise 1.3.8(vi), each \(\Delta(\cdot,h)\), an unsigned measurable function. By the Claim, \(S_{1/n}\) and \(I_{1/n}\) are countable suprema and infima of such functions, hence unsigned measurable by Exercise 1.3.3(iii); one more application gives measurability of \(D^+F = \inf_n S_{1/n}\) and \(D_+F = \sup_n I_{1/n}\), and with it of all four Dini derivatives. \(\square\)

The Vitali-type covering lemma gives the bound with \(C=3\): for every monotone non-decreasing \(F : [a,b] \to \mathbb{R}\), continuous or not, and every \(\lambda>0\),

\begin{equation*} m\big(\{x \in [a,b] : D^+F(x) \geq \lambda\}\big) \leq 3\,\frac{F(b)-F(a)}{\lambda}, \end{equation*}

and likewise for the other three Dini derivatives. Extend \(F\) by \(F(a)\) on \((-\infty,a)\) and \(F(b)\) on \((b,\infty)\): it remains non-decreasing, the Dini derivatives on \((a,b)\) are unchanged, \(\{a,b\}\) is null, and \(D^+F\) is measurable by Exercise 1.6.30. It suffices to prove the strict form

\begin{equation*} m\big(\{x \in (a,b) : D^+F(x) > \lambda\}\big) \leq 3\,\frac{F(b)-F(a)}{\lambda}, \tag{$\ast$} \end{equation*}

since applying \((\ast)\) with \(\lambda^{\prime}<\lambda\) and letting \(\lambda^{\prime}\uparrow\lambda\) recovers the non-strict bound.

By inner regularity it suffices to bound \(m(K)\) for compact \(K\) inside that set. For \(x \in K\), \(D^+F(x)>\lambda\) supplies \(0<h_x<b-x\) with \(F(x+h_x)-F(x) > \lambda h_x\); strictness lets us pick \(\eta_x \in (0, x-a+1)\) so small that the open interval \(I^{\prime}_x := (x-\eta_x, x+h_x) =: (\alpha_x,\beta_x)\) satisfies, using \(F(x-\eta_x) \leq F(x)\),

\begin{equation*} F(\beta_x) - F(\alpha_x) \geq F(x+h_x)-F(x) > \lambda(h_x+\eta_x) = \lambda\,|I^{\prime}_x|. \tag{$\dagger$} \end{equation*}

The \(I^{\prime}_x\) are open and cover the compact \(K\), so finitely many \(J_1,\dots,J_n\) do; these are finitely many bounded intervals, i.e. balls in \(\mathbb{R}\), so Lemma 1.6.22 yields pairwise disjoint \(J^{\prime}_1,\dots,J^{\prime}_m\) among them with \(\bigcup_i J_i \subseteq \bigcup_k 3J^{\prime}_k\). Writing \(J^{\prime}_k = (\alpha_k,\beta_k)\) and relabelling so that \(\alpha_1 < \beta_1 \leq \cdots \leq \alpha_m < \beta_m\) (possible by disjointness), all endpoints lie in \([a-1,b]\) and monotonicity telescopes the sum:

\begin{equation*} \begin{aligned} m(K) &\leq \sum_{k=1}^m 3|J^{\prime}_k| < \frac{3}{\lambda}\sum_{k=1}^m \big(F(\beta_k)-F(\alpha_k)\big) \\ &\leq \frac{3}{\lambda}\big(F(b)-F(a-1)\big) = \frac{3(F(b)-F(a))}{\lambda}. \end{aligned} \leq \frac{3}{\lambda}\big(F(b)-F(a-1)\big) = \frac{3(F(b)-F(a))}{\lambda}. \end{equation*}

Since \(D_+F \leq D^+F\) pointwise, the same bound holds for \(D_+F\); and for the left-hand derivatives apply the above to \(\tilde F(x) := -F(-x)\) on \([-b,-a]\), which is non-decreasing with the same total increment and has \(D^-F(x) = D^+\tilde F(-x)\), \(D_-F(x) = D_+\tilde F(-x)\) (Exercise 1.6.30), the reflection \(y \mapsto -y\) preserving Lebesgue measure. \(\square\)

Method (2): \(C=2\). Keep the \(I^{\prime}_x\) obeying \((\dagger)\) and the finite subcover \(J_1,\dots,J_n\). The one-dimensional Besicovitch lemma (Exercise 1.6.21) extracts \(J^{\prime}_1,\dots,J^{\prime}_m\) with \(\bigcup_k J^{\prime}_k = \bigcup_i J_i \supseteq K\) and every point in at most two of them. Order \(\alpha_1 \leq \cdots \leq \alpha_m\) and two-colour greedily: the earlier intervals meeting \(J^{\prime}_k = (\alpha_k,\beta_k)\) are those with \(\alpha_j \leq \alpha_k < \beta_j\), and all of them contain \(\alpha_k+\varepsilon\) together with \(J^{\prime}_k\) for small \(\varepsilon>0\), so multiplicity at most two leaves at most one colour forbidden. Each colour class is pairwise disjoint, so telescoping over each class separately gives \(\sum_k (F(\beta_k)-F(\alpha_k)) \leq 2(F(b)-F(a))\), whence by \((\dagger)\)

\begin{equation*} m(K) \leq \sum_{k=1}^m |J^{\prime}_k| < \frac1\lambda\sum_{k=1}^m\big(F(\beta_k)-F(\alpha_k)\big) \leq \frac{2(F(b)-F(a))}{\lambda}. \qquad \square \leq \frac{2(F(b)-F(a))}{\lambda}. \qquad \square \end{equation*}

Both decompositions carry the same jump data, which pins them down. Let \(F = F_c + F_{pp} = G_c + G_{pp}\) with \(F_c,G_c\) continuous non-decreasing and \(F_{pp},G_{pp}\) jump functions, and let \(J = \sum_n c_n J_n\) be any jump function in the sense of Definition 1.6.30, with \(c_n>0\), \(\sum_n c_n<\infty\), and \(J_n\) basic with discontinuity \(x_n\) and fraction \(\theta_n\). We may take the \(x_n\) distinct: a subfamily \(S\) sharing a point \(x_0\) has \(\sum_{n\in S}c_nJ_n = cJ^{\prime}\) with \(c := \sum_{n\in S}c_n\), \(J^{\prime}\) basic at \(x_0\) with fraction \(\theta := c^{-1}\sum_{n\in S}c_n\theta_n \in [0,1]\) (evaluate at \(x<x_0\), \(x=x_0\), \(x>x_0\)).

(i) \(J(-\infty)=0\): given \(\varepsilon>0\) pick \(N\) with \(\sum_{n>N}c_n<\varepsilon\); for \(x < \min_{n\leq N}x_n\) all terms with \(n\leq N\) vanish, so \(0 \leq J(x) < \varepsilon\).

(ii) For \(a<b\), absolute convergence gives \(J(b)-J(a) = \sum_n c_n(J_n(b)-J_n(a))\), and Definition 1.6.30 gives \(J_n(b)-J_n(a) = 1, 1-\theta_n, \theta_n, 0\) according as \(a<x_n<b\), \(x_n=a\), \(x_n=b\), or \(x_n \notin [a,b]\). Hence

\begin{equation*} J(b)-J(a) = \sum_{n\,:\,x_n\in(a,b)}c_n \ +\ (1-\theta_m)c_m\mathbf{1}_{a\in\{x_n\}} \ +\ \theta_k c_k \mathbf{1}_{b\in\{x_n\}}, \tag{1} \end{equation*}

where \(x_m=a\) and \(x_k=b\) when such indices exist.

(iii) In (1) let \(a \uparrow x\): the tail \(\sum_{n:x_n\in[a,x)}c_n\) tends to \(0\) (pick \(N\) with \(\sum_{n>N}c_n<\varepsilon\), then \(a\) close enough to \(x\) that no \(x_n\) with \(n \leq N\) lies in \([a,x)\)), so \(J(x)-J_-(x) = \theta_kc_k\); letting \(b \downarrow x\) likewise gives \(J_+(x)-J(x) = (1-\theta_k)c_k\), where \(x=x_k\), both being \(0\) if \(x \notin \{x_n\}\). Adding,

\begin{equation*} J_+(x)-J_-(x) = c_k, \qquad J(x)-J_-(x) = \theta_k c_k. \tag{2} \end{equation*}

(iv) Continuity of \(F_c\) gives \(F_+-F_- = (F_{pp})_+-(F_{pp})_-\) and \(F-F_- = F_{pp}-(F_{pp})_-\) at every point, so by (2) the discontinuity set of \(F_{pp}\) is exactly the discontinuity set \(A\) of \(F\), with data

\begin{equation*} c_x = F_+(x)-F_-(x), \qquad \theta_x = \frac{F(x)-F_-(x)}{F_+(x)-F_-(x)} \qquad (x \in A), \end{equation*}

depending on \(F\) alone. The identical computation applies to \(G_{pp}\), so by (1) \(F_{pp}-G_{pp}\) is constant, and by (i) that constant is \(0\). Hence \(F_{pp}=G_{pp}\) and \(F_c = F - F_{pp} = G_c\). \(\square\)

The right notion is that of a locally jump function: a function \(F = \sum_n c_n \tilde J_n\) over an at most countable index set, where \(J_n\) is the basic jump function with discontinuity \(x_n\) (the \(x_n\) distinct) and fraction \(\theta_n\), \(\tilde J_n := J_n - \mathbf{1}_{x_n<0}\) is its recentring, \(c_n>0\), and summability is only local:

\begin{equation*} \sum_{n\,:\,|x_n| \leq R} c_n < \infty \qquad \text{for every } R>0 . \end{equation*}

Global summability must be dropped because \(F(x) := \lfloor x\rfloor\) jumps by \(1\) at every integer, and the recentring is what keeps the series convergent to the left of \(0\).

On \([-R,R]\) every term with \(|x_n|>R\) vanishes (if \(x_n>R\) then \(\tilde J_n = J_n = 0\) there; if \(x_n<-R\) then \(\tilde J_n = 1-1 = 0\)), so the series there reduces to \(\sum_{|x_n|\leq R}c_n\tilde J_n\), absolutely and uniformly convergent since \(|\tilde J_n| \leq 1\); hence \(F\) is well defined, and monotone non-decreasing as a pointwise limit of such. Every jump function is a locally jump function up to an additive constant: after the merging normalisation of Exercise 1.6.32, \(\sum_n c_nJ_n = \sum_n c_n\tilde J_n + \sum_{n:x_n<0}c_n\).

Since \(\tilde J_n - J_n\) is constant, the increment formula of Exercise 1.6.32 holds verbatim, all sums being finite by local summability:

\begin{equation*} F(b)-F(a) = \sum_{n\,:\,x_n\in(a,b)}c_n + (1-\theta_m)c_m\mathbf{1}_{a\in\{x_n\}} + \theta_kc_k\mathbf{1}_{b\in\{x_n\}}. \tag{$1^{\prime}$} \end{equation*}

Fixing \(R>|x|+1\) and discarding the terms with \(|x_n|>R\), which vanish near \(x\), leaves a globally summable family, so part (iii) of Exercise 1.6.32 applies and gives

\begin{equation*} F_+(x)-F_-(x) = c_k, \qquad F(x)-F_-(x) = \theta_kc_k \tag{$2^{\prime}$} \end{equation*}

when \(x=x_k\), and \(F_+(x)=F(x)=F_-(x)\) otherwise. The normalisation \(J(-\infty)=0\) is replaced by

\begin{equation*} \lim_{x\to0^-}F(x) = 0, \tag{$3^{\prime}$} \end{equation*}

since for \(x<0\) only terms with \(x \leq x_n<0\) contribute and \(|F(x)| \leq \sum_{x_n\in[x,0)}c_n \to 0\) by the same tail argument.

Theorem. Every monotone non-decreasing \(F : \mathbb{R}\to\mathbb{R}\) is uniquely \(F = F_c + F_{pp}\) with \(F_c\) continuous non-decreasing and \(F_{pp}\) a locally jump function.

Existence. Parts (i),(ii) of Lemma 1.6.31 do not use boundedness, so the discontinuity set \(A\) of \(F\) is at most countable and consists of jump discontinuities; for \(x \in A\) put \(c_x := F_+(x)-F_-(x)>0\) and \(\theta_x := (F(x)-F_-(x))/c_x \in [0,1]\), with \(J_x\) basic at \(x\) with fraction \(\theta_x\). For \(x \in A\cap[-R,R]\) the intervals \((F_-(x),F_+(x))\) have length \(c_x\), are pairwise disjoint by monotonicity, and lie in \([F(-R-1),F(R+1)]\), so

\begin{equation*} \sum_{x \in A\cap[-R,R]}c_x \leq F(R+1)-F(-R-1) < \infty \end{equation*}

and \(F_{pp} := \sum_{x\in A}c_x\tilde J_x\) is a locally jump function. By \((2^{\prime})\), \(F_{pp}\) satisfies at every \(x\) exactly the relations \(F_{pp}(x) = (F_{pp})_-(x)+\theta_xc_x\), \((F_{pp})_+(x) = (F_{pp})_-(x)+c_x\) that \(F\) satisfies by definition of \(c_x,\theta_x\) (with \(c_x := 0\) off \(A\)); subtracting, \((F_c)_-(x) = (F_c)_+(x) = F_c(x)\) for \(F_c := F-F_{pp}\), so \(F_c\) is continuous. And \(F_c\) is non-decreasing, because by \((1^{\prime})\) the increment \(F_{pp}(b)-F_{pp}(a)\) is the total length of the intervals \((F_-(x),F_+(x))\) for \(x \in A\cap(a,b)\) together with \((F(a),F_+(a))\) and \((F_-(b),F(b))\), which are pairwise disjoint and contained in \((F(a),F(b))\), whence \(F_{pp}(b)-F_{pp}(a) \leq F(b)-F(a)\).

Uniqueness. As in part (iv) of Exercise 1.6.32, which used only continuity of the continuous part together with \((2^{\prime})\), any two such decompositions give locally jump functions with the same jump data; so \((1^{\prime})\) makes \(F_{pp}-G_{pp}\) constant and \((3^{\prime})\) makes that constant \(0\). \(\square\)

We may assume \(F\) monotone non-decreasing, since replacing \(F\) by \(-F\) changes neither \(\|F\|_{TV}\) (the summands \(|F(x_i)-F(x_{i-1})|\) are unchanged), nor \(|F(b)-F(a)|\), nor boundedness. For \(a<b\), every admissible sequence \(a \leq x_0 < \cdots < x_n \leq b\) in Definition 1.6.33 has non-negative increments, so the absolute values drop and the sum telescopes:

\begin{equation*} \sum_{i=1}^n |F(x_i)-F(x_{i-1})| = F(x_n)-F(x_0) \leq F(b)-F(a), \end{equation*}

with equality for the sequence \(a<b\). Hence \(\|F\|_{TV([a,b])} = F(b)-F(a) = |F(b)-F(a)|\), and both sides vanish when \(a=b\).

Since every finite increasing sequence of reals lies in some \([-N,N]\),

\begin{equation*} \|F\|_{TV(\mathbb{R})} = \sup_{N\geq1}\|F\|_{TV([-N,N])} = \sup_{N\geq1}\big(F(N)-F(-N)\big). \end{equation*}

If \(|F| \leq M\) this is at most \(2M < \infty\). Conversely, if \(L := \|F\|_{TV(\mathbb{R})} < \infty\), then for any \(x\) and any \(N \geq |x|\) monotonicity gives

\begin{equation*} F(0) \leq F(x) \leq F(N) \leq F(-N) + L \leq F(0)+L \qquad (x \geq 0), \end{equation*}

and symmetrically \(F(0) \geq F(x) \geq F(-N) \geq F(N)-L \geq F(0)-L\) for \(x \leq 0\); so \(|F| \leq |F(0)|+L\). \(\square\)

All three properties already hold partition by partition. Write

\begin{equation*} V(H;P) := \sum_{i=1}^n |H(x_i)-H(x_{i-1})| \end{equation*}

for a finite increasing sequence \(P : x_0<\cdots<x_n\) (\(n \geq 0\)), so that \(\|H\|_{TV(\mathbb{R})} = \sup_P V(H;P) \in [0,+\infty]\) by Definition 1.6.33. Applying the triangle inequality in \(\mathbb{R}\) termwise,

\begin{equation*} V(F+G;P) \leq V(F;P)+V(G;P) \leq \|F\|_{TV(\mathbb{R})}+\|G\|_{TV(\mathbb{R})}, \end{equation*}

and taking the supremum over \(P\) gives the triangle property. Next \(V(cF;P) = |c|V(F;P)\); for \(c \neq 0\) multiplication by \(|c|\) is an increasing bijection of \([0,+\infty]\), hence commutes with suprema, giving \(\|cF\|_{TV} = |c|\|F\|_{TV}\), while for \(c=0\) both sides are \(0\) under the convention \(0\cdot(+\infty)=0\). Finally, a constant \(F\) has \(V(F;P)=0\) for every \(P\); conversely \(\|F\|_{TV(\mathbb{R})}=0\) applied to the two-point sequence \(x<y\) gives \(|F(y)-F(x)| \leq 0\), so \(F\) is constant. \(\square\)

Both inequalities come from two facts about the sums \(S(P) := \sum_{i=1}^n|F(x_i)-F(x_{i-1})|\), where \(P = \{x_0<\cdots<x_n\}\) is a finite set, so that \(\|F\|_{TV([a,b])} = \sup_{P \subseteq [a,b]}S(P)\).

(i) Refinement increases \(S\): \(P \subseteq P^{\prime}\) implies \(S(P) \leq S(P^{\prime})\). By induction it suffices to add one point \(y\): if \(y<x_0\) or \(y>x_n\), one extra non-negative term appears; if \(x_{i-1}<y<x_i\), the term \(|F(x_i)-F(x_{i-1})|\) is replaced by \(|F(y)-F(x_{i-1})|+|F(x_i)-F(y)|\), which is at least as large.

(ii) Sums split at a common point: if \(b \in R\) and \(R_1 := R\cap(-\infty,b]\), \(R_2 := R\cap[b,+\infty)\), then \(S( R) = S(R_1)+S(R_2)\), since listing \(R\) as \(z_0<\cdots<z_m\) with \(b = z_k\), the consecutive pairs with \(j \leq k\) are exactly those of \(R_1\) and the rest exactly those of \(R_2\).

For \(\geq\): given finite \(P \subseteq [a,b]\) and \(Q \subseteq [b,c]\), put \(R := P\cup Q\cup\{b\}\), so \(R_1 = P\cup\{b\}\) and \(R_2 = Q\cup\{b\}\), and

\begin{equation*} S(P)+S(Q) \leq S(P\cup\{b\})+S(Q\cup\{b\}) = S( R) \leq \|F\|_{TV([a,c])}; \end{equation*}

take suprema over \(P\) then \(Q\). For \(\leq\): given finite \(Z \subseteq [a,c]\), apply (i) and (ii) to \(R := Z\cup\{b\}\), whose pieces satisfy \(R_1 \subseteq [a,b]\), \(R_2 \subseteq [b,c]\):

\begin{equation*} S(Z) \leq S(R_1)+S(R_2) \leq \|F\|_{TV([a,b])}+\|F\|_{TV([b,c])} . \end{equation*}

Taking the supremum over \(Z\) gives the identity in \([0,+\infty]\). \(\square\)

(i). With \(V := \|f\|_{TV(\mathbb{R})} < \infty\) and \(S(P)\) as in Exercise 1.6.36, the two-point set \(\{0,x\}\) gives \(|f(x)-f(0)| \leq V\), so \(|f| \leq |f(0)|+V\) and \(f\) is bounded. Hence \(L := \limsup_{x\to+\infty}f(x)\) and \(\ell := \liminf_{x\to+\infty}f(x)\) are finite. If \(\ell<L\), put \(\varepsilon := (L-\ell)/3>0\) and choose recursively \(0 < z_1 < z_2 < \cdots < z_{2N}\) with \(f(z_j) > L-\varepsilon\) for \(j\) odd and \(f(z_j) < \ell+\varepsilon\) for \(j\) even (possible by the definitions of \(\limsup\) and \(\liminf\)). Then each consecutive increment is at least \((L-\varepsilon)-(\ell+\varepsilon) = \varepsilon\), so

\begin{equation*} \|f\|_{TV(\mathbb{R})} \geq S(\{z_1,\dots,z_{2N}\}) \geq (2N-1)\varepsilon \xrightarrow[N\to\infty]{} \infty, \end{equation*}

a contradiction. So \(\ell=L\) and \(\lim_{x\to+\infty}f(x)\) exists finitely; the same construction with a decreasing sequence \(z_1>\cdots>z_{2N} \to -\infty\), read in increasing order, handles \(-\infty\).

(ii). Take \(f(x) := x\sin(\pi/x)\) for \(0<x\leq1\) and \(f(x) := 0\) otherwise. It is bounded by \(1\), supported in \([0,1]\), and continuous (\(|f(x)| \leq |x| \to 0\) at \(0^+\), and \(x\sin(\pi/x) \to \sin\pi = 0\) at \(1^-\)). With \(x_k := 2/(2k+1) \in (0,1)\) we have \(\sin(\pi/x_k) = \sin\big(\tfrac{(2k+1)\pi}{2}\big) = (-1)^k\), so \(f(x_k) = (-1)^k\tfrac{2}{2k+1}\) alternates in sign and

\begin{equation*} |f(x_{k+1})-f(x_k)| = \frac{2}{2k+3}+\frac{2}{2k+1} \geq \frac{2}{2k+3}. \end{equation*}

The partition \(x_{N+1}<x_N<\cdots<x_1\) therefore has \(S \geq \sum_{k=1}^N \tfrac{2}{2k+3} \to \infty\), so \(\|f\|_{TV(\mathbb{R})} = +\infty\). \(\square\)

Write \(I_g(x) := \int_{(-\infty,x]}g\) and \(T(g) := \|I_g\|_{TV(\mathbb{R})}\), so that \(I_g(v)-I_g(u) = \int_{(u,v]}g\) for \(u<v\) by additivity over \((-\infty,v] = (-\infty,u]\sqcup(u,v]\). For any finite increasing sequence \(x_0<\cdots<x_n\), the triangle inequality for integrals (Lemma 1.3.19) and the disjointness of the intervals \((x_{i-1},x_i]\) give

\begin{equation*} \sum_{i=1}^n |I_g(x_i)-I_g(x_{i-1})| = \sum_{i=1}^n\Big|\int_{(x_{i-1},x_i]}g\Big| \leq \sum_{i=1}^n \int_{(x_{i-1},x_i]}|g| \leq \|g\|_{L^1(\mathbb{R})}, \end{equation*}

so \(T(g) \leq \|g\|_{L^1(\mathbb{R})}\) for every \(g \in L^1\) (1); in particular \(F = I_f\) is of bounded variation. Since \(I_h = I_g + I_{h-g}\), the triangle property of Exercise 1.6.35 together with (1) gives \(T(h) \leq T(g)+\|h-g\|_{L^1}\), i.e.

\begin{equation*} T(g) \geq T(h) - \|h-g\|_{L^1(\mathbb{R})}, \tag{2} \end{equation*}

all quantities being finite by (1).

For a step function we may discard null pieces and write, almost everywhere,

\begin{equation*} g = \sum_{j=1}^k c_j 1_{[a_j,b_j)}, \qquad c_j \neq 0, \quad a_1<b_1\leq a_2<b_2\leq\cdots\leq a_k<b_k . \end{equation*}

Along the partition listing \(a_1,b_1,\dots,a_k,b_k\) in increasing order we have \(I_g(b_j)-I_g(a_j) = c_j(b_j-a_j)\) and \(I_g(a_{j+1})-I_g(b_j) = 0\), so

\begin{equation*} T(g) \geq \sum_{j=1}^k |c_j|(b_j-a_j) = \|g\|_{L^1(\mathbb{R})}, \end{equation*}

and with (1), \(T(g) = \|g\|_{L^1(\mathbb{R})}\) for every step function \(g\).

Now let \(f \in L^1\) and \(\varepsilon>0\), and use Theorem 1.3.20(ii) to pick a step function \(g\) with \(\|f-g\|_{L^1} \leq \varepsilon\). By (2),

\begin{equation*} T(f) \geq T(g) - \|g-f\|_{L^1} = \|g\|_{L^1} - \|g-f\|_{L^1} \geq \|f\|_{L^1(\mathbb{R})} - 2\varepsilon . \end{equation*}

Letting \(\varepsilon \to 0\) and combining with (1) gives \(\|F\|_{TV(\mathbb{R})} = \|f\|_{L^1(\mathbb{R})}\). \(\square\)

For a finite \(P = \{x_0<\cdots<x_n\}\) write \(S^+(P), S^-(P), S(P)\) for the sums of \(\max(F(x_i)-F(x_{i-1}),0)\), \(\max(F(x_{i-1})-F(x_i),0)\), \(|F(x_i)-F(x_{i-1})|\). From \(\max(t,0)+\max(-t,0)=|t|\) and \(\max(t,0)-\max(-t,0)=t\),

\begin{equation*} S^+(P)+S^-(P) = S(P), \qquad S^+(P)-S^-(P) = F(x_n)-F(x_0), \tag{a} \end{equation*}

the second by telescoping. Also \(S^-\) is the \(S^+\)-sum of \(-F\), so \(F^- = (-F)^+\) and every statement proved for positive variations transfers to negative ones (b). Since \(S^\pm \leq S \leq \|F\|_{TV(\mathbb{R})}<\infty\), the functions \(F^\pm\) are finite, non-negative and non-decreasing. Three facts:

(A) Refinement: \(P \subseteq P^{\prime}\) gives \(S^\pm(P) \leq S^\pm(P^{\prime})\). Inserting one \(y\) outside \([x_0,x_n]\) adds a non-negative term; inserting \(y \in (x_{i-1},x_i)\) replaces \(\max(F(x_i)-F(x_{i-1}),0)\) by \(\max(F(y)-F(x_{i-1}),0)+\max(F(x_i)-F(y),0)\), which is at least as large since \(\max(s+t,0) \leq \max(s,0)+\max(t,0)\).

(B) Splitting: for finite \(R \ni c\), \(S^\pm( R) = S^\pm(R\cap(-\infty,c]) + S^\pm(R\cap[c,+\infty))\), since the consecutive pairs of \(R\) are exactly those of the two pieces.

(C) Common refinement: for an interval \(J\), writing \(V^\pm(J) := \sup\{S^\pm(P): P\subset J \text{ finite}\}\),

\begin{equation*} \sup_{P \subset J}S(P) = V^+(J)+V^-(J), \end{equation*}

where \(\leq\) is (a) and \(\geq\) follows by taking \(P^\pm\) with \(S^\pm(P^\pm)\) within \(\varepsilon\) of \(V^\pm(J)\) and applying (A) to \(P^{*} := P^+\cup P^-\): \(S(P^{*}) = S^+(P^{*})+S^-(P^{*}) \geq V^+(J)+V^-(J)-2\varepsilon\).

Identity 1. \(F(-\infty)\) exists finitely by Exercise 1.6.37(i). Fix \(x\) and \(\varepsilon>0\); choose finite \(P^\pm \subset (-\infty,x]\) with \(S^\pm(P^\pm) \geq F^\pm(x)-\varepsilon\), choose \(z < \min(P^+\cup P^-\cup\{x\})\) with \(|F(z)-F(-\infty)| \leq \varepsilon\), and put \(P := P^+\cup P^-\cup\{z,x\} \subset (-\infty,x]\). By (A) and the definition of \(F^\pm\), \(|S^\pm(P)-F^\pm(x)| \leq \varepsilon\), while \(F(x)-F(z) = S^+(P)-S^-(P)\) by (a); hence

\begin{equation*} \big|\big(F(x)-F(-\infty)\big) - \big(F^+(x)-F^-(x)\big)\big| \leq 3\varepsilon \end{equation*}

for every \(\varepsilon>0\), i.e. \(F(x) = F(-\infty)+F^+(x)-F^-(x)\).

Identity 2. We claim \(V^+([a,b]) = F^+(b)-F^+(a)\). For \(\leq\): let \(P \subset [a,b]\) be finite, with \(a \in P\) by (A), and pick finite \(Q \subset (-\infty,a]\) with \(a \in Q\) and \(S^+(Q) \geq F^+(a)-\varepsilon\); then \(R := Q\cup P \subset (-\infty,b]\) splits at \(a\) into \(Q\) and \(P\), so by (B)

\begin{equation*} F^+(b) \geq S^+( R) = S^+(Q)+S^+(P) \geq F^+(a)-\varepsilon+S^+(P). \end{equation*}

For \(\geq\): pick finite \(R \subset (-\infty,b]\) with \(a \in R\) and \(S^+( R) \geq F^+(b)-\varepsilon\); splitting at \(a\),

\begin{equation*} F^+(b)-\varepsilon \leq S^+(R\cap(-\infty,a]) + S^+(R\cap[a,b]) \leq F^+(a)+V^+([a,b]). \end{equation*}

By (b) also \(V^-([a,b]) = F^-(b)-F^-(a)\), so (C) with \(J=[a,b]\) gives

\begin{equation*} \|F\|_{TV([a,b])} = V^+([a,b])+V^-([a,b]) = F^+(b)-F^+(a)+F^-(b)-F^-(a). \end{equation*}

Identity 3. \(G := F^++F^-\) is bounded non-decreasing, and Identity 2 gives \(\|F\|_{TV([-N,N])} = G(N)-G(-N)\), so \(\|F\|_{TV} = \sup_N \|F\|_{TV([-N,N])} = G(+\infty)-G(-\infty)\). It remains to show \(F^+(-\infty)=0\), the case of \(F^-\) following by (b). If \(c := F^+(-\infty) > 0\), then \(F^+ \geq c\) everywhere; set \(m_0 := 0\) and, given \(m_{k-1}\), choose finite \(P_k \subset (-\infty,m_{k-1}]\) with \(m_{k-1} \in P_k\) (by (A)) and \(S^+(P_k) \geq c/2\), then \(m_k := \min P_k\). Splitting \(P_1\cup\cdots\cup P_k\) at \(m_{k-1}\) by (B) and inducting,

\begin{equation*} S^+(P_1\cup\cdots\cup P_k) \geq \sum_{j=1}^k S^+(P_j) \geq \frac{kc}{2} \xrightarrow[k\to\infty]{} \infty, \end{equation*}

contradicting \(S^+ \leq \|F\|_{TV}<\infty\). So \(G(-\infty)=0\) and \(\|F\|_{TV} = F^+(+\infty)+F^-(+\infty)\). \(\square\)

Truncate and apply Corollary 1.6.35. With \(\pi_n(x) := \max(-n,\min(n,x))\) put \(F_n := F\circ\pi_n\), so \(F_n = F\) on \([-n,n]\) and \(F_n\) is constant on each side. Then

\begin{equation*} \|F_n\|_{TV(\mathbb{R})} = \|F\|_{TV([-n,n])} < \infty : \end{equation*}

\(\geq\) because \(\pi_n\) is the identity on \([-n,n]\), and \(\leq\) because for any \(x_0<\cdots<x_m\) the values \(\pi_n(x_i)\) are non-decreasing in \([-n,n]\), the repeats contribute zero terms, and the distinct values \(y_0<\cdots<y_k\) lie in \([-n,n]\), so

\begin{equation*} \sum_{i=1}^m |F_n(x_i)-F_n(x_{i-1})| = \sum_{j=1}^k |F(y_j)-F(y_{j-1})| \leq \|F\|_{TV([-n,n])}. \end{equation*}

So each \(F_n\) has bounded variation on \(\mathbb{R}\) and is, by Corollary 1.6.35, differentiable outside a null set \(E_n\). Differentiability is a local property and \(F_n = F\) on the open interval \((-n,n)\), so \(F\) is differentiable on \((-n,n)\setminus E_n\). Hence \(F\) is differentiable outside the null set \(E := \bigcup_n E_n\): given \(x \notin E\), take \(n>|x|\). \(\square\)

The Lipschitz constant \(C_F\) is itself a Lipschitz constant: taking admissible \(C_k \to C_F\) and letting \(k \to \infty\) in \(|F(x)-F(y)| \leq C_k|x-y|\) gives \(|F(x)-F(y)| \leq C_F|x-y|\). Hence for any \(a \leq x_0<\cdots<x_n\leq b\) the increments telescope,

\begin{equation*} \sum_{i=1}^n |F(x_i)-F(x_{i-1})| \leq C_F\sum_{i=1}^n (x_i-x_{i-1}) = C_F(x_n-x_0) \leq C_F(b-a), \end{equation*}

so \(\|F\|_{TV([a,b])} \leq C_F(b-a) < \infty\) for every compact interval, i.e. \(F\) is locally of bounded variation, and therefore differentiable almost everywhere by Exercise 1.6.40. Finally, wherever \(F^{\prime}(x)\) exists,

\begin{equation*} |F^{\prime}(x)| = \lim_{h\to0}\Big|\frac{F(x+h)-F(x)}{h}\Big| \leq C_F, \end{equation*}

since each difference quotient is bounded by \(C_F|h|/|h| = C_F\). \(\square\)

Everything follows from the three-chord inequality: writing \(s(u,v) := (f(v)-f(u))/(v-u) = s(v,u)\) for chord slopes,

\begin{equation*} s(x,y) \leq s(x,z) \leq s(y,z) \qquad (x<y<z). \end{equation*}

Indeed \(y = (1-t)x+tz\) with \(t := \frac{y-x}{z-x} \in (0,1)\), and convexity gives \(f(y) \leq (1-t)f(x)+tf(z)\); subtracting \(f(x)\) and dividing by \(y-x = t(z-x)>0\) gives the first inequality, and subtracting the same from \(f(z)\) and dividing by \(z-y = (1-t)(z-x)>0\) gives the second.

Fix a compact \([b,c]\) and pick \(a<b\), \(d>c\). For \(b \leq u < v \leq c\), repeated use of the three-chord inequality (steps where \(u=b\) or \(v=c\) being identities) gives

\begin{equation*} s(a,b) \leq s(a,u) \leq s(a,v) \leq s(u,v) \leq s(u,d) \leq s(v,d) \leq s(c,d), \end{equation*}

so with \(K := \max(|s(a,b)|,|s(c,d)|)\) we get \(|f(v)-f(u)| \leq K|v-u|\) on \([b,c]\). In particular \(f\) is continuous, and exactly as in Exercise 1.6.41 \(\|f\|_{TV([b,c])} \leq K(c-b) < \infty\), so \(f\) is locally of bounded variation and hence differentiable almost everywhere by Exercise 1.6.40.

For fixed \(x\), the map \(h \mapsto s(x,x+h)\) is non-decreasing on \(h>0\) by the three-chord inequality and bounded below by \(s(x-1,x)\), so

\begin{equation*} D^+f(x) := \lim_{h\to0^+}\frac{f(x+h)-f(x)}{h} = \inf_{h>0}s(x,x+h) \in \mathbb{R} \end{equation*}

exists at every \(x\). It is non-decreasing: for \(x<y\), taking \(h := y-x\) gives \(D^+f(x) \leq s(x,y)\), while \(s(x,y) \leq s(y,y+k)\) for every \(k>0\) gives \(s(x,y) \leq D^+f(y)\). At every point of differentiability of \(f\) (that is, almost everywhere) the one-sided limit agrees with the two-sided one, so \(f^{\prime} = D^+f\) there. Thus \(f^{\prime}\) equals almost everywhere the monotone non-decreasing function \(D^+f\), which by Theorem 1.6.25 is itself differentiable almost everywhere. \(\square\)

In fact \(\int_{\mathbb{R}}|F^{\prime}| \leq \|F\|_{TV(\mathbb{R})} < \infty\). By Exercise 1.6.39,

\begin{equation*} F = F(-\infty) + F^+ - F^-, \qquad F^+(+\infty)+F^-(+\infty) = \|F\|_{TV(\mathbb{R})}, \end{equation*}

with \(F^\pm\) monotone non-decreasing and bounded by \(\|F\|_{TV(\mathbb{R})}\). By Theorem 1.6.25 both are differentiable outside a null set \(N\), and off \(N\)

\begin{equation*} F^{\prime} = (F^+)^{\prime} - (F^-)^{\prime} . \end{equation*}

Monotone functions are measurable (Exercise 1.6.29) and their almost everywhere defined derivatives are measurable (the modification of Exercise 1.6.1 noted before Proposition 1.6.37), so \(F^{\prime}\) is measurable. For each \(M>0\), \(F^+\) restricted to \([-M,M]\) is monotone non-decreasing, so Proposition 1.6.37 gives

\begin{equation*} \int_{[-M,M]}(F^+)^{\prime} \leq F^+(M)-F^+(-M) \leq F^+(+\infty), \end{equation*}

and since \((F^+)^{\prime} \geq 0\), the functions \((F^+)^{\prime}\mathbf{1}_{[-M,M]}\) increase to \((F^+)^{\prime}\), so monotone convergence (Theorem 1.4.44) gives \(\int_{\mathbb{R}}(F^+)^{\prime} \leq F^+(+\infty)\), and likewise for \(F^-\). As \(|F^{\prime}| \leq (F^+)^{\prime}+(F^-)^{\prime}\) almost everywhere,

\begin{equation*} \int_{\mathbb{R}}|F^{\prime}| \leq F^+(+\infty)+F^-(+\infty) = \|F\|_{TV(\mathbb{R})} < \infty . \qquad \square \end{equation*}

Assume \(a<b\) and let \(C\) be a Lipschitz constant for \(F\). Extend \(F\) by \(F(a)\) on \((-\infty,a)\) and \(F(b)\) on \((b,\infty)\); the extension is \(F\circ\pi\) with \(\pi(x) := \min(\max(x,a),b)\) the \(1\)-Lipschitz retraction onto \([a,b]\), hence Lipschitz with the same constant \(C\), so by Exercise 1.6.41 it is differentiable almost everywhere with \(|F^{\prime}| \leq C\); as \(\{a,b\}\) is null, \(\int_{[a,b]}F^{\prime}\) is unambiguous. Put

\begin{equation*} f_n(x) := n\big(F(x+\tfrac1n)-F(x)\big), \end{equation*}

a continuous function with \(|f_n| \leq C\) everywhere and \(f_n \to F^{\prime}\) almost everywhere. The constant \(C\) is absolutely integrable on the bounded interval \([a,b]\), so dominated convergence (Theorem 1.4.49) gives

\begin{equation*} \int_{[a,b]}F^{\prime}(x)\,dx = \lim_{n\to\infty}\int_{[a,b]}f_n(x)\,dx . \end{equation*}

For \(n\) with \(1/n \leq b-a\), translation invariance turns \(\int_{[a,b]}F(x+\tfrac1n)dx\) into \(\int_{[a+1/n,b+1/n]}F\), and the common piece \([a+\tfrac1n,b]\) cancels:

\begin{equation*} \int_{[a,b]}f_n(x)\,dx = n\int_{[b,\,b+1/n]}F(y)\,dy \ -\ n\int_{[a,\,a+1/n]}F(y)\,dy . \end{equation*}

The first term equals \(F(b)\) exactly, since \(F \equiv F(b)\) on \([b,b+\tfrac1n]\); and since \(|F(y)-F(a)| \leq C/n\) on \([a,a+\tfrac1n]\), the second differs from \(F(a)\) by at most \(n\cdot\tfrac1n\cdot\tfrac Cn = \tfrac Cn \to 0\). Hence \(\int_{[a,b]}F^{\prime} = F(b)-F(a)\). \(\square\)

Apply Exercise 1.6.44 to \(H := FG\). Let \(C_F, C_G\) be Lipschitz constants; on the compact \([a,b]\) both functions are bounded, say \(|F| \leq M_F := |F(a)|+C_F(b-a)\) and \(|G| \leq M_G\), so for \(x,y \in [a,b]\)

\begin{equation*} |H(x)-H(y)| \leq |F(x)||G(x)-G(y)| + |G(y)||F(x)-F(y)| \leq (M_FC_G+M_GC_F)|x-y|, \end{equation*}

i.e. \(H\) is Lipschitz. By Exercise 1.6.41, \(F\) and \(G\) are differentiable outside a null set \(N\), and for \(x \notin N\)

\begin{equation*} \frac{H(x+h)-H(x)}{h} = F(x+h)\frac{G(x+h)-G(x)}{h} + G(x)\frac{F(x+h)-F(x)}{h} \end{equation*}

tends, by continuity of \(F\), to \(F^{\prime}(x)G(x)+F(x)G^{\prime}(x)\); so \(H^{\prime} = F^{\prime}G+FG^{\prime}\) almost everywhere. Both summands are measurable (\(F,G\) are continuous and \(F^{\prime},G^{\prime}\) are almost everywhere limits of the continuous quotients \(n(F(\cdot+\tfrac1n)-F(\cdot))\)) and bounded by \(C_FM_G\) and \(M_FC_G\) using \(|F^{\prime}| \leq C_F\), \(|G^{\prime}| \leq C_G\) (Exercise 1.6.41), hence absolutely integrable on the finite-measure set \([a,b]\). So Exercise 1.6.44 applied to \(H\), followed by linearity, gives

\begin{equation*} F(b)G(b)-F(a)G(a) = \int_{[a,b]}H^{\prime} = \int_{[a,b]}F^{\prime}G + \int_{[a,b]}FG^{\prime}, \end{equation*}

which rearranges to the stated formula. \(\square\)

Split off a finite piece and let Proposition 1.6.37 kill the tail. Write \(F = \sum_{n\geq1}c_nJ_n\) as in Definition 1.6.30 (\(c_n>0\), \(\sum_n c_n<\infty\), \(J_n\) basic with discontinuity \(x_n\)); the case of a finite index set is contained in what follows. Being a sum of monotone non-decreasing functions with summable coefficients, \(F\) is bounded monotone non-decreasing, hence differentiable almost everywhere with \(F^{\prime} \geq 0\) (Theorem 1.6.25). Put

\begin{equation*} F_N := \sum_{n=1}^N c_nJ_n, \qquad R_N := F - F_N = \sum_{n>N}c_nJ_n, \qquad \varepsilon_N := \sum_{n>N}c_n \to 0 . \end{equation*}

The remainder \(R_N\) is again a jump function, monotone non-decreasing with values in \([0,\varepsilon_N]\), so for any compact \([a,b]\) Proposition 1.6.37 gives

\begin{equation*} \int_{[a,b]}R_N^{\prime}(x)\,dx \leq R_N(b)-R_N(a) \leq \varepsilon_N . \end{equation*}

Off the finite set \(\{x_1,\dots,x_N\}\) the function \(F_N\) is constant near \(x\), so \(F\) and \(R_N\) have the same difference quotients near \(x\) and hence \(F^{\prime} = R_N^{\prime}\) almost everywhere. Therefore

\begin{equation*} \int_{[a,b]}F^{\prime}(x)\,dx \leq \varepsilon_N \qquad \text{for every } N, \end{equation*}

and letting \(N \to \infty\) gives \(\int_{[a,b]}F^{\prime} = 0\). Since \(F^{\prime} \geq 0\) is measurable, Exercise 1.4.36(viii) forces \(F^{\prime} = 0\) almost everywhere on \([a,b]\); taking \([a,b] = [-M,M]\) for \(M \in \mathbb{N}\) and using countable subadditivity, \(F^{\prime} = 0\) almost everywhere on \(\mathbb{R}\). \(\square\)

Write \(C := \bigcap_n I_n\) for the Cantor set of Exercise 1.2.9, \(I_n\) the union of the \(2^n\) closed intervals \([\sum_{i\leq n}a_i3^{-i},\ \sum_{i\leq n}a_i3^{-i}+3^{-n}]\) with \(a_i \in \{0,2\}\), and recall \(m( C)=0\).

(i). Each \(F_n\) is the piecewise-linear staircase with slope \((3/2)^n\) on each of the \(2^n\) intervals of \(I_n\) and constant on each of the \(2^n-1\) removed middle thirds: \(F_0\) is the diagonal; \(F_1\) has slope \(3/2\) on \([0,\tfrac13]\) and \([\tfrac23,1]\) and value \(\tfrac12\) between; \(F_2\) has slope \(9/4\) on \([0,\tfrac19],[\tfrac29,\tfrac13],[\tfrac23,\tfrac79],[\tfrac89,1]\) and values \(\tfrac14,\tfrac12,\tfrac34\) on the removed intervals; \(F_3\) has slope \(27/8\) on the eight intervals of \(I_3\) and values \(\tfrac18,\tfrac28,\dots,\tfrac78\) on the seven removed intervals, left to right. Superimposed, they converge to the devil’s staircase.

(ii). Induct: \(F_0(x)=x\) is continuous, non-decreasing, \(0 \mapsto 0\), \(1 \mapsto 1\), with range \([0,1]\). If \(F_{n-1}\) has these properties then on \([0,\tfrac13]\) the map \(x \mapsto \tfrac12F_{n-1}(3x)\) is continuous non-decreasing from \(0\) to \(\tfrac12\), on \((\tfrac13,\tfrac23)\) the value is \(\tfrac12\), and on \([\tfrac23,1]\) the map \(x\mapsto \tfrac12+\tfrac12F_{n-1}(3x-2)\) is continuous non-decreasing from \(\tfrac12\) to \(1\). The pieces agree at \(\tfrac13\) and \(\tfrac23\) and their ranges are ordered \([0,\tfrac12],\{\tfrac12\},[\tfrac12,1]\), so \(F_n\) is continuous and non-decreasing with \(F_n(0)=0\), \(F_n(1)=1\) and range in \([0,1]\).

(iii). Induct again: \(|F_1-F_0| \leq 1 = 2^0\) since both lie in \([0,1]\), and if \(\sup|F_n-F_{n-1}| \leq 2^{-(n-1)}\) then the recursion gives \(|F_{n+1}-F_n| = \tfrac12|F_n(3x)-F_{n-1}(3x)| \leq 2^{-n}\) on \([0,\tfrac13]\), \(0\) on \((\tfrac13,\tfrac23)\), and \(\tfrac12|F_n(3x-2)-F_{n-1}(3x-2)| \leq 2^{-n}\) on \([\tfrac23,1]\). Hence for \(m>n\),

\begin{equation*} |F_m(x)-F_n(x)| \leq \sum_{k\geq n}2^{-k} = 2^{-n+1}, \end{equation*}

so \((F_n)\) is uniformly Cauchy; it converges to a limit \(F\) with \(\sup_x|F-F_n| \leq 2^{-n+1}\), i.e. uniformly.

(iv). A uniform limit of continuous functions is continuous; \(F_n(x) \leq F_n(y)\) for \(x\leq y\) passes to the limit; and \(F(0)=0\), \(F(1)=1\), \(0 \leq F \leq 1\). Passing to the limit in the recursion,

\begin{equation*} F(x) = \tfrac12F(3x) \ \ \text{on } [0,\tfrac13], \qquad F \equiv \tfrac12 \ \ \text{on } [\tfrac13,\tfrac23], \qquad F(x) = \tfrac12+\tfrac12F(3x-2) \ \ \text{on } [\tfrac23,1]. \tag{$*$} \end{equation*}

(v). \(F\) is constant on every removed interval, by induction on the level: at level \(1\), \(F \equiv \tfrac12\) on \((\tfrac13,\tfrac23)\) by \((*)\); and an interval \(J\) removed at level \(n+1\) is \(\tfrac13 J^{\prime}\) or \(\tfrac{2+J^{\prime}}{3}\) for some \(J^{\prime}\) removed at level \(n\), on which \((*)\) expresses \(F|_J\) as an affine function of \(F|_{J^{\prime}}\). If \(x \notin C\) then \(x \notin I_n\) for some \(n\), so \(x\) lies in such an open interval and \(F\) is constant near \(x\); hence \(F^{\prime}(x)=0\) for \(x \in (0,1)\setminus C\). As \(m( C)=0\),

\begin{equation*} \int_{[0,1]}F^{\prime}(x)\,dx = 0 \neq 1 = F(1)-F(0). \end{equation*}

(vi). Put \(\sigma(a) := \sum_{n\geq1}a_{n+1}3^{-n}\). If \(a_1=0\) then \(x = \tfrac13\sigma(a) \in [0,\tfrac13]\) and \(3x = \sigma(a)\); if \(a_1=2\) then \(x = \tfrac23+\tfrac13\sigma(a) \in [\tfrac23,1]\) and \(3x-2 = \sigma(a)\). Either way \((*)\) gives \(F(x) = \tfrac{a_1}{2}2^{-1}+\tfrac12F(\sigma(a))\), so iterating \(N\) times,

\begin{equation*} F(x) = \sum_{n=1}^N \frac{a_n}{2}2^{-n} + 2^{-N}F(\sigma^N(a)), \end{equation*}

where the error is at most \(2^{-N} \to 0\) because \(0 \leq F \leq 1\).

(vii). \(|I| = 3^{-n}\) is immediate. Both endpoints have ternary digits in \(\{0,2\}\): \(\alpha = \sum_{i\leq n}a_i3^{-i}\) padded with zeros, and \(\beta = \alpha+3^{-n} = \sum_{i\leq n}a_i3^{-i}+\sum_{i>n}2\cdot3^{-i}\). So (vi) gives

\begin{equation*} F(\alpha) = \sum_{i\leq n}\frac{a_i}{2}2^{-i}, \qquad F(\beta) = F(\alpha)+\sum_{i>n}2^{-i} = F(\alpha)+2^{-n}. \end{equation*}

Since \(F\) is continuous and non-decreasing, \(F(I) = [F(\alpha),F(\beta)]\), an interval of length \(2^{-n}\).

(viii). Let \(x \in C\) and let \([\alpha_n,\beta_n]\) be the interval of \(I_n\) containing it, so \(\beta_n-\alpha_n = 3^{-n}\) and \(F(\beta_n)-F(\alpha_n) = 2^{-n}\) by (vii). Some endpoint \(y_n \neq x\) satisfies

\begin{equation*} \frac{F(y_n)-F(x)}{y_n-x} \geq \Big(\frac32\Big)^n : \end{equation*}

if \(x\) is an endpoint, take the other one, the quotient being exactly \(2^{-n}/3^{-n}\); otherwise set \(p_1 := F(x)-F(\alpha_n)\), \(p_2 := F(\beta_n)-F(x)\), \(q_1 := x-\alpha_n\), \(q_2 := \beta_n-x\), so that \(p_1+p_2 = 2^{-n}\) and \(q_1+q_2 = 3^{-n}\), and if both \(p_i/q_i < (3/2)^n\) then \(p_1+p_2 < (3/2)^n3^{-n} = 2^{-n}\), a contradiction; the endpoint realising \(p_i/q_i \geq (3/2)^n\) is the difference quotient above. Since \(|y_n-x| \leq 3^{-n} \to 0\) while these quotients exceed \((3/2)^n \to \infty\), \(F\) is not differentiable at \(x\). \(\square\)

(i). Definition 1.6.39 applied to the single interval \((a,b)\) gives \(|F(b)-F(a)| \leq \varepsilon\) whenever \(b-a \leq \delta\), which is uniform continuity, hence continuity.

(ii). Take \(\delta\) for \(\varepsilon = 1\). For \([c,d] \subset [a,b]\) with \(d-c\leq\delta\), any \(c \leq x_0<\cdots<x_n\leq d\) yields disjoint intervals \((x_{i-1},x_i)\) of total length \(x_n-x_0 \leq \delta\), so \(\sum_i|F(x_i)-F(x_{i-1})| \leq 1\) and \(\|F\|_{TV([c,d])} \leq 1\). Cutting \([a,b]\) into \(N := \lceil (b-a)/\delta\rceil\) consecutive pieces of length at most \(\delta\), Exercise 1.6.36 gives

\begin{equation*} \|F\|_{TV([a,b])} = \sum_{k=1}^N \|F\|_{TV([c_{k-1},c_k])} \leq N < \infty, \end{equation*}

so \(F\) is locally of bounded variation and hence differentiable almost everywhere by Exercise 1.6.40.

(iii). With \(C>0\) a Lipschitz constant (the case \(C=0\) being trivial), \(\delta := \varepsilon/C\) works:

\begin{equation*} \sum_{j=1}^n|F(b_j)-F(a_j)| \leq C\sum_{j=1}^n (b_j-a_j) \leq C\delta = \varepsilon . \end{equation*}

(iv). \(\sqrt{\cdot}\) is not Lipschitz on \([0,1]\), since \((\sqrt x - \sqrt 0)/x = 1/\sqrt x \to \infty\) as \(x \to 0^+\). It is absolutely continuous because

\begin{equation*} \sqrt{a+h}-\sqrt a = \frac{h}{\sqrt{a+h}+\sqrt a} \end{equation*}

is non-increasing in \(a\): ordering disjoint intervals as \(a_1<b_1\leq\cdots\leq a_n<b_n\) and setting \(h_j := b_j-a_j\), \(s_j := h_1+\cdots+h_j\), disjointness gives \(a_j \geq s_{j-1}\), so \(\sqrt{b_j}-\sqrt{a_j} \leq \sqrt{s_j}-\sqrt{s_{j-1}}\) and the sum telescopes:

\begin{equation*} \sum_{j=1}^n |F(b_j)-F(a_j)| \leq \sqrt{s_n} = \sqrt{\textstyle\sum_j (b_j-a_j)} . \end{equation*}

So \(\delta := \varepsilon^2\) works.

(v). The Cantor function is continuous and monotone by Exercise 1.6.47(iv), hence uniformly continuous on the compact \([0,1]\) (Heine-Borel). It is not absolutely continuous: given \(\delta>0\) choose \(n\) with \((2/3)^n \leq \delta\); the \(2^n\) intervals \([\alpha_k,\beta_k]\) of \(I_n\) are disjoint of total length \(2^n3^{-n} \leq \delta\), yet each has \(F(\beta_k)-F(\alpha_k) = 2^{-n}\) by Exercise 1.6.47(vii), so

\begin{equation*} \sum_{k=1}^{2^n}|F(\beta_k)-F(\alpha_k)| = 2^n\cdot2^{-n} = 1 > \tfrac12 . \end{equation*}

(vi). The single function \(f\) is uniformly integrable (Exercise 1.5.10), so Exercise 1.5.13 supplies, for each \(\varepsilon>0\), a \(\delta>0\) with \(\int_E|f| \leq \varepsilon\) whenever \(m(E) \leq \delta\). For disjoint \((a_j,b_j)\) of total length at most \(\delta\), put \(E := \bigcup_j(a_j,b_j)\), so \(m(E) \leq \delta\), and use \(F(b_j)-F(a_j) = \int_{(a_j,b_j]}f\):

\begin{equation*} \sum_{j=1}^n |F(b_j)-F(a_j)| \leq \sum_{j=1}^n\int_{(a_j,b_j)}|f| = \int_E |f| \leq \varepsilon . \end{equation*}

That \(F\) is differentiable almost everywhere with \(F^{\prime} = f\) is Theorem 1.6.11 applied to \(f\).

(vii). For the sum, take \(\delta := \min(\delta_F,\delta_G)\) for tolerance \(\varepsilon/2\) apiece and apply the triangle inequality termwise. For the product, \(F\) and \(G\) are continuous by (i), hence bounded by some \(M \geq 1\) on the compact \([a,b]\), and

\begin{equation*} \sum_j |(FG)(b_j)-(FG)(a_j)| \leq M\Big(\sum_j|G(b_j)-G(a_j)| + \sum_j|F(b_j)-F(a_j)|\Big), \end{equation*}

so a \(\delta\) making each sum at most \(\varepsilon/(2M)\) suffices. On \(\mathbb{R}\) the sum argument is unchanged, but the product can fail, boundedness being what compactness supplied: \(F(x)=G(x)=x\) is Lipschitz, hence absolutely continuous by (iii), while \(x \mapsto x^2\) is not uniformly continuous on \(\mathbb{R}\) (\(|(x+\tfrac1x)^2-x^2| > 2\) at distance \(\tfrac1x \to 0\)) and so is not absolutely continuous by (i). \(\square\)

(i). It suffices to prove \(m^{*}(F(E)) = 0\), and since \(E = \bigcup_k (E\cap[k,k+1])\) with countable unions of null sets null, we may assume \(E \subset [a,b]\). Let \(\varepsilon>0\) and take \(\delta>0\) from Definition 1.6.39. Choose a bounded open \(U \supset E\) with \(m(U)<\delta\) and write it (Exercise 1.6.10) as a countable disjoint union \(U = \bigcup_j (a_j,b_j)\), so \(\sum_j (b_j-a_j) = m(U) < \delta\). Put \(M_j := \sup_{(a_j,b_j)}F\) and \(m_j := \inf_{(a_j,b_j)}F\), finite because \(F\) is continuous (Exercise 1.6.48(i)) on the compact \([a_j,b_j]\); then \(F((a_j,b_j)) \subset [m_j,M_j]\) gives \(m^{*}(F((a_j,b_j))) \leq M_j-m_j\).

Claim: \(\sum_j (M_j-m_j) \leq \varepsilon\). Fix \(\eta>0\) and \(N\), and for \(j \leq N\) choose \(c_j,d_j \in (a_j,b_j)\) with \(F(c_j) < m_j+\tfrac\eta2 2^{-j}\) and \(F(d_j) > M_j - \tfrac\eta2 2^{-j}\), so that

\begin{equation*} |F(d_j)-F(c_j)| > (M_j-m_j) - \eta 2^{-j}. \end{equation*}

For \(j \leq N\) with \(c_j = d_j\) this forces \(M_j-m_j < \eta2^{-j}\); the remaining indices \(S\) give finitely many disjoint open intervals with endpoints \(c_j, d_j\) (disjoint because the \((a_j,b_j)\) are) of total length \(<\delta\), so \(\sum_{j\in S}|F(d_j)-F(c_j)| \leq \varepsilon\) by the choice of \(\delta\). Hence

\begin{equation*} \sum_{j\leq N}(M_j-m_j) < \varepsilon + \eta\sum_{j\geq1}2^{-j} = \varepsilon+\eta, \end{equation*}

and \(\eta \to 0\), \(N \to \infty\) prove the claim. Since \(F(E) \subset \bigcup_j F((a_j,b_j))\), countable subadditivity gives \(m^{*}(F(E)) \leq \sum_j(M_j-m_j) \leq \varepsilon\) for every \(\varepsilon>0\).

(ii). For the Cantor function \(F( C) = [0,1]\) while \(m( C)=0\). Given \(y \in [0,1]\), take a binary expansion \(y = \sum_n b_n2^{-n}\) with \(b_n \in \{0,1\}\), set \(a_n := 2b_n \in \{0,2\}\) and \(x := \sum_n a_n3^{-n}\). For each \(n\),

\begin{equation*} x = \sum_{i\leq n}a_i3^{-i} + t_n, \qquad t_n \in \Big[0,\ \sum_{i>n}2\cdot3^{-i}\Big] = [0,3^{-n}], \end{equation*}

so \(x\) lies in a constituent interval of \(I_n\), whence \(x \in \bigcap_n I_n = C\); and \(F(x) = \sum_n \tfrac{a_n}{2}2^{-n} = y\) by Exercise 1.6.47(vi). With \(0 \leq F \leq 1\) this gives \(F( C) = [0,1]\), of measure \(1 \neq 0\). \(\square\)

If. Let \(F(x) := \int_{[a,x]}f + C\) with \(f\) absolutely integrable, so \(F\) is finite. The constant sequence \(f\) is uniformly integrable (Exercise 1.5.10(i)), so Exercise 1.5.13 gives, for each \(\varepsilon>0\), a \(\delta>0\) with \(\int_E|f| \leq \varepsilon\) whenever \(m(E) \leq \delta\). For disjoint \((a_1,b_1),\dots,(a_n,b_n) \subset [a,b]\) of total length at most \(\delta\), additivity over \([a,b_j] = [a,a_j]\cup(a_j,b_j]\) gives \(F(b_j)-F(a_j) = \int_{(a_j,b_j)}f\), so with \(E := \bigcup_j (a_j,b_j)\), of measure at most \(\delta\),

\begin{equation*} \sum_{j=1}^n |F(b_j)-F(a_j)| \leq \sum_{j=1}^n \int_{(a_j,b_j)}|f| = \int_E |f| \leq \varepsilon . \end{equation*}

Only if. By Exercise 1.6.48(ii), \(F\) is of bounded variation on \([a,b]\), hence differentiable almost everywhere (Corollary 1.6.35) with \(F^{\prime}\) absolutely integrable (Exercise 1.6.43). Let \(f := F^{\prime}\) where the derivative exists and \(f := 0\) elsewhere: this is measurable by Exercise 1.6.1 and agrees with \(F^{\prime}\) off a null set. For \(x \in (a,b]\) the restriction \(F|_{[a,x]}\) is again absolutely continuous (the same \(\delta\) works, disjoint intervals in \([a,x]\) being disjoint intervals in \([a,b]\)), so Theorem 1.6.40 on \([a,x]\) gives

\begin{equation*} \int_{[a,x]}f(y)\,dy = \int_{[a,x]}F^{\prime}(y)\,dy = F(x)-F(a), \end{equation*}

i.e. \(F(x) = \int_{[a,x]}f + C\) with \(C := F(a)\), the case \(x=a\) being trivial. \(\square\)

Apply Theorem 1.6.40 to \(F\varphi\) and use \(\varphi(a) = \varphi(b) = 0\), which holds since \(\varphi\) vanishes outside a compact subset of \((a,b)\).

The function \(\varphi\) is Lipschitz on \([a,b]\) (its continuous derivative is bounded there, so use the mean value theorem), hence absolutely continuous by Exercise 1.6.48(iii), and therefore \(F\varphi\) is absolutely continuous by Exercise 1.6.48(vii). At every point \(x\) where \(F\) is differentiable — all but a null set, by Exercise 1.6.48(ii) and Corollary 1.6.35 — the product rule gives \((F\varphi)^{\prime}(x) = F^{\prime}(x)\varphi(x) + F(x)\varphi^{\prime}(x)\). Both summands are absolutely integrable, since \(F^{\prime}\) is (Exercise 1.6.43) and \(\varphi, \varphi^{\prime}, F\) are bounded on \([a,b]\). Hence

\begin{align*} 0 &= F(b)\varphi(b) - F(a)\varphi(a) = \int_{[a,b]} (F\varphi)^{\prime}(x)\, dx \\ &= \int_{[a,b]} F^{\prime}\varphi(x)\, dx + \int_{[a,b]} F\varphi^{\prime}(x)\, dx, \end{align*}

the middle equality by Theorem 1.6.40. Rearranging gives the claim.

The chain and product rules give, for \(x \neq 0\),

\begin{align*} F^{\prime}(x) &= 2x \sin\!\left(\frac{1}{x^3}\right) - \frac{3}{x^2}\cos\!\left(\frac{1}{x^3}\right) \\ &=: G(x) + H(x), \end{align*}

while at the origin \(|(F(h)-F(0))/h| = |h||\sin(1/h^3)| \leq |h| \to 0\), so \(F^{\prime}(0) = 0\); set \(G(0) := H(0) := 0\). Thus \(F\) is everywhere differentiable on \([-1,1]\). Since \(|G| \leq 2|x| \leq 2\), \(G\) is absolutely integrable, so it suffices to show \(\int_{(0,1]}|H| = +\infty\).

For \(n \geq 1\) put

\begin{equation*} I_n := [\alpha_n, \beta_n], \quad \alpha_n := \left(2\pi n + \tfrac{\pi}{3}\right)^{-1/3}, \quad \beta_n := \left(2\pi n - \tfrac{\pi}{3}\right)^{-1/3}. \end{equation*}

As \(t \mapsto t^{-1/3}\) is a decreasing bijection of \((0,\infty)\) and the intervals \([2\pi n - \pi/3, 2\pi n + \pi/3]\) are pairwise disjoint, the \(I_n\) are pairwise disjoint subsets of \((0,1]\) (indeed \(\beta_1 < 1\)). For \(x \in I_n\) we have \(1/x^3\) within \(\pi/3\) of \(2\pi n\), so \(|\cos(1/x^3)| \geq \cos(\pi/3) = 1/2\); also \(3/x^2 \geq 3\beta_n^{-2} \geq 3(\pi n)^{2/3}\), using \(2\pi n - \pi/3 \geq \pi n\). The mean value theorem applied to \(t \mapsto t^{-1/3}\) gives

\begin{align*} \beta_n - \alpha_n = \frac{2\pi}{3}\cdot\frac{1}{3}\xi_n^{-4/3} &\geq \frac{2\pi}{9}\left(2\pi n + \tfrac{\pi}{3}\right)^{-4/3} \\ &\geq \frac{2\pi}{9}(3\pi n)^{-4/3}. \end{align*}

Multiplying the three bounds, \(\int_{I_n}|H| \geq c\,n^{-2/3}\) with \(c := \frac{\pi}{3}\pi^{2/3}(3\pi)^{-4/3} > 0\), whence by disjointness

\begin{align*} \int_{(0,1]}|H(x)|\, dx &\geq \sum_{n=1}^\infty \int_{I_n}|H(x)|\, dx \\ &\geq c\sum_{n=1}^\infty n^{-2/3} = +\infty . \end{align*}

So \(F^{\prime} = G + H\) is not absolutely integrable, and the hypothesis of Proposition 1.6.41 fails: \(\int_{[-1,1]}F^{\prime}(x)\, dx\) does not exist as an absolutely convergent Lebesgue integral.

Write \(\Delta_j := t_j - t_{j-1}\) and \(\Sigma(f;\mathcal{P}) := \sum_j f(t_j^*)\Delta_j\) for a tagged partition \(\mathcal{P}\), called \(\delta\)-fine when \(\Delta_j \leq \delta(t_j^*)\) for all \(j\). Cousin’s theorem (Exercise 1.6.23) supplies a \(\delta\)-fine tagged partition for every gauge \(\delta\), and is used silently throughout.

(i) If \(L, L^{\prime}\) both work, take gauges \(\delta, \delta^{\prime}\) for \(\varepsilon\) and a \(\min(\delta,\delta^{\prime})\)-fine \(\mathcal{P}\) (Cousin); it is both \(\delta\)-fine and \(\delta^{\prime}\)-fine, so

\begin{align*} |L - L^{\prime}| &\leq |L - \Sigma(f;\mathcal{P})| \\ &\quad + |\Sigma(f;\mathcal{P}) - L^{\prime}| \ \leq\ 2\varepsilon . \end{align*}

Let \(\varepsilon \to 0\).

(ii) Let \(I\) be the Riemann integral and \(M := 1 + \sup_{[a,b]}|f| < \infty\) (Riemann integrability forces boundedness). Given \(\varepsilon > 0\) pick a partition \(P : a = s_0 < \dots < s_m = b\) whose Darboux sums satisfy \(U(f,P) - L(f,P) \leq \varepsilon\), so that \(I - \varepsilon \leq L(f,P) \leq U(f,P) \leq I + \varepsilon\), and take the constant gauge \(\delta := \varepsilon/(4mM)\). For a \(\delta\)-fine \(\mathcal{P}\), split \(\{1,\dots,k\}\) into

\begin{equation*} A := \{j : (t_{j-1},t_j) \cap P = \emptyset\}, \qquad B := \{1,\dots,k\} \setminus A . \end{equation*}

(a) The open intervals \((t_{j-1},t_j)\) are disjoint, so \(\#B \leq m-1\) and \(\sum_{j \in B}\Delta_j \leq m\delta = \varepsilon/(4M)\), whence \(|\sum_{j\in B} f(t_j^*)\Delta_j| \leq \varepsilon/4\). (b) For \(j \in A\) the interval \([t_{j-1},t_j]\) is one of the intervals of the common refinement \(R := P \cup \{t_0,\dots,t_k\}\), and \(\inf_{[t_{j-1},t_j]}f \leq f(t_j^*) \leq \sup_{[t_{j-1},t_j]}f\); the terms of the Darboux sums of \(R\) not arising this way subdivide the \(B\)-intervals and so contribute at most \(M \cdot \varepsilon/(4M)\) in absolute value. Hence, \(R\) refining \(P\),

\begin{align*} I - \tfrac{5\varepsilon}{4} \ \leq\ L(f,P) - \tfrac{\varepsilon}{4} &\ \leq\ \sum_{j\in A} f(t_j^*)\Delta_j \\ &\ \leq\ U(f,P) + \tfrac{\varepsilon}{4} \ \leq\ I + \tfrac{5\varepsilon}{4}. \end{align*}

Adding, \(|\Sigma(f;\mathcal{P}) - I| \leq 2\varepsilon\) for every \(\delta\)-fine \(\mathcal{P}\); as \(\varepsilon\) was arbitrary, \(f\) is Henstock-Kurzweil integrable with integral \(I\).

(iii) Let \(I := \int_{[a,b]}f\) (Lebesgue) and let \(\varepsilon > 0\). The single function \(f\) is uniformly integrable (Exercise 1.5.10(i)), so by Exercise 1.5.13 there is \(0 < \kappa \leq \varepsilon\) with \(\int_U |f| \leq \varepsilon\) whenever \(m(U) \leq \kappa\). Let \(E\) consist of \(a\), \(b\), and the non-Lebesgue points of \(f\); \(E\) is null by Theorem 1.6.19, so by outer regularity choose open \(U_m \supset E\) with \(m(U_m) \leq \kappa/4^m\), giving \(m(\bigcup_m U_m) \leq \kappa\) and \(\int_{\bigcup_m U_m}|f| \leq \varepsilon\). Define a gauge:

• \(x \in E\): pick the least \(m(x) \geq 1\) with \(|f(x)| \leq 2^{m(x)}\) (possible as \(f(x)\) is finite) and \(\delta(x) > 0\) with \((x-\delta(x),x+\delta(x)) \subset U_{m(x)}\).
• \(x \notin E\): any interval \(Q \ni x\) of length \(h\) lies in \([x-h,x+h]\), so \(\frac{1}{|Q|}\int_Q|f(y)-f(x)|\,dy \leq \frac{2}{2h}\int_{[x-h,x+h]}|f(y)-f(x)|\,dy \to 0\); pick \(\delta(x) > 0\) so that \(\int_Q f = |Q|f(x) + O(\varepsilon|Q|)\) for all such \(Q\) with \(h \leq \delta(x)\).

For a \(\delta\)-fine \(\mathcal{P}\), split by \(J_1 := \{j : t_j^* \notin E\}\), \(J_2 := \{j : t_j^* \in E\}\), and set \(S := \bigcup_{j\in J_1}[t_{j-1},t_j]\). Summing the second bullet over \(J_1\), and noting that for \(j \in J_2\) we have \([t_{j-1},t_j] \subset (t_j^*-\delta(t_j^*), t_j^*+\delta(t_j^*)) \subset U_{m(t_j^*)}\), so \([a,b]\setminus S \subset \bigcup_m U_m\),

\begin{align*} \sum_{j\in J_1} f(t_j^*)\Delta_j &= \int_S f(y)\, dy + O(\varepsilon(b-a)) \\ &= I + O\Big(\int_{[a,b]\setminus S}|f|\Big) + O(\varepsilon(b-a)) \\ &= I + O(\varepsilon(1 + b-a)). \end{align*}

For \(J_2\), group by \(m = m(t_j^*)\): the corresponding intervals are non-overlapping subsets of \(U_m\), of total length \(\leq \kappa/4^m\), with \(|f(t_j^*)| \leq 2^m\), so

\begin{align*} \Big|\sum_{j\in J_2} f(t_j^*)\Delta_j\Big| \leq \sum_{m=1}^\infty 2^m\frac{\kappa}{4^m} = \kappa \leq \varepsilon . \end{align*}

Thus \(|\Sigma(f;\mathcal{P}) - I| \leq C(1 + b-a)\varepsilon\); rerunning with \(\varepsilon\) replaced by \(\varepsilon^{\prime}/(C(1+b-a))\) gives the Henstock-Kurzweil integrability of \(f\), with integral \(I\).

(iv) Given \(\varepsilon > 0\), put \(\varepsilon^{\prime} := \varepsilon/(b-a)\) and use differentiability at each \(x\) to pick \(\delta(x) > 0\) with

\begin{align*} |F(y) - F(x) - (y-x)F^{\prime}(x)| \leq \varepsilon^{\prime}|y-x| \\ (y \in [a,b],\ |y-x| \leq \delta(x)), \end{align*}

trivially true at \(y = x\). For a \(\delta\)-fine \(\mathcal{P}\), both \(y = t_j\) and \(y = t_{j-1}\) are within \(\Delta_j \leq \delta(t_j^*)\) of \(x = t_j^*\); subtracting the two resulting inequalities and using \((t_j - t_j^*) + (t_j^* - t_{j-1}) = \Delta_j\) gives \(|F(t_j) - F(t_{j-1}) - F^{\prime}(t_j^*)\Delta_j| \leq \varepsilon^{\prime}\Delta_j\). Summing and telescoping,

\begin{align*} \Big|\sum_{j=1}^k F^{\prime}(t_j^*)\Delta_j - (F(b) - F(a))\Big| \leq \varepsilon^{\prime}(b-a) = \varepsilon . \end{align*}

(v) The zero function has Henstock-Kurzweil integral \(0\) (every Riemann sum vanishes; use \(\delta \equiv 1\)).

Exercise 1.6.4. If \(F^{\prime} = G^{\prime}\) everywhere, then \(H := F - G\) is everywhere differentiable with \(H^{\prime} \equiv 0\); applying (iv) on \([a,x]\) for \(x \in (a,b]\) makes \(H(x) - H(a)\) the Henstock-Kurzweil integral of \(0\), so \(H(x) = H(a) =: C\) by (i). The converse is immediate.

Proposition 1.6.41. If \(F\) is everywhere differentiable with \(F^{\prime}\) absolutely integrable, then \(F^{\prime}\) is everywhere defined and finite, so (iii) gives it Henstock-Kurzweil integral \(\int_{[a,b]}F^{\prime}(x)\,dx\) while (iv) gives it \(F(b)-F(a)\); by (i) the two agree.

Monotonicity kills the first term: \(\mu^*(A \cap E) \le \mu^*(E) = 0\). Hence for every \(A \subset X\),

\begin{equation*} \mu^*(A \cap E) + \mu^*(A \setminus E) = \mu^*(A \setminus E) \le \mu^*(A) \le \mu^*(A \cap E) + \mu^*(A \setminus E), \end{equation*}

the outer inequalities by monotonicity and by countable subadditivity applied to \(A = (A\cap E) \cup (A \setminus E) \cup \emptyset \cup \dots\) (axioms (ii), (iii), (i) of Definition 1.7.1). So equality holds throughout, which is Carathéodory measurability (Definition 1.7.2).

Both directions run through Exercise 1.2.17, \(m^*\) being an outer measure by Exercise 1.2.3.

Carathéodory implies Lebesgue. Testing the Carathéodory identity against elementary \(A\) and using \(m^*(A) = m(A)\) there (Lemma 1.2.6) gives condition (ii) of Exercise 1.2.17, which is equivalent to Lebesgue measurability.

Lebesgue implies Carathéodory. By Exercise 1.2.17 ((i) implies (iii)), \(|B| = m^*(B \cap E) + m^*(B\setminus E)\) for every box \(B\). Fix \(A \subset \mathbb{R}^d\); the inequality \(m^*(A) \le m^*(A\cap E) + m^*(A \setminus E)\) is subadditivity, so assume \(m^*(A) < \infty\) and let \(\varepsilon > 0\). By the definition of \(m^*\) choose boxes \(B_n\) with \(A \subset \bigcup_n B_n\) and \(\sum_n |B_n| \le m^*(A) + \varepsilon\). Since \(A \cap E \subset \bigcup_n (B_n \cap E)\) and \(A \setminus E \subset \bigcup_n (B_n \setminus E)\), countable subadditivity and the box identity give

\begin{align*} m^*(A \cap E) + m^*(A \setminus E) &\le \sum_{n=1}^\infty \big(m^*(B_n \cap E) + m^*(B_n \setminus E)\big) \\ &= \sum_{n=1}^\infty |B_n| \le m^*(A) + \varepsilon, \end{align*}

the regrouping being legitimate as all terms are non-negative. Let \(\varepsilon \to 0\).

Disjointify: given \(E_1, E_2, \dots \in \mathcal{B}\), set

\begin{equation*} F_n := E_n \setminus \bigcup_{m=1}^{n-1} E_m \qquad (n \ge 1), \end{equation*}

the empty union being \(\emptyset\). Each \(F_n \in \mathcal{B}\), a Boolean algebra being closed under finite unions and under differences (\(A \setminus B = (A^c \cup B)^c\)). They are disjoint, since for \(n < n^{\prime}\) we have \(F_n \subset E_n\) while \(F_{n^{\prime}} \cap E_n = \emptyset\). And \(\bigcup_n F_n = \bigcup_n E_n\): the inclusion \(\subset\) is \(F_n \subset E_n\), and conversely each \(x \in \bigcup_n E_n\) lies in \(F_n\) for the least \(n\) with \(x \in E_n\). So the hypothesis puts \(\bigcup_n E_n = \bigcup_n F_n\) in \(\mathcal{B}\); with \(\emptyset \in \mathcal{B}\) and closure under complements, \(\mathcal{B}\) is a \(\sigma\)-algebra (Definition 1.4.12).

The converse is immediate: a \(\sigma\)-algebra is closed under all countable unions.

Write (P) for the countable additivity condition of Definition 1.7.7: \(\mu_0(\bigcup_n E_n) = \sum_n \mu_0(E_n)\) whenever the \(E_n \in \mathcal{B}_0\) are disjoint with union in \(\mathcal{B}_0\). In (i) and (ii) only one implication needs proof, the other being contained in Definition 1.7.7.

(i) Finite additivity follows from (P). For disjoint \(E, F \in \mathcal{B}_0\), apply (P) to \(E, F, \emptyset, \emptyset, \dots\) (disjoint, with union \(E \cup F \in \mathcal{B}_0\)):

\begin{align*} \mu_0(E \cup F) = \mu_0(E) + \mu_0(F) + \sum_{n\ge 3}\mu_0(\emptyset) = \mu_0(E) + \mu_0(F). \end{align*}

(ii) The reverse inequality is automatic for a finitely additive measure. With \(E := \bigcup_n E_n \in \mathcal{B}_0\) and \(N \ge 1\), finite additivity and monotonicity (Exercise 1.4.20(i),(ii)) give

\begin{equation*} \sum_{n=1}^{N} \mu_0(E_n) = \mu_0\Big(\bigcup_{n=1}^{N} E_n\Big) \le \mu_0(E), \end{equation*}

and letting \(N \to \infty\) yields \(\sum_n \mu_0(E_n) \le \mu_0(E)\).

(iii) Take \(X := \{1,2\}\), \(\mathcal{B}_0 := 2^X\), and

\begin{equation*} \mu_0(\emptyset) := 0, \qquad \mu_0(\{1\}) = \mu_0(\{2\}) = \mu_0(X) := 1 . \end{equation*}

Discarding empty terms, a disjoint sequence in \(\mathcal{B}_0\) has at most two non-empty terms, so the relaxed inequality reduces to three cases: union \(\emptyset\) (\(0 \le 0\)); one non-empty term \(F\) (equality); the two terms \(\{1\},\{2\}\) (\(1 \le 2\)). Yet \(\mu_0\) is not finitely additive, since \(\mu_0(\{1\}\cup\{2\}) = 1 \ne 2\), so it is not a pre-measure.

On the elementary Boolean algebra \(\mathcal{E}[\mathbb{R}^d]\) of Exercise 1.4.1 — sets that are elementary or co-elementary — put \(\mu_0(E) := m(E)\) for \(E\) elementary and \(\mu_0(E) := +\infty\) for \(E\) co-elementary. No set is both (elementary sets are bounded, co-elementary ones are not), so this is well defined, and boundedness gives two facts used below:

(A) Two co-elementary sets are never disjoint: \(E \cap F = \emptyset\) would force \(\mathbb{R}^d = E^c \cup F^c\), a union of two bounded sets. (B) A member of \(\mathcal{E}[\mathbb{R}^d]\) containing a co-elementary set is unbounded, hence co-elementary.

Finite additivity. \(\mu_0(\emptyset) = 0\), and for disjoint \(E, F\) at most one is co-elementary by (A). If both are elementary, \(\mu_0(E \cup F) = m(E) + m(F)\) by finite additivity of elementary measure (Lemma 1.1.2). If \(E\) is co-elementary, so is \(E \cup F\) by (B), and \(\mu_0(E\cup F) = +\infty = \mu_0(E) + \mu_0(F)\).

Countable additivity. Let the \(E_n \in \mathcal{E}[\mathbb{R}^d]\) be disjoint with \(E := \bigcup_n E_n \in \mathcal{E}[\mathbb{R}^d]\). Throughout we use only that \(m^* = m\) on elementary sets (Lemma 1.2.6) and that \(m^*\) is monotone and countably subadditive (Exercise 1.2.3); no property of Lebesgue measure is invoked.

(i) Some \(E_{n_0}\) is co-elementary. Then \(\sum_n \mu_0(E_n) = +\infty\), and \(E \supset E_{n_0}\) is co-elementary by (B), so \(\mu_0(E) = +\infty\).

(ii) All \(E_n\) elementary, \(E\) elementary. For each \(N\), finite additivity and monotonicity of \(m\) give \(\sum_{n\le N} m(E_n) = m(\bigcup_{n \le N}E_n) \le m(E)\); let \(N \to \infty\). Conversely

\begin{align*} m(E) = m^*\Big(\bigcup_{n=1}^\infty E_n\Big) \le \sum_{n=1}^\infty m^*(E_n) = \sum_{n=1}^\infty m(E_n). \end{align*}

(iii) All \(E_n\) elementary, \(E\) co-elementary. Then \(\mu_0(E) = +\infty\), and we must show \(S := \sum_n m(E_n) = +\infty\). Pick \(R\) with \(E^c \subset [-R,R]^d\), so \(Q_k := [-k,k]^d \setminus [-R,R]^d \subset E\) for \(k > R\), with \(m(Q_k) = (2k)^d - (2R)^d\) by finite additivity. Then

\begin{equation*} (2k)^d - (2R)^d = m^*(Q_k) \le \sum_{n=1}^\infty m^*(E_n) = S \end{equation*}

for every \(k\), so \(S = +\infty\).

These cases are exhaustive, since \(E \in \mathcal{E}[\mathbb{R}^d]\) is elementary or co-elementary.

Take \(X := \mathbb{N}\), \(\mathcal{B}_0 := 2^{\mathbb{N}}\), and

\begin{equation*} \mu_0(E) := 0 \text{ if } E \text{ is finite}, \qquad \mu_0(E) := +\infty \text{ if } E \text{ is infinite}. \end{equation*}

This is a finitely additive measure: \(\mu_0(\emptyset) = 0\), and for disjoint \(E, F\) either both are finite, so \(\mu_0(E\cup F) = 0 = 0 + 0\), or one is infinite, so both sides are \(+\infty\).

It is not a pre-measure: the singletons \(E_n := \{n\}\) are disjoint with union \(\mathbb{N} \in \mathcal{B}_0\), yet

\begin{align*} \mu_0\Big(\bigcup_{n=1}^{\infty}E_n\Big) = \mu_0(\mathbb{N}) = +\infty \ne 0 = \sum_{n=1}^{\infty}\mu_0(E_n), \end{align*}

contradicting Definition 1.7.7.

The inequality \(\mu^{\prime} \le \mu\) plus \(\sigma\)-finiteness forces equality, by a complementation argument on sets of finite measure. Here \(\mu\) is the restriction to \(\mathcal{B}\) of the outer measure

\begin{equation*} \mu^*(E) := \inf\Big\{ \sum_{n=1}^{\infty} \mu_0(F_n) : E \subset \bigcup_{n=1}^{\infty} F_n, \ F_n \in \mathcal{B}_0 \Big\} \end{equation*}

built in Theorem 1.7.8.

The inequality. Let \(E \in \mathcal{B}^{\prime}\) and let \(F_n \in \mathcal{B}_0 \subset \mathcal{B}^{\prime}\) cover \(E\). Being countably additive, \(\mu^{\prime}\) is monotone and countably subadditive (Exercises 1.4.20(i), 1.4.23(i)), so

\begin{align*} \mu^{\prime}(E) &\le \mu^{\prime}\Big(\bigcup_{n=1}^{\infty}F_n\Big) \\ &\le \sum_{n=1}^{\infty}\mu^{\prime}(F_n) = \sum_{n=1}^{\infty}\mu_0(F_n), \end{align*}

and taking the infimum over covers gives \(\mu^{\prime}(E) \le \mu^*(E)\); in particular \(\mu^{\prime} \le \mu\) on \(\mathcal{B}\cap\mathcal{B}^{\prime}\).

Reduction. Write \(X = \bigcup_n E_n\) with \(E_n \in \mathcal{B}_0\) of finite \(\mu_0\)-measure and disjointify as in Exercise 1.7.3: \(X_n := E_n \setminus \bigcup_{m<n}E_m \in \mathcal{B}_0\) are disjoint with union \(X\) and \(\mu_0(X_n) \le \mu_0(E_n) < \infty\). For \(E \in \mathcal{B}\cap\mathcal{B}^{\prime}\) the sets \(E \cap X_n\) lie in \(\mathcal{B}\cap\mathcal{B}^{\prime}\), are disjoint, and have union \(E\), so by countable additivity of both measures it suffices to treat \(F \in \mathcal{B}\cap\mathcal{B}^{\prime}\) with \(F \subset Y := X_n\).

Finite-measure case. Put \(G := Y \setminus F \in \mathcal{B}\cap\mathcal{B}^{\prime}\). Since both measures extend \(\mu_0\) and \(Y \in \mathcal{B}_0\),

\begin{equation*} \mu(F) + \mu(G) = \mu_0(Y) = \mu^{\prime}(F) + \mu^{\prime}(G) < \infty . \end{equation*}

All four terms are finite, and \(\mu^{\prime}(F) \le \mu(F)\), \(\mu^{\prime}(G) \le \mu(G)\); a strict inequality in either would make the two sides differ. So \(\mu^{\prime}(F) = \mu(F)\), and summing over \(n\) gives \(\mu(E) = \mu^{\prime}(E)\).

As in the proof of Theorem 1.7.9 we allow infinite endpoints in \([a,b)\); without this \(\mathcal{A}\) is not closed under complements, since \([0,1)^c = (-\infty,0)\cup[1,+\infty)\). With it \(\mathcal{A}\) is a Boolean algebra (Check!: use \([a,b)\cap[c,d) = [\max(a,c),\min(b,d))\) and De Morgan). Note that a non-empty \(E \in \mathcal{A}\) contains some \([a,b)\) with \(a<b\), hence is uncountable.

(i) Let the \(E_n \in \mathcal{A}\) be disjoint with \(E := \bigcup_n E_n \in \mathcal{A}\) (finite additivity is the case where all but finitely many are empty). If every \(E_n\) is empty then both sides of the additivity identity vanish; otherwise some \(E_{n_0} \ne \emptyset\), so \(E \supset E_{n_0}\) is non-empty and \(\mu_0(E) = +\infty = \sum_n \mu_0(E_n)\). So \(\mu_0\) is a pre-measure.

(ii) Each \([a,b) = [a,+\infty) \cap (-\infty,b)\) is Borel, so \(\langle\mathcal{A}\rangle \subset \mathcal{B}[\mathbb{R}]\). Conversely \((a,b) = \bigcup_n [a_n,b)\) for any reals \(a_n \downarrow a\) in \((a,b)\), and every open \(U \subset \mathbb{R}\) is a countable union of open intervals (those with rational endpoints inside \(U\)); as the open sets generate \(\mathcal{B}[\mathbb{R}]\), the reverse inclusion follows.

(iii) In the outer measure of Theorem 1.7.8, any cover of a non-empty \(E\) by sets of \(\mathcal{A}\) has a non-empty member, so \(\sum_n \mu_0(E_n) = +\infty\); hence

\begin{equation*} \mu^*(E) = \begin{cases} 0, & E = \emptyset, \\ +\infty, & E \neq \emptyset. \end{cases} \end{equation*}

Every \(A \subset \mathbb{R}\) is Carathéodory measurable for this \(\mu^*\): for \(A = \emptyset\) all three terms vanish, and otherwise one of \(A \cap E\), \(A \setminus E\) is non-empty, making both sides \(+\infty\). So \(\mathcal{B} = 2^{\mathbb{R}} \supset \mathcal{B}[\mathbb{R}]\) and \(\mu(E) = +\infty\) for every non-empty Borel \(E\).

(iv) For \(c \in (0,+\infty]\), \(c\#\) is countably additive on \(\mathcal{B}[\mathbb{R}]\) by Exercise 1.4.22(i), and \(c\#(\emptyset) = 0\); a non-empty \(E \in \mathcal{A}\) is uncountable, so \(c\#(E) = +\infty = \mu_0(E)\). Thus \(c\#\) extends \(\mu_0\) to \(\langle\mathcal{A}\rangle = \mathcal{B}[\mathbb{R}]\), while \(\#(\{0\}) = 1 \ne +\infty = \mu(\{0\})\): uniqueness fails, \(\mu_0\) not being \(\sigma\)-finite.

Here \(\mu^*\) is the outer measure of Theorem 1.7.8, \(\mathcal{B}\) its Carathéodory \(\sigma\)-algebra, \(\mu = \mu^*|_{\mathcal{B}}\), with \(\mathcal{B}_0 \subset \mathcal{B}\), \(\mu^* = \mu_0\) on \(\mathcal{B}_0\), and hence \(\langle\mathcal{B}_0\rangle \subset \mathcal{B}\). Call a countable union of sets of \(\mathcal{B}_0\) a \(\mathcal{B}_0\)-union; two such intersect in a \(\mathcal{B}_0\)-union, as \((\bigcup_m A_m)\cap(\bigcup_l B_l) = \bigcup_{m,l}(A_m \cap B_l)\). By \(\sigma\)-finiteness write \(X = \bigcup_k X_k\) with \(X_k \in \mathcal{B}_0\), \(\mu_0(X_k) < \infty\), disjointified as in Exercise 1.7.3.

(i) Finite measure. For each \(n\), pick \(F_{n,m} \in \mathcal{B}_0\) with \(E \subset F_n := \bigcup_m F_{n,m}\) and \(\sum_m \mu_0(F_{n,m}) \le \mu^*(E) + \frac1n\), so \(\mu^*(F_n) \le \mu(E) + \frac1n\) by countable subadditivity. Then \(F := \bigcap_n F_n \in \langle\mathcal{B}_0\rangle\) contains \(E\), and monotonicity in both directions gives \(\mu(F) = \mu(E) < \infty\), whence

\begin{equation*} \mu(F \setminus E) = \mu(F) - \mu(E) = 0 . \end{equation*}

Replacing \(F_n\) by \(F_1 \cap \dots \cap F_n\) makes the \(F_n\) decreasing without changing \(F\), and replacing \(F_n\) by \(F_n \cap Y = \bigcup_m (F_{n,m}\cap Y)\) for a prescribed \(Y \in \mathcal{B}_0\) with \(E \subset Y\) forces \(F \subset Y\); both keep each \(F_n\) a \(\mathcal{B}_0\)-union.

General case. Put \(E^{(k)} := E \cap X_k\), of measure at most \(\mu_0(X_k) < \infty\). The previous paragraph (with \(Y = X_k\)) supplies decreasing \(\mathcal{B}_0\)-unions \(G^{(k)}_n \subset X_k\) with \(F^{(k)} := \bigcap_n G^{(k)}_n \supset E^{(k)}\) and \(\mu(F^{(k)}\setminus E^{(k)}) = 0\). Set \(F := \bigcup_k F^{(k)} \supset E\); then \(F \setminus E \subset \bigcup_k (F^{(k)}\setminus E^{(k)})\), so \(\mu(F\setminus E) = 0\) by countable subadditivity. Finally

\begin{equation*} F = \bigcap_{n=1}^\infty F_n, \qquad F_n := \bigcup_{k=1}^\infty G^{(k)}_n, \end{equation*}

each \(F_n\) being a countable union of \(\mathcal{B}_0\)-unions, hence a \(\mathcal{B}_0\)-union. Indeed \(\subset\) is clear; conversely if \(x \in \bigcap_n F_n\), choose \(k_n\) with \(x \in G^{(k_n)}_n \subset X_{k_n}\) — disjointness of the \(X_k\) forces all \(k_n\) equal to one \(k\), so \(x \in F^{(k)} \subset F\).

(ii) Take \(F := E_1 \cup \dots \cup E_N \in \mathcal{B}_0\), where \(E_n \in \mathcal{B}_0\) cover \(E\) with \(\sum_n \mu_0(E_n) \le \mu(E) + \varepsilon/2 < \infty\) and \(N\) is chosen so that \(\sum_{n>N}\mu_0(E_n) \le \varepsilon/2\). Then \(E \setminus F \subset \bigcup_{n>N}E_n\) gives \(\mu(E\setminus F) \le \varepsilon/2\), while with \(G := \bigcup_n E_n \supset E\) of finite measure,

\begin{equation*} \mu(F \setminus E) \le \mu(G\setminus E) = \mu(G) - \mu(E) \le \tfrac{\varepsilon}{2}, \end{equation*}

the subtraction being legitimate as \(\mu(E) < \infty\). Adding, \(\mu(E \Delta F) \le \varepsilon\).

(iii) Fix \(A \subset X\) with \(\mu^*(A) < \infty\) (otherwise nothing to prove); the inequality \(\le\) in the Carathéodory identity is subadditivity. Given \(\varepsilon > 0\) choose \(F \in \mathcal{B}_0\) with \(\mu^*(E \Delta F) \le \varepsilon\), so \(\mu^*(E \setminus F), \mu^*(F\setminus E) \le \varepsilon\). Since

\begin{align*} A \cap E &\subset (A\cap F) \cup (E \setminus F), \\ A \setminus E &\subset (A \setminus F) \cup (F\setminus E), \end{align*}

subadditivity and the Carathéodory measurability of \(F \in \mathcal{B}_0 \subset \mathcal{B}\) give

\begin{align*} \mu^*(A\cap E) + \mu^*(A\setminus E) &\le \mu^*(A\cap F) + \mu^*(A\setminus F) + 2\varepsilon \\ &= \mu^*(A) + 2\varepsilon . \end{align*}

Let \(\varepsilon \to 0\); thus \(E \in \mathcal{B}\).

All four claims collapse once the \(F\)-volume is written as a difference of endpoint heights. For a non-empty interval \(I\) with endpoints \(a \le b\) (possibly infinite), set

\begin{equation*} \ell(I) := \begin{cases} F_-(a) & a \in I \\ F_+(a) & a \notin I \end{cases} \qquad r(I) := \begin{cases} F_+(b) & b \in I \\ F_-(b) & b \notin I, \end{cases} \end{equation*}

so that comparison with the four definitions of \(|I|_F\) (including \(|\{a\}|_F = F_+(a) - F_-(a)\)) gives \(|I|_F = r(I) - \ell(I)\) in every case, with \(|\emptyset|_F := 0\). Here \(\ell(I) \le r(I)\): for \(a < b\) pick \(a<y<b\) and use \(\ell(I) \le F_+(a) \le F(y) \le F_-(b) \le r(I)\), while for \(a = b\) only \(I = \{a\}\) occurs. So \(|I|_F \in [0,+\infty]\), and since \(F(x-1) \le F_-(x) \le F_+(x) \le F(x+1)\) for finite \(x\), infinities enter only through infinite endpoints. Two limit facts, both from monotonicity of \(F\):

• Lemma A: \(\inf_{c>a}F_-( c) = F_+(a)\). For \(c > a\) and \(a<y<c\), \(F_-( c) \ge F(y) \ge F_+(a)\); conversely each \(y > a\) admits \(a<c<y\) with \(F_-( c) \le F(y)\).
• Lemma B: \(\sup_{d<b}F_+(d) = F_-(b)\), symmetrically.

(a) Assume both intervals non-empty (else trivial) and relabel so \(I\) is left of \(J\), sharing the finite endpoint \(c\), which lies in exactly one of them since \(I \cup J\) is an interval. Either way \(r(I) = \ell(J) =: t\), a real number: if \(c \in I\) both equal \(F_+( c)\), if \(c \in J\) both equal \(F_-( c)\). Also \(\ell(I\cup J) = \ell(I)\) and \(r(I \cup J) = r(J)\), so

\begin{equation*} |I|_F + |J|_F = (t - \ell(I)) + (r(J) - t) = r(J) - \ell(I) = |I \cup J|_F, \end{equation*}

the cancellation of the finite \(t\) being legitimate even if \(\ell(I) = -\infty\) or \(r(J) = +\infty\). By induction, \(|I|_F = \sum_{j=1}^k |I_j|_F\) whenever disjoint intervals \(I_j\) partition an interval \(I\): order them left to right, so that each \(I_1 \cup \dots \cup I_j\) is an interval adjacent to \(I_{j+1}\).

(b) Every \(E \in \mathcal{B}_0\) is a finite disjoint union of intervals: the finite set \(P\) of finite endpoints of a representation \(E = I_1 \cup \dots \cup I_k\) cuts \(\mathbb{R}\) into finitely many atoms (points of \(P\) and the open intervals between them), and no atom straddles an endpoint, so each is inside or disjoint from every \(I_j\), hence from \(E\). Given two such decompositions \(E = \bigsqcup_j I_j = \bigsqcup_l J_l\), the intervals \(I_j \cap J_l\) partition \(I_j\) for fixed \(j\) and \(J_l\) for fixed \(l\), so by (a)

\begin{equation*} \sum_{j=1}^k |I_j|_F = \sum_{j,l}|I_j \cap J_l|_F = \sum_{l=1}^{k^{\prime}}|J_l|_F, \end{equation*}

all terms being non-negative. So \(\mu_0\) is well defined; finite additivity follows by concatenating the interval decompositions of \(E_1, \dots, E_n\) into one for \(E = E_1 \sqcup \dots \sqcup E_n\). In particular \(\mu_0(\emptyset) = 0\) and \(\mu_0\) is monotone, since \(\mu_0(B) = \mu_0(A) + \mu_0(B\setminus A)\) for \(A \subset B\) in \(\mathcal{B}_0\).

(c) Only \(\le\) needs proof (\(\ge\) is monotonicity), and we may take \(E \ne \emptyset\) with endpoints \(a \le b\). Choose \(c_n \in E\) with \(c_n := a\) if \(a \in E\) and \(c_n \downarrow a\), \(c_n > a\) otherwise; symmetrically \(d_n\) at \(b\). For large \(n\), \(K_n := [c_n,d_n] \subset E\) is compact with \(|K_n|_F = F_+(d_n) - F_-(c_n)\). If \(a \in E\) then \(F_-(c_n) = F_-(a) = \ell(E)\); if not (this includes \(a = -\infty\)) then \(F_-(c_n) \to F_+(a) = \ell(E)\) by Lemma A, and dually for \(d_n\) by Lemma B. Hence \(|K_n|_F \to r(E) - \ell(E) = |E|_F\).

(d) Dually, take \(c_n := a\) if \(a \notin E\) and \(c_n \uparrow a\), \(c_n < a\) otherwise; symmetrically \(d_n\). Then \(U_n := (c_n,d_n)\) is an open interval containing \(E\), with \(|U_n|_F = F_-(d_n) - F_+(c_n)\), and the same two lemmas give \(F_+(c_n) \to \ell(E)\), \(F_-(d_n) \to r(E)\), so \(|U_n|_F \to |E|_F\).

(Monotone means non-decreasing, as in Theorem 1.7.9.)

Part 1: \(\mu_F\) is Radon. Since \(\mu_F\) restricts the Hahn-Kolmogorov extension of the pre-measure \(\mu_0\) of Theorem 1.7.9, and splitting each covering set of \(\mathcal{B}_0\) into its component intervals leaves the sum unchanged by finite additivity,

\begin{equation*} \mu_F(E) = \inf\Big\{\sum_{n=1}^\infty |I_n|_F : E \subset \bigcup_{n=1}^\infty I_n,\ I_n \text{ intervals}\Big\}. \end{equation*}

(i) A compact \(K\) lies in some \([-N,N]\), so by (1.33) \(\mu_F(K) \le F_+(N) - F_-(-N) < \infty\), these being real numbers.

(iii) Only \(\inf_{U \supset E}\mu_F(U) \le \mu_F(E)\) needs proof, and we may assume \(\mu_F(E) < \infty\). Given \(\varepsilon > 0\), cover \(E\) by intervals \(I_n\) with \(\sum_n |I_n|_F \le \mu_F(E) + \varepsilon/2\); each \(|I_n|_F\) is then finite, so Exercise 1.7.10(d) supplies open intervals \(U_n \supset I_n\) with \(|U_n|_F \le |I_n|_F + \varepsilon 2^{-n-1}\). Then \(U := \bigcup_n U_n\) is open, contains \(E\), and

\begin{equation*} \mu_F(U) \le \sum_{n=1}^\infty |U_n|_F \le \mu_F(E) + \varepsilon . \end{equation*}

(ii) Again only \(\sup_{K\subset E}\mu_F(K) \ge \mu_F(E)\) is at issue. First let \(E \subset [-N,N]\) and put \(E^{\prime} := [-N,N]\setminus E\), of finite measure by (i). Given \(\varepsilon > 0\), outer regularity gives open \(U \supset E^{\prime}\) with \(\mu_F(U) \le \mu_F(E^{\prime}) + \varepsilon\); then \(K := [-N,N]\setminus U\) is compact with \(K \subset E\), and since \(U \cap [-N,N] = E^{\prime} \sqcup (E \cap U)\),

\begin{equation*} \mu_F(E \setminus K) = \mu_F(E \cap U) \le \mu_F(U) - \mu_F(E^{\prime}) \le \varepsilon, \end{equation*}

all quantities finite, so \(\mu_F(K) \ge \mu_F(E) - \varepsilon\). In general \(E \cap [-N,N] \uparrow E\), so \(\mu_F(E \cap [-N,N]) \to \mu_F(E)\) by Exercise 1.4.23; given \(t < \mu_F(E)\), pick \(N\) with \(\mu_F(E\cap[-N,N]) > t\) and apply the bounded case.

Part 2: every Radon measure is some \(\mu_F\). Only local finiteness is used. Define

\begin{equation*} F(x) := \begin{cases} \mu([0,x)) & \text{if } x \geq 0 \\ -\mu([x,0)) & \text{if } x < 0. \end{cases} \end{equation*}

These are finite (both sets lie in compact intervals), and \(F\) is non-decreasing (Check! on the three sign cases). Splitting \([0,b)\), \([a,0)\) or \([a,b)\) at \(0\) according as \(0 \le a\), \(b \le 0\) or \(a < 0 \le b\) gives, all measures being finite,

\begin{equation*} \mu([a,b)) = F(b) - F(a) \qquad (-\infty < a < b < \infty). \tag{$\ast$} \end{equation*}

As \(y \uparrow x\) we have \([y,x) \downarrow \emptyset\) with \(\mu([y_0,x)) < \infty\), so \(\mu([y,x)) \to 0\) by Exercise 1.4.23 and \(F_-(x) = F(x)\) by \((\ast)\); as \(y \downarrow x\) we have \([x,y) \downarrow \{x\}\), so likewise

\begin{equation*} F_+(x) = F(x) + \mu(\{x\}) = F_-(x) + \mu(\{x\}). \end{equation*}

This last display is (1.34), and with \((\ast)\) it gives \(\mu([a,b)) = F_-(b) - F_-(a)\), whence by additivity on sets of finite measure

\begin{align*} \mu([a,b]) &= \mu([a,b)) + \mu(\{b\}) = F_+(b) - F_-(a), \\ \mu((a,b]) &= \mu([a,b]) - \mu(\{a\}) = F_+(b) - F_+(a), \\ \mu((a,b)) &= \mu([a,b)) - \mu(\{a\}) = F_-(b) - F_+(a), \end{align*}

which is (1.33). By the uniqueness in Theorem 1.7.9, \(\mu = \mu_F\).

Both directions rest on (1.33)-(1.34), in which every \(F_\pm(x)\) at finite \(x\) is real (being squeezed between \(F(x-1)\) and \(F(x+1)\)), so ordinary arithmetic applies.

Sufficiency. If \(F_\pm = F^{\prime}_\pm + C\) pointwise, the constant cancels in each of the four differences of (1.33) and in \(F_+(a) - F_-(a)\) of (1.34); so \(\mu_{F^{\prime}}\) obeys (1.33)-(1.34) for \(F\), and the uniqueness in Theorem 1.7.9 gives \(\mu_{F^{\prime}} = \mu_F\).

Necessity. Evaluating both measures on \([a,b]\) and writing \(G_\pm := F_\pm - F^{\prime}_\pm\),

\begin{equation*} G_+(b) = G_-(a) \qquad \text{whenever } a < b . \end{equation*}

Fix \(b_0\) and set \(C := G_+(b_0)\). For any \(x\), taking \(a := \min(x,b_0) - 1\) gives \(G_+(x) = G_-(a) = G_+(b_0) = C\), and taking \(b := \max(x,b_0)+1\) gives \(G_-(x) = G_+(b) = G_-(b_0-1) = C\). So \(G_\pm \equiv C\).

Consequences. For \(\mu_F = \mu_{F_+} = \mu_{F_-}\), apply the criterion with \(C = 0\): \((F_-)_+ = F_+\) and \((F_+)_- = F_-\) are Lemmas A and B of Exercise 1.7.10, while \((F_+)_+ = F_+\) and \((F_-)_- = F_-\) follow by squeezing \(F_+(y)\) between \(F_+(x)\) and \(F(z)\) for \(x<y<z\), and dually.

If \(G\) is monotone with \(G = F\) on the (co-countable, hence dense) continuity set \(S\) of \(F\), then \(F_-(x) = \sup_{y<x,\, y\in S}F(y) = G_-(x)\) since each \(y<x\) is dominated by some \(y^{\prime} \in S \cap (y,x)\), and dually \(F_+ = G_+\); so \(\mu_F = \mu_G\), and the values of \(F\) at its discontinuities are irrelevant.

Finally, when \(F_+(-\infty) = 0\) and \(F_-(+\infty) = 1\): \(\mu_F(\mathbb{R}) = 1\) and \(\mu_F((-\infty,x]) = F_+(x)\).

(i) For \(F(x) = x\), continuity gives \(F_- = F_+ = F\), so (1.33)-(1.34) read: every interval with endpoints \(a<b\) has measure \(b-a\), and singletons have measure \(0\). Lebesgue measure restricted to \(\mathcal{B}[\mathbb{R}]\) is a Borel measure with exactly these values, so \(\mu_F = m\) by the uniqueness in Theorem 1.7.9.

(ii) Absolute continuity makes \(F\) continuous (Exercise 1.6.48(i)), of bounded variation on compacts (Exercise 1.6.48(ii)), hence differentiable a.e. with locally absolutely integrable derivative (Exercise 1.6.43). Exercise 1.4.49 will be applied on \((\mathbb{R},\mathcal{B}[\mathbb{R}],m)\), so fix a Borel representative: with \(Q_h(x) := (F(x+h)-F(x))/h\) continuous in \(x\) and in \(h \ne 0\), each \(\sup_{0<|h|<1/k}Q_h\) is a countable (rational \(h\)) supremum of continuous functions, so \(\overline{D} := \limsup_{h\to 0}Q_h\) and \(\underline{D}\) are Borel, the differentiability set \(D := \{\overline{D} = \underline{D} \in \mathbb{R}\}\) is Borel, and \(F^{\prime} := \overline{D}1_D\) is Borel, finite, non-negative (monotone difference quotients), and equal to the derivative off a null set. So

\begin{equation*} m_{F^{\prime}}(E) := \int_{\mathbb{R}} 1_E F^{\prime}\, dm \end{equation*}

is a Borel measure (Exercise 1.4.49). It satisfies (1.34), singletons being null, and Theorem 1.6.40 on \([a,b]\) gives

\begin{equation*} m_{F^{\prime}}([a,b]) = \int_{[a,b]}F^{\prime}(x)\, dx = F(b) - F(a) = F_+(b) - F_-(a), \end{equation*}

with the same value for \([a,b), (a,b], (a,b)\) since endpoints are null — this is (1.33), as \(F_\pm = F\). Uniqueness in Theorem 1.7.9 gives \(\mu_F(E) = \int_E F^{\prime}(x)\, dx\).

The integral identity now follows by the standard ladder. For \(f = 1_E\) it is that identity; for unsigned simple \(f = \sum_j c_j 1_{E_j}\) it follows by linearity of both integrals (Exercise 1.4.34(iii),(iv)); and for general unsigned Borel \(f\), take the standard simple \(f_n \uparrow f\) and apply monotone convergence (Theorem 1.4.44) on both \((\mathbb{R},\mathcal{B}[\mathbb{R}],\mu_F)\) and \((\mathbb{R},\mathcal{B}[\mathbb{R}],m)\), using \(f_n F^{\prime} \uparrow f F^{\prime}\):

\begin{equation*} \int_{\mathbb{R}} f \, d\mu_F = \lim_{n\to\infty}\int_{\mathbb{R}} f_n\, d\mu_F = \lim_{n\to\infty}\int_{\mathbb{R}} f_n F^{\prime}\, dm = \int_{\mathbb{R}} f F^{\prime}\, dm . \end{equation*}

Both parts verify (1.33)-(1.34) and invoke the uniqueness in Theorem 1.7.9; recall \(\delta_x(E) = 1_E(x)\) (Example 1.4.22). Part (i) is the case of part (ii) with a single jump, \(c = 1\), \(x_0 = 0\), \(\theta = 1\), so we prove (ii) and read off (i) — there \(H_- = 1_{(0,+\infty)}\), \(H_+ = 1_{[0,+\infty)}\) and \(\nu = \delta_0\).

Let \(F = \sum_n c_n J_n\) be as in Definition 1.6.30: \(c_n > 0\), \(\sum_n c_n < \infty\), and \(J_n = 0\) below \(x_n\), \(= \theta_n\) at \(x_n\), \(= 1\) above. The series converges uniformly, so \(F\) is real-valued and non-decreasing, and \((J_n)_- = 1_{(x_n,+\infty)}\), \((J_n)_+ = 1_{[x_n,+\infty)}\).

One-sided limits pass through the sum.

\begin{equation*} F_-(x) = \sum_{n\,:\,x_n < x} c_n, \qquad F_+(x) = \sum_{n\,:\,x_n \le x} c_n . \end{equation*}

For the first: \(J_n(y) \le (J_n)_-(x)\) for \(y<x\) gives \(\le\). Conversely, given \(\varepsilon > 0\), pick a finite \(N\) with \(\sum_{n\notin N}c_n \le \varepsilon\) and then \(y < x\) exceeding every \(x_n\) (\(n \in N\)) that is \(< x\); for \(n \in N\) this forces \(J_n(y) = (J_n)_-(x)\), so

\begin{equation*} F_-(x) \ge F(y) \ge \sum_{n\in N} c_n (J_n)_-(x) \ge \sum_n c_n (J_n)_-(x) - \varepsilon . \end{equation*}

The second is the mirror image, with \(y > x\) below every \(x_n > x\) (\(n \in N\)).

Comparison. By Exercise 1.4.22(i),(ii), \(\nu := \sum_n c_n \delta_{x_n}\) is a Borel measure, finite since \(\nu(\mathbb{R}) = \sum_n c_n < \infty\), and by the display above \(\nu((-\infty,x)) = F_-(x)\), \(\nu((-\infty,x]) = F_+(x)\). Finiteness licenses subtraction, so

\begin{align*} \nu(\{a\}) &= F_+(a) - F_-(a), & \nu([a,b]) &= F_+(b) - F_-(a), \\ \nu([a,b)) &= F_-(b) - F_-(a), & \nu((a,b]) &= F_+(b) - F_+(a), \end{align*}

and \(\nu((a,b)) = F_-(b) - F_+(a)\); these are (1.34) and (1.33). Hence \(\mu_F = \sum_n c_n \delta_{x_n}\).

(i) Monotonicity sandwiches \(F_-(x) \le F(x) \le F_+(x)\), with \(F_\mp(x)\) the one-sided limits; so \(F\) is continuous at \(x\) iff \(F_+(x) - F_-(x) = 0\), which by (1.34) is \(\mu_F(\{x\}) = 0\). Quantify over \(x\).

(ii) Extend the Cantor function \(F\) of Exercise 1.6.47 by \(0\) left of \(0\) and \(1\) right of \(1\); it is continuous and non-decreasing on \(\mathbb{R}\) (Exercise 1.6.47(iv)), so \(F_\pm = F\) by (i), with \(F_+(-\infty) = 0\) and \(F_-(+\infty) = 1\). Hence \(\mu_F(\mathbb{R}) = 1 - 0 = 1\) by (1.33), a probability measure. The two unbounded pieces of \(\mathbb{R}\setminus C\) have measure \(F(0) - 0 = 0\) and \(1 - F(1) = 0\); and \([0,1]\setminus C\) is the disjoint countable union of the deleted middle thirds \((a_j,b_j)\), on each of which \(F\) is constant (Exercise 1.6.47(v) makes \(F\) locally constant there, and each interval is connected), so \(\mu_F((a_j,b_j)) = F(b_j) - F(a_j) = 0\). Countable additivity gives \(\mu_F(\mathbb{R}\setminus C) = 0\), i.e. \(\mu_F( C) = 1\).

(iii) Write \(\mu := \mu_F\). Passing to the uniform limit (Exercise 1.6.47(iii)) in the recursion of Exercise 1.6.47 gives, for \(x \in [0,1]\),

\begin{equation*} F\big(\tfrac{x}{3}\big) = \tfrac12 F(x), \qquad F\big(\tfrac{x+2}{3}\big) = \tfrac12 + \tfrac12 F(x). \tag{$\star$} \end{equation*}

For Borel \(E \subset [0,1]\) put \(\nu(E) := 2\mu(\frac13 \cdot E)\) and \(\tau(E) := 2\mu(\frac13\cdot E + \frac23)\); these are Borel sets (preimages under the continuous maps \(x \mapsto 3x\), \(x \mapsto 3x-2\)), and \(\nu, \tau\) are countably additive since those maps are bijections carrying disjoint unions to disjoint unions. On a closed subinterval, \(\frac13\cdot[c,d] = [\frac{c}{3},\frac{d}{3}]\) and \(\frac13\cdot[c,d]+\frac23 = [\frac{c+2}{3},\frac{d+2}{3}]\), so \((\star)\) and \(\mu([c,d]) = F(d)-F( c)\) give

\begin{align*} \nu([c,d]) = 2\big(\tfrac12 F(d) - \tfrac12 F( c)\big) = F(d) - F( c) = \tau([c,d]) = \mu([c,d]); \end{align*}

and since all three measures are atomless (by (i), \(\nu(\{t\}) = 2\mu(\{t/3\}) = 0\), etc.), the other three interval types agree too.

Let \(\mathcal{A}\) be the Boolean algebra of finite unions of subintervals of \([0,1]\), so \(\langle\mathcal{A}\rangle\) is the Borel \(\sigma\)-algebra of \([0,1]\). By finite additivity, \(\mu, \nu, \tau\) share a common restriction \(\mu_0\) to \(\mathcal{A}\), which is a pre-measure (being the restriction of a countably additive measure) and finite, hence \(\sigma\)-finite, since \(\mu_0([0,1]) = 1\). All three are countably additive extensions of \(\mu_0\) to \(\langle\mathcal{A}\rangle\), so Exercise 1.7.7 forces \(\mu = \nu = \tau\) there, which is the pair of self-similarity identities.

Take \(\delta := \eta/2\), where \(|f(s)-f(t)| \le \varepsilon^{\prime} := \varepsilon/(2L+1)\) for \(|s-t| \le \eta\) by uniform continuity of \(f\) on \([a,b]\), and \(L := \mu_F([a,b]) = F(b) - F(a) < \infty\) by (1.33) and the continuity of \(F\) at \(a, b\) (Exercise 1.7.15(i)). The same continuity gives \(G(a) = F(a)\), \(G(b) = F(b)\) for \(G := F_+ \ge F\). Note \(f\) is bounded and Borel, so \(\int_{[a,b]}f\,d\mu_F\) is well defined.

Fix a partition of mesh \(\le \delta\) with tags, write \(S\) for its Riemann-Stieltjes sum and \(S^{\prime} := \sum_i f(t_i^*)(G(t_i) - G(t_{i-1}))\).

Replacing \(F\) by \(G\) costs \(\varepsilon^{\prime}L\). With \(h := F - G\), which vanishes at \(t_0 = a\) and \(t_n = b\), summation by parts gives

\begin{align*} S - S^{\prime} &= \sum_{i=1}^n f(t_i^*)(h(t_i) - h(t_{i-1})) \\ &= \sum_{i=1}^{n-1} h(t_i)\big(f(t_i^*) - f(t_{i+1}^*)\big). \end{align*}

Here \(t_i^*, t_{i+1}^* \in [t_{i-1},t_{i+1}]\), of length \(\le 2\delta = \eta\), so the bracket is at most \(\varepsilon^{\prime}\); and \(|h(t_i)| \le F_+(t_i) - F_-(t_i) = \mu_F(\{t_i\})\) by (1.34), while \(\sum_{i=1}^{n-1}\mu_F(\{t_i\}) = \mu_F(\{t_1,\dots,t_{n-1}\}) \le L\).

\(S^{\prime}\) is a step-function integral. With \(I_1 := [t_0,t_1]\) and \(I_i := (t_{i-1},t_i]\) for \(i \ge 2\) — disjoint Borel sets covering \([a,b]\) — (1.33) gives \(\mu_F(I_i) = G(t_i) - G(t_{i-1})\) for all \(i\), using \(F_-(t_0) = G(t_0)\) for \(i = 1\). So \(g := \sum_i f(t_i^*)1_{I_i}\) has integral \(S^{\prime}\).

\(g\) approximates \(f\). Each \(x \in I_i\) lies within \(\delta \le \eta\) of \(t_i^*\), so \(|f - g| \le \varepsilon^{\prime}\) on \([a,b]\) and

\begin{align*} \Big|\int_{[a,b]}f\,dF - S^{\prime}\Big| \le \int_{[a,b]}|f-g|\,d\mu_F \le \varepsilon^{\prime}L . \end{align*}

Adding the first and third estimates, \(|S - \int_{[a,b]}f\,dF| \le 2\varepsilon^{\prime}L \le \varepsilon\), with \(\delta\) independent of the partition and tags.

Tag the right endpoint in one Riemann-Stieltjes sum and the left in the other, so that the pair telescopes exactly. For \(a < b\) (the case \(a = b\) being trivial, both measures being atomless by Exercise 1.7.15(i)) take \(t_i := a + \frac{i}{n}(b-a)\) and

\begin{equation*} S_n := \sum_{i=1}^n F(t_i)\big(G(t_i) - G(t_{i-1})\big), \quad T_n := \sum_{i=1}^n G(t_{i-1})\big(F(t_i) - F(t_{i-1})\big), \end{equation*}

legitimate tagged sums with \(t_i^* = t_i\) and \(t_i^* = t_{i-1}\) respectively. The cross terms cancel:

\begin{align*} S_n + T_n &= \sum_{i=1}^n \big(F(t_i)G(t_i) - F(t_{i-1})G(t_{i-1})\big) \\ &= F(b)G(b) - F(a)G(a), \end{align*}

exactly, for every \(n\). Exercise 1.7.16 applies to both pairings (each integrand is continuous on \([a,b]\), each integrator is non-decreasing and continuous everywhere, in particular at \(a\) and \(b\)), so given \(\varepsilon > 0\), all \(n\) with mesh \((b-a)/n\) below the two resulting thresholds satisfy

\begin{align*} \Big|\int_{[a,b]} F\,dG + \int_{[a,b]} G\,dF - \big(F(b)G(b) - F(a)G(a)\big)\Big| \le 2\varepsilon . \end{align*}

The left side is independent of \(\varepsilon\), so it vanishes; rearrange.

Write \(\mathcal{P} := \{E \times F : E \in \mathcal{B}_X, F \in \mathcal{B}_Y\}\) for the measurable rectangles; by definition \(\mathcal{B}_X\times\mathcal{B}_Y = \langle \pi_X^*(\mathcal{B}_X)\cup\pi_Y^*(\mathcal{B}_Y)\rangle\).

(i) \(\langle\mathcal{P}\rangle = \mathcal{B}_X\times\mathcal{B}_Y\). Indeed \(\pi_X^*(\mathcal{B}_X)\cup\pi_Y^*(\mathcal{B}_Y) \subset \mathcal{P}\) (take the second factor \(Y\), resp. the first \(X\)), giving \(\subset\) one way; and \(E \times F = (E\times Y)\cap(X\times F)\) puts \(\mathcal{P}\) inside \(\mathcal{B}_X\times\mathcal{B}_Y\), giving the other. Hence any \(\sigma\)-algebra containing \(\mathcal{P}\) contains \(\langle\mathcal{P}\rangle = \mathcal{B}_X\times\mathcal{B}_Y\), which is the “coarsest” formulation.

(ii) By Remark 1.4.33, \(\pi_X\) is a measurable morphism from \((X\times Y,\mathcal{B})\) exactly when \(\pi_X^*(\mathcal{B}_X) \subset \mathcal{B}\), and likewise for \(\pi_Y\); so both are morphisms exactly when \(\mathcal{B}\) contains \(\pi_X^*(\mathcal{B}_X)\cup\pi_Y^*(\mathcal{B}_Y)\), i.e. contains its generated \(\sigma\)-algebra \(\mathcal{B}_X\times\mathcal{B}_Y\) — which itself qualifies.

(iii) \(\mathcal{S} := \{E \subset X \times Y : E_x \in \mathcal{B}_Y\}\) is a \(\sigma\)-algebra, slicing commuting with complements and countable unions, and it contains \(\mathcal{P}\), since \((E\times F)_x\) is \(F\) or \(\emptyset\). By (i), \(\mathcal{B}_X\times\mathcal{B}_Y \subset \mathcal{S}\). The claim for \(E^y\) is symmetric.

(iv) For Borel \(U \subset [0,+\infty]\),

\begin{equation*} f_x^{-1}(U) = \big(f^{-1}(U)\big)_x \in \mathcal{B}_Y \end{equation*}

by (iii), since \(f^{-1}(U) \in \mathcal{B}_X\times\mathcal{B}_Y\). Symmetrically for \(f^y\).

(v) Let \(\mathcal{S}\) be the collection of \(E \subset X\times Y\) for which some at most countable \(\mathcal{A}_E \subset 2^Y\) has \(\{E_x : x \in X\} \subset \langle\mathcal{A}_E\rangle\). It contains \(\mathcal{P}\) (take \(\mathcal{A}_{E\times F} := \{F\}\), since every slice is \(F\) or \(\emptyset\)), is closed under complements (slices complement, \(\mathcal{A}_E\) unchanged) and under countable unions (take \(\mathcal{A}_E := \bigcup_n \mathcal{A}_{E_n}\), still countable, and \(E_x = \bigcup_n (E_n)_x\)). So \(\mathcal{B}_X\times\mathcal{B}_Y \subset \mathcal{S}\) by (i).

For the cardinality, \(|\{E_x : x \in X\}| \le |\langle\mathcal{A}_E\rangle|\). A finite \(\mathcal{A}_E\) generates a finite \(\sigma\)-algebra; a countably infinite one gives, by Exercise 1.4.16,

\begin{equation*} \big|\langle\mathcal{A}_E\rangle\big| \le \aleph_0^{\aleph_0} = 2^{\aleph_0} = \mathfrak{c}, \end{equation*}

since \(2^{\aleph_0} \le \aleph_0^{\aleph_0} \le (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0}\). Either way at most \(\mathfrak{c}\) distinct slices.

Throughout, Exercise 1.7.18(i) identifies \(\mathcal{B}_X\times\mathcal{B}_Y\) as the \(\sigma\)-algebra generated by the measurable rectangles.

(i) The only rectangles are \(\emptyset\) and \(X\times Y\), and \(\{\emptyset, X\times Y\}\) is already a \(\sigma\)-algebra.

(ii) As printed this is false, part (vii) being a counterexample: with \(X, Y\) as there, \(2^X\times 2^Y\) atomic with non-empty atoms \((C_\gamma)\) would force each singleton \(\{x\}\times\{y\}\) — a rectangle, so in the product — to be a union of disjoint non-empty atoms, hence to be an atom; the atoms would then be exactly the singletons and \(2^X\times2^Y = 2^{X\times Y}\), contradiction.

The corrected claim: if \(\mathcal{B}_X = \mathcal{A}((A_\alpha)_{\alpha\in I})\) and \(\mathcal{B}_Y = \mathcal{A}((B_\beta)_{\beta\in J})\) with non-empty atoms and \(I\) at most countable, then \(\mathcal{B}_X\times\mathcal{B}_Y = \mathcal{C} := \mathcal{A}((A_\alpha\times B_\beta))\), a \(\sigma\)-algebra by Exercise 1.4.10 since the \(A_\alpha\times B_\beta\) partition \(X \times Y\). For \(\subset\): \(E \times F = \bigcup_{I^{\prime}\times J^{\prime}} A_\alpha \times B_\beta \in \mathcal{C}\) whenever \(E = \bigcup_{I^{\prime}}A_\alpha\), \(F = \bigcup_{J^{\prime}}B_\beta\). For \(\supset\): given \(S = \bigcup_{(\alpha,\beta)\in K}A_\alpha\times B_\beta\), put \(F_\alpha := \bigcup_{\beta : (\alpha,\beta)\in K}B_\beta \in \mathcal{B}_Y\); as the \(A_\alpha\) partition \(X\),

\begin{equation*} S = \bigcup_{\alpha\in I}\big(A_\alpha \times F_\alpha\big), \end{equation*}

an at most countable union of rectangles.

(iii) Finite \(\sigma\)-algebras are atomic (Exercise 1.4.4), with finitely many atoms \(A_1,\dots,A_n\) and \(B_1,\dots,B_m\); (ii) applies, so \(\mathcal{B}_X\times\mathcal{B}_Y\) is atomic with the \(nm\) atoms \(A_i \times B_j\), hence has at most \(2^{nm}\) elements.

(iv) \(\subset\): the projections \(\pi_1, \pi_2\) are continuous, hence Borel morphisms (the sets with Borel preimage form a \(\sigma\)-algebra containing the open sets; Exercise 1.4.14(i)), so \(E \times F = \pi_1^{-1}(E)\cap\pi_2^{-1}(F)\) is Borel in \(\mathbb{R}^{d+d^{\prime}}\). \(\supset\): every open \(U \subset \mathbb{R}^{d+d^{\prime}}\) is a countable union of rational boxes \(B \times B^{\prime}\), each a measurable rectangle, so the product \(\sigma\)-algebra contains all open sets; apply Exercise 1.4.14(i).

(v) Slicing a non-measurable set out of a null hyperplane contradicts Exercise 1.7.18(iii). The proof of Proposition 1.2.18, run with the countable dense subgroup \(\mathbb{Q}^d \le \mathbb{R}^d\) — choose one representative in \([0,1]^d\) per coset, so the disjoint translates \(A + q\), \(q \in \mathbb{Q}^d\cap[-1,1]^d\), cover \([0,1]^d\) inside \([-1,2]^d\), making \(m^d(A)\) neither \(0\) nor positive — gives a non-measurable \(A \subset [0,1]^d\). The hyperplane \(H := \mathbb{R}^d\times\{0\}\) is null (cover \([-n,n]^d\times\{0\}\) by boxes of volume \((2n)^d(2\varepsilon)^{d^{\prime}}\) and use countable subadditivity), so every subset of \(H\) is Lebesgue measurable (Lemma 1.2.13(iii)); in particular \(E := A \times \{0\}\). Were the product \(\sigma\)-algebra all of \(\mathcal{L}[\mathbb{R}^{d+d^{\prime}}]\), Exercise 1.7.18(iii) at \(y = 0\) would make \(E^0 = A\) Lebesgue measurable.

(vi) Write \(\mathcal{P} := \mathcal{L}[\mathbb{R}^d]\times\mathcal{L}[\mathbb{R}^{d^{\prime}}]\). By Exercise 1.2.22(ii), rectangles of Lebesgue sets are Lebesgue measurable with \(m^{d+d^{\prime}}(E\times F) = m^d(E)m^{d^{\prime}}(F)\), so \(\mathcal{P} \subset \mathcal{L}[\mathbb{R}^{d+d^{\prime}}]\) and \(m^{d+d^{\prime}}|_{\mathcal{P}}\) obeys (1.36); \(\sigma\)-finiteness of both factors and the uniqueness in Proposition 1.7.11 then give

\begin{equation*} m^d\times m^{d^{\prime}} = m^{d+d^{\prime}}\big|_{\mathcal{P}} . \tag{$\dagger$} \end{equation*}

Let \(\overline{\mathcal{P}}\) be the completion. Any \(S \Delta N \in \overline{\mathcal{P}}\) (Exercise 1.4.26) has \(N\) contained in a \(\mathcal{P}\)-null set, which by \((\dagger)\) is Lebesgue null, so \(N\) and hence \(S\Delta N\) are Lebesgue measurable. Conversely, given Lebesgue measurable \(E\), outer regularity supplies Borel \(F \subset E \subset G\) with \(m^{d+d^{\prime}}(G\setminus F) = 0\) (take \(G := \bigcap_n U_n\) for open \(U_n \supset E\) with \((m^{d+d^{\prime}})^*(U_n\setminus E) \le 1/n\), and \(F\) the analogous set for \(E^c\)); by (iv) both lie in \(\mathcal{B}[\mathbb{R}^{d+d^{\prime}}] \subset \mathcal{P}\), and \(E = F \cup (E\setminus F)\) with \(E \setminus F \subset G\setminus F\) a \(\mathcal{P}\)-subnull set, so \(E \in \overline{\mathcal{P}}\). The measures agree since the completed measure of \(E\) is \((m^d\times m^{d^{\prime}})(F) = m^{d+d^{\prime}}(E)\).

(vii) Take \(X = Y := 2^{\mathbb{R}}\), so \(|X| = 2^{\mathfrak{c}} > \mathfrak{c}\), with the discrete \(\sigma\)-algebras. The diagonal \(E := \{(x,x)\}\) has slices \(E_x = \{x\}\), pairwise distinct, so it has \(2^{\mathfrak{c}}\) slices; Exercise 1.7.18(v) caps sets in \(2^X\times2^Y\) at \(\mathfrak{c}\) slices, so \(E \notin 2^X\times2^Y\) although \(E \in 2^{X\times Y}\).

(viii) Say \(X\) is at most countable. Any \(E \subset X\times Y\) is \(E = \bigcup_{x\in X}\{x\}\times E_x\), an at most countable union of rectangles, so \(2^X\times2^Y = 2^{X\times Y}\).

In both parts, exhibit a measure obeying the base-times-height law (1.36) and invoke the uniqueness in Proposition 1.7.11.

(i) \(\delta_{x_0}\times\delta_{y_0} = \delta_{(x_0,y_0)}\). Both factors are probability measures, hence \(\sigma\)-finite, and \((x_0,y_0) \in E\times F\) iff \(x_0 \in E\) and \(y_0 \in F\), so

\begin{equation*} \delta_{(x_0,y_0)}(E\times F) = \delta_{x_0}(E)\,\delta_{y_0}(F), \end{equation*}

which is (1.36).

(ii) \(\#_X \times \#_Y = \#_{X\times Y}|_{\mathcal{B}_X\times\mathcal{B}_Y}\). We read the exercise with the singletons measurable — countability of \(X\) alone does not give \(\sigma\)-finiteness (take \(X = \mathbb{N}\) with \(\mathcal{B}_X = \{\emptyset, X\}\), where every measure value is \(0\) or \(+\infty\)) — so that \(X = \bigcup_{x}\{x\}\) exhibits \(\sigma\)-finiteness, and likewise for \(Y\). Then \(|E \times F| = |E||F|\), with the convention \(0\cdot\infty = 0\):

• (a) \(E\) or \(F\) empty: both sides \(0\).
• (b) both non-empty and finite: elementary counting.
• (c) both non-empty, one infinite: that one is countably infinite, so \(E\times F\) contains a copy of it and both sides are \(+\infty\).

Both bracketings generate the same \(\sigma\)-algebra \(\mathcal{D} := \langle\mathcal{R}\rangle\), where \(\mathcal{R}\) is the family of boxes \(E\times F\times G\) (\(E \in \mathcal{B}_X\), \(F\in\mathcal{B}_Y\), \(G\in\mathcal{B}_Z\)), and they then carry the same measure.

The \(\sigma\)-algebras. Write \(\mathcal{C} := (\mathcal{B}_X\times\mathcal{B}_Y)\times\mathcal{B}_Z\), generated by the \(S \times G\) with \(S \in \mathcal{B}_X\times\mathcal{B}_Y\) (Exercise 1.7.18(i)). Each box is such a generator, so \(\mathcal{D} \subset \mathcal{C}\). Conversely, fix \(G\) and set \(\mathcal{S}_G := \{S \subset X\times Y : S \times G \in \mathcal{D}\}\). This is a \(\sigma\)-algebra, since

\begin{align*} \big((X\times Y)\setminus S\big)\times G &= (X\times Y\times G)\setminus(S\times G), \\ \Big(\bigcup_n S_n\Big)\times G &= \bigcup_n (S_n\times G), \end{align*}

and it contains every rectangle \(E \times F\) because \((E\times F)\times G \in \mathcal{R}\); so \(\mathcal{B}_X\times\mathcal{B}_Y \subset \mathcal{S}_G\) by Exercise 1.7.18(i), i.e. \(\mathcal{C} \subset \mathcal{D}\). By symmetry \(\mathcal{B}_X\times(\mathcal{B}_Y\times\mathcal{B}_Z) = \mathcal{D}\) too.

The measures. With \(X = \bigcup_n X_n\), \(Y = \bigcup_m Y_m\), \(Z = \bigcup_k Z_k\) of finite measure, (1.36) makes \(X \times Y = \bigcup_{n,m}X_n\times Y_m\) \(\sigma\)-finite, so Proposition 1.7.11 applies twice and \(\mu^{\prime} := (\mu_X\times\mu_Y)\times\mu_Z\), \(\mu^{\prime\prime} := \mu_X\times(\mu_Y\times\mu_Z)\) are both defined on \(\mathcal{D}\). Two applications of (1.36) give

\begin{align*} \mu^{\prime}(E\times F\times G) = \mu_X(E)\mu_Y(F)\mu_Z(G) = \mu^{\prime\prime}(E\times F\times G). \end{align*}

Let \(\mathcal{B}_0\) be the finite unions of boxes: a Boolean algebra, since boxes are closed under intersection and

\begin{align*} (E\times F\times G)^c &= \big((X\setminus E)\times Y\times Z\big) \\ &\quad \cup\ \big(E\times(Y\setminus F)\times Z\big) \ \cup\ \big(E\times F\times(Z\setminus G)\big), \end{align*}

with \(\langle\mathcal{B}_0\rangle = \mathcal{D}\). Every \(S = \bigcup_{i\le k}E_i\times F_i\times G_i\) in \(\mathcal{B}_0\) is a finite disjoint union of boxes: take the atoms \(a_u, b_v, c_w\) of the finite algebras generated by the \(E_i\), \(F_i\), \(G_i\) (Exercise 1.4.4); the boxes \(a_u\times b_v\times c_w\) are disjoint and each \(E_i\times F_i\times G_i\), hence \(S\), is a union of some of them. So finite additivity makes \(\mu^{\prime}\) and \(\mu^{\prime\prime}\) agree on \(\mathcal{B}_0\); their common restriction \(\mu_0\) is a pre-measure (restriction of a countably additive measure) and \(\sigma\)-finite, since \(X\times Y\times Z = \bigcup_{n,m,k}X_n\times Y_m\times Z_k\) with \(\mu_0(X_n\times Y_m\times Z_k) < \infty\). Both \(\mu^{\prime}\) and \(\mu^{\prime\prime}\) are countably additive extensions of \(\mu_0\) to \(\langle\mathcal{B}_0\rangle = \mathcal{D}\), so Exercise 1.7.7 gives \(\mu^{\prime} = \mu^{\prime\prime}\).

Write \(\mathcal{B} := \mathcal{L}([0,1])\times 2^{[0,1]}\). Counting measure on \(Y\) is not \(\sigma\)-finite: sets of finite counting measure are finite, and \([0,1]\) is uncountable.

(i) \(E = \bigcap_{n\ge 1}E_n\), where \(E_n := \bigcup_{j=1}^n I_{n,j}\times I_{n,j}\) for the partition of \([0,1]\) into the \(n\) intervals \(I_{n,j}\) of length \(1/n\). Each \(I_{n,j}\times I_{n,j} \in \mathcal{B}\), so \(E \in \mathcal{B}\). Indeed \((x,x) \in E_n\) always, while \((x,y) \in E_n\) forces \(|x-y| \le 1/n\).

(ii) \(f_x = 1_{\{x\}}\) has \(\int_Y f_x\, d\# = 1\), so the iterated integral is \(\int_X 1\, dm = 1\).

(iii) \(f^y = 1_{\{y\}}\) has \(\int_X f^y\, dm = 0\), so the iterated integral is \(0\). No contradiction with Tonelli (Theorem 1.7.15): \(Y\) is not \(\sigma\)-finite.

(iv) Take \(\mu_2(S) := \sum_{y\in[0,1]} m(S^y)\) — the second iterated integral, the sum over the index set agreeing with the counting-measure integral by Exercise 1.4.40 — and \(\mu_1 :=\) the Hahn-Kolmogorov extension of the elementary product pre-measure. Both obey (1.36), and \(\mu_2(E) = 0 \ne +\infty = \mu_1(E)\).

\(\mu_2\). It is well defined: \(S^y \in \mathcal{L}([0,1])\) by Exercise 1.7.18(iii), and every function on \(Y\) is measurable. It is a measure: for disjoint \(S_n\) with union \(S\), each \(S^y = \bigsqcup_n S_n^y\), so countable additivity of \(m\) and interchange of two unsigned sums (Exercise 0.0.2) give

\begin{equation*} \mu_2(S) = \sum_{y}\sum_{n} m(S_n^y) = \sum_{n}\sum_{y} m(S_n^y) = \sum_n \mu_2(S_n). \end{equation*}

And \((A\times B)^y\) is \(A\) for \(y \in B\), else \(\emptyset\), so \(\mu_2(A\times B) = m(A)\#(B)\) (correct in the degenerate cases too, with \(0\cdot\infty = 0\)). Part (iii) gives \(\mu_2(E) = 0\).

\(\mu_1\). Let \(\mathcal{B}_0\) be the Boolean algebra of finite unions of products \(A\times B\), with \(\mu_0(\bigsqcup_j A_j\times B_j) := \sum_j m(A_j)\#(B_j)\). The proof in Proposition 1.7.11 that \(\mu_0\) is a well-defined pre-measure uses only finite additivity of \(m\) and \(\#\) plus two applications of monotone convergence (Theorem 1.4.44); \(\sigma\)-finiteness enters there only at the final uniqueness step, via Exercise 1.7.7. So Theorem 1.7.8 extends \(\mu_0\) to a measure on a \(\sigma\)-algebra containing \(\langle\mathcal{B}_0\rangle = \mathcal{B}\) (Exercise 1.7.18(i)); restrict it to \(\mathcal{B}\) to get \(\mu_1\), which obeys (1.36) by construction.

To compute \(\mu_1(E) = \mu_0^*(E)\): refining covers into countable covers by products (which does not raise the total), take \(E \subset \bigcup_n A_n\times B_n\). Every \((x,x)\) lies in some \(A_n\times B_n\), so \([0,1] = \bigcup_n (A_n \cap B_n)\) and countable subadditivity of \(m^*\) gives some \(n\) with \(m^*(A_n \cap B_n) > 0\) (the sets \(A_n\cap B_n\) need not be measurable). For that \(n\), \(m(A_n) \ge m^*(A_n\cap B_n) > 0\), and \(A_n\cap B_n\) is uncountable (countable sets are null), so \(\#(B_n) = +\infty\) and the cover has total mass \(\mu_0(A_n\times B_n) = +\infty\). Hence \(\mu_1(E) = +\infty\).

Take the disjoint intervals \(I_n := \left( \tfrac{1}{n+1}, \tfrac{1}{n} \right]\) partitioning \((0,1]\), the normalised bumps \(g_n := \tfrac{1}{|I_n|} 1_{I_n}\) (so \(\int_{[0,1]} g_n = 1\) for \(n \geq 1\)), \(g_0 := 0\), and \[ f(x,y) := \sum_{n=1}^{\infty} \big( g_n(x) - g_{n+1}(x) \big) g_n(y) . \] At any point at most one term is nonzero, since \(y\) lies in at most one \(I_M\); so \(f\) is real valued, and being the pointwise limit of the partial sums – finite combinations of indicators of the Borel rectangles \(I_j \times I_n\) – it is Borel measurable (Exercise 1.4.29(vi)). Explicitly \(f(x,y) = 0\) when \(x = 0\) or \(y = 0\), while

\begin{equation*} \begin{aligned} f(x,y) &= \tfrac{1}{|I_N|}\big( g_N(y) - g_{N-1}(y) \big) \qquad (x \in I_N), \\ f(x,y) &= \tfrac{1}{|I_M|}\big( g_M(x) - g_{M+1}(x) \big) \qquad (y \in I_M), \end{aligned} \end{equation*}

since \(g_n(x) \neq 0\) only for \(n = N\) and \(g_{n+1}(x) \neq 0\) only for \(n = N-1\). Every slice is thus bounded and Borel, hence absolutely integrable, and using \(\int g_n = 1\) for \(n \geq 1\) together with \(g_0 = 0\), \[ \int_{[0,1]} f(x,y)\, dy = 2 \cdot 1_{I_1}(x) , \qquad \int_{[0,1]} f(x,y)\, dx = 0 . \] Both outer integrands are bounded and Borel, hence absolutely integrable, and \(|I_1| = \tfrac12\), so

\begin{equation*} \int_{[0,1]} \left( \int_{[0,1]} f(x,y)\, dy \right) dx = 2 |I_1| = 1 \neq 0 = \int_{[0,1]} \left( \int_{[0,1]} f(x,y)\, dx \right) dy . \end{equation*}

Both directions are slicing arguments; write \(\Sigma_{\leq} := \{(x,t) : 0 \leq t \leq f(x)\}\) and \(\Sigma_{<} := \{(x,t) : 0 \leq t < f(x)\}\). Since \((\mathbb{R},\mathcal{B}[\mathbb{R}],m)\) is \(\sigma\)-finite (\(\mathbb{R} = \bigcup_n [-n,n]\)) and \((X,\mathcal{B},\mu)\) is so by hypothesis, \(\mu \times m\) exists (Proposition 1.7.11) and Tonelli’s theorem (Theorem 1.7.15) is available.

(i) \(f\) measurable \(\Rightarrow\) \(\Sigma_{\leq}, \Sigma_{<}\) product measurable. By density of \(\mathbb{Q}\) (for the first, pick a rational \(q\) with \(f(x) < q < t\), legitimate as \(f(x) < t < +\infty\); for the second, \(t < q < f(x)\), legitimate whether or not \(f(x) = +\infty\)),

\begin{equation*} \{ t > f(x) \} = \bigcup_{q \in \mathbb{Q}} \{ f < q \} \times (q, +\infty) , \qquad \{ t < f(x) \} = \bigcup_{q \in \mathbb{Q}} \{ f > q \} \times (-\infty, q) , \end{equation*}

countable unions of measurable rectangles, since \(\{f<q\}, \{f>q\} \in \mathcal{B}\) (Exercise 1.4.29(i),(iii)). Complementing the first, then intersecting each with \(X \times [0,+\infty)\), puts \(\Sigma_{\leq}\) and \(\Sigma_{<}\) in \(\mathcal{B} \times \mathcal{B}[\mathbb{R}]\).

(ii) \(\Sigma_{\leq}\) or \(\Sigma_{<}\) product measurable \(\Rightarrow\) \(f\) measurable. Horizontal slices of a product measurable set lie in \(\mathcal{B}\) (Exercise 1.7.18(iii)), and for \(\lambda \geq 0\) \[ (\Sigma_{\leq})^{\lambda} = \{ f \geq \lambda \} , \qquad (\Sigma_{<})^{\lambda} = \{ f > \lambda \} . \] In the second case this gives \(\{f > \lambda\} \in \mathcal{B}\) directly for \(\lambda \geq 0\); in the first, \(\{f > \lambda\} = \bigcup_{n} \{ f \geq \lambda + \tfrac1n \} \in \mathcal{B}\). As \(\{f > \lambda\} = X\) for \(\lambda < 0\), \(f\) is measurable by Exercise 1.4.29(i).

The area formula. Tonelli applied to \(1_{\Sigma}\) integrates the vertical slices,

\begin{equation*} (\mu \times m)(\Sigma) = \int_X \left( \int_{\mathbb{R}} 1_{\Sigma}(x,t)\, dm(t) \right) d\mu(x) = \int_X m(\Sigma_x) \, d\mu(x) , \end{equation*}

and \((\Sigma_{\leq})_x\) is \([0,f(x)]\) or \([0,+\infty)\), and \((\Sigma_{<})_x\) is \([0,f(x))\) or \([0,+\infty)\), according as \(f(x) < +\infty\) or not; in all four cases \(m(\Sigma_x) = f(x)\). Hence \[ (\mu \times m)(\Sigma_{\leq}) = (\mu \times m)(\Sigma_{<}) = \int_X f(x)\, d\mu(x) . \]

Slice the regions \(\Sigma_{\leq} := \{0 \leq t \leq f(x)\}\) and \(\Sigma_{<} := \{0 \leq t < f(x)\}\) of Exercise 1.7.24 horizontally rather than vertically. Both factor spaces are \(\sigma\)-finite, so \(\mu \times m\) exists (Proposition 1.7.11) and Tonelli’s theorem (Theorem 1.7.15) applied to \(1_{\Sigma}\), integrating in \(x\) first, gives

\begin{equation*} (\mu \times m)(\Sigma) = \int_{\mathbb{R}} \left( \int_X 1_{\Sigma}(x,t)\, d\mu(x) \right) dm(t) = \int_{\mathbb{R}} \mu(\Sigma^t) \, dm(t) , \end{equation*}

Tonelli also supplying measurability of \(t \mapsto \mu(\Sigma^t)\). The horizontal slices are empty for \(t < 0\), while for \(t \geq 0\) \[ (\Sigma_{\leq})^t = \{ x \in X : f(x) \geq t \} , \qquad (\Sigma_{<})^t = \{ x \in X : f(x) > t \} , \] and \((\mu \times m)(\Sigma_{\leq}) = (\mu \times m)(\Sigma_{<}) = \int_X f \, d\mu\) by Exercise 1.7.24. Hence

\begin{equation*} \int_X f\, d\mu = \int_{[0,+\infty)} \mu(\{ f \geq \lambda \})\, d\lambda = \int_{[0,+\infty)} \mu(\{ f > \lambda \})\, d\lambda , \end{equation*}

and since \(\{+\infty\}\) is Lebesgue null, \([0,+\infty)\) may be replaced by \([0,+\infty]\) throughout, which is the claim.

The density argument runs on the contraction bound \(\|h * P_t\|_{L^1} \leq \|h\|_{L^1}\). Throughout, the linear change of variables \(x \mapsto tx\) (Exercise 1.3.16, \(|\det| = t^d\)) and translation and reflection invariance (Exercises 1.3.15, 1.3.16, \(|\det(-\mathrm{id})| = 1\)) give

\begin{equation*} \begin{aligned} \int_{\mathbb{R}^d} P_t &= \int_{\mathbb{R}^d} P = 1, \qquad \int_{\mathbb{R}^d} P_t(x-y)\, dx = 1, \\ \int_{|y|>\delta} P_t(y)\, dy &= \int_{|s| > \delta/t} P(s)\, ds \longrightarrow 0 \qquad (\ast) \end{aligned} \end{equation*}

as \(t \to 0\), for each fixed \(\delta > 0\): indeed \(\int_{|s| \leq n} P \uparrow \int P = 1 < \infty\) by monotone convergence (Theorem 1.4.44), so \(\int_{|s|>n} P \to 0\), and \(R \mapsto \int_{|s|>R} P\) is non-increasing.

(1) The bound. For absolutely integrable \(h\), the function \((x,y) \mapsto |h(y)| P_t(x-y)\) is unsigned and \(\mathcal{L}[\mathbb{R}^d] \times \mathcal{L}[\mathbb{R}^d]\)-measurable: \(P\) is Borel, its superlevel sets \(\{P > \lambda\}\) being balls (or \(\emptyset\), or \(\mathbb{R}^d\)) by radiality and radial monotonicity, so \(P_t(x-y)\) is Borel on \(\mathbb{R}^{2d}\) as a composition with the continuous \((x,y) \mapsto x-y\), and \(\mathcal{B}[\mathbb{R}^{2d}] = \mathcal{B}[\mathbb{R}^d] \times \mathcal{B}[\mathbb{R}^d]\) (Exercise 1.7.19(iv)); while \((x,y) \mapsto |h(y)|\) is measurable through the projection (Exercise 1.7.18(ii)). Both factors being \(\sigma\)-finite, Tonelli’s theorem (Theorem 1.7.15) gives

\begin{equation*} \int \left( \int |h(y)| P_t(x-y) \, dy \right) dx = \int |h(y)| \left( \int P_t(x-y)\, dx \right) dy = \|h\|_{L^1} < \infty . \end{equation*}

So the inner integral is finite for a.e. \(x\), i.e. \(h * P_t\) is defined a.e.; \((x,y) \mapsto h(y)P_t(x-y)\) is absolutely integrable, so Fubini’s theorem (Theorem 1.7.21; both factors are complete and \(\sigma\)-finite) makes \(h * P_t\) measurable and absolutely integrable, and by the triangle inequality \[ \| h * P_t \|_{L^1} \leq \|h\|_{L^1} . \qquad (1) \]

(2) Convergence for \(g\) continuous and supported in \(\overline{B(0,R)}\). Here \(g\) is bounded, so \(g * P_t\) is defined everywhere; changing variables \(y = x - z\) (splitting into real, imaginary, positive and negative parts to apply the unsigned statements) and using \(\int P_t = 1\), \[ g*P_t(x) - g(x) = \int_{\mathbb{R}^d} \big( g(x-z) - g(x) \big) P_t(z)\, dz . \] With \(\omega(\delta) := \sup_{|z| \leq \delta} \sup_{x} |g(x-z)-g(x)|\), which tends to \(0\) by uniform continuity, splitting at \(|z| = \delta\) and using \(P_t \geq 0\) gives, uniformly in \(x\),

\begin{equation*} |g*P_t(x) - g(x)| \leq \omega(\delta) + 2\|g\|_{L^\infty} \int_{|z|>\delta} P_t(z)\, dz \longrightarrow 0 \qquad (2) \end{equation*}

as \(t \to 0\), by choosing \(\delta\) first and then invoking \((\ast)\). For the tail, \(g\) vanishes on \(|x| > 2R\), and \(|y| \leq R\) with \(|x| > 2R\) forces \(|x-y| > R\), so Tonelli gives

\begin{equation*} \int_{|x|>2R} |g*P_t(x) - g(x)| \, dx \leq \|g\|_{L^1} \int_{|z|>R} P_t(z)\, dz \longrightarrow 0 , \qquad (3) \end{equation*}

again by \((\ast)\). Since \(m(B(0,2R)) < \infty\), adding (2) over \(B(0,2R)\) to (3) yields \(\|g*P_t - g\|_{L^1} \to 0\).

(3) Density. Given \(\eta > 0\), pick \(g\) continuous and compactly supported with \(\|f-g\|_{L^1} \leq \eta\) (Theorem 1.3.20(iii)). Convolution is linear, so by the triangle inequality, (1) applied to \(f-g\), and (2),

\begin{equation*} \|f*P_t - f\|_{L^1} \leq \|(f-g)*P_t\|_{L^1} + \|g*P_t - g\|_{L^1} + \|g-f\|_{L^1} \leq 3\eta \end{equation*}

for all sufficiently small \(t > 0\). As \(\eta\) was arbitrary, \(f * P_t \to f\) in \(L^1\) norm.

Given \(\varepsilon > 0\), the Lipschitz bound \(d_Y(f(x),f(x^{\prime})) \leq C\, d_X(x,x^{\prime})\) of Definition 2.2.1 makes the single choice \(\delta := \varepsilon / C\) work at every pair of points at once, since \(d_X(x,x^{\prime}) < \delta\) forces \(d_Y(f(x),f(x^{\prime})) < C\delta = \varepsilon\); so \(f\) is uniformly continuous, and a fortiori continuous at each \(x\).

A uniformly continuous non-Lipschitz example is \(f(x) := \sqrt{x}\) on \([0,1]\), which indeed maps into \([0,1]\). For \(x \geq y \geq 0\),

\begin{equation*} (\sqrt{x} - \sqrt{y})^2 \leq (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) = x - y , \end{equation*}

so \(|\sqrt x - \sqrt y| \leq \sqrt{|x-y|}\) and \(\delta := \varepsilon^2\) witnesses uniform continuity. But testing a putative Lipschitz constant \(C\) at \(x = 1/n\), \(x^{\prime} = 0\) gives

\begin{equation*} \frac{1}{\sqrt{n}} = \left| f\!\left(\tfrac{1}{n}\right) - f(0) \right| \leq \frac{C}{n}, \qquad \text{i.e.} \qquad \sqrt{n} \leq C \end{equation*}

for every \(n \geq 1\), which is absurd.

Substitute \(k = hv\) into the total-differentiability estimate. Definition 2.2.2 says exactly that

\begin{equation*} f(x_0 + k) - f(x_0) = k \cdot \nabla f(x_0) + |k| \, \varepsilon(k) \quad (k \neq 0), \qquad \varepsilon(k) \to 0 \ \text{ as } k \to 0 . \end{equation*}

For \(v \neq 0\) and \(h \neq 0\) the vector \(k := hv\) is nonzero with \(|k| = |h|\,|v|\), so

\begin{equation*} \frac{f(x_0 + hv) - f(x_0)}{h} = v \cdot \nabla f(x_0) + \frac{|h|}{h} |v| \, \varepsilon(hv), \end{equation*}

whose error term has modulus \(|v|\,|\varepsilon(hv)| \to 0\) as \(h \to 0\), since \(hv \to 0\) through nonzero vectors (this is where \(v \neq 0\) is used). Hence \(D_v f(x_0)\) exists and equals \(v \cdot \nabla f(x_0)\), which is (2.2); for \(v = 0\) the difference quotient is identically \(0 = 0 \cdot \nabla f(x_0)\).

Taking \(v = e_i\) gives existence of \(\frac{\partial f}{\partial x_i}(x_0) = D_{e_i}f(x_0) = e_i \cdot \nabla f(x_0)\), the \(i\)-th coordinate of \(\nabla f(x_0)\), for each \(i = 1,\dots,d\); this is (2.3).

Fix \(x_0\) and put \(g := \bigl( \frac{\partial f}{\partial x_1}(x_0), \dots, \frac{\partial f}{\partial x_d}(x_0) \bigr)\); we show \(f\) is totally differentiable at \(x_0\) with \(\nabla f(x_0) = g\), whence (2.2) and (2.3) by Exercise 2.2.2.

Given \(\varepsilon > 0\), continuity of the \(d\) partials at \(x_0\) supplies (minimise over \(i\)) a common \(\delta > 0\) with \(\bigl| \frac{\partial f}{\partial x_i}(y) - \frac{\partial f}{\partial x_i}(x_0) \bigr| \leq \varepsilon\) for all \(i\) and all \(|y - x_0| \leq \delta\). Fix \(h\) with \(0 < |h| \leq \delta\) and walk from \(x_0\) to \(x_0 + h\) one coordinate at a time: with \(x^{(i)} := x_0 + \sum_{j \leq i} h_j e_j\),

\begin{equation*} f(x_0 + h) - f(x_0) = \sum_{i=1}^{d} \left[ f(x^{(i)}) - f(x^{(i-1)}) \right] = \sum_{i=1}^{d} h_i \frac{\partial f}{\partial x_i}(y_i) \end{equation*}

for suitable \(y_i\) on the segment from \(x^{(i-1)}\) to \(x^{(i)}\). Indeed, for \(h_i \neq 0\) the map \(\varphi_i(t) := f(x^{(i-1)} + t h_i e_i)\) is differentiable on \([0,1]\) with \(\varphi_i^{\prime}(t) = h_i \frac{\partial f}{\partial x_i}(x^{(i-1)} + t h_i e_i)\), since \(\partial f / \partial x_i\) exists everywhere, so the mean value theorem gives such a \(y_i = x^{(i-1)} + t_i h_i e_i\); for \(h_i = 0\) both sides vanish and we take \(y_i := x^{(i-1)}\). Each such point satisfies \(|y_i - x_0|^2 = \sum_{j<i} h_j^2 + t_i^2 h_i^2 \leq |h|^2 \leq \delta^2\), so the estimate above applies to it. Subtracting \(h \cdot g = \sum_i h_i \frac{\partial f}{\partial x_i}(x_0)\) and using \(|h_i| \leq |h|\),

\begin{equation*} \begin{aligned} \bigl| f(x_0 + h) - f(x_0) - h \cdot g \bigr| &= \Bigl| \sum_{i=1}^{d} h_i \Bigl[ \tfrac{\partial f}{\partial x_i}(y_i) - \tfrac{\partial f}{\partial x_i}(x_0) \Bigr] \Bigr| \\ &\leq \sum_{i=1}^{d} |h_i| \, \varepsilon \leq d \, \varepsilon \, |h| . \end{aligned} \end{equation*}

As \(\varepsilon > 0\) was arbitrary and \(d\) is fixed, the quotient by \(|h|\) tends to \(0\), i.e. \(f\) is totally differentiable at \(x_0\) with gradient \(g\).

Away from the origin, \(t \mapsto f(x+tv)\) is a quotient of polynomials in \(t\) whose denominator \(|x+tv|^2\) equals \(|x|^2 > 0\) at \(t = 0\), hence is nonzero on some \((-\rho,\rho)\); by the quotient rule it is differentiable at \(t = 0\), and that derivative is \(D_v f(x)\).

At the origin, \(f\) is homogeneous of degree \(1\): for \(\lambda \neq 0\) and \(y \neq 0\),

\begin{equation*} f(\lambda y) = \frac{\lambda^3 y_1 y_2^2}{\lambda^2 (y_1^2 + y_2^2)} = \lambda f(y), \end{equation*}

and \(f(\lambda y) = 0 = \lambda f(y)\) when \(\lambda = 0\) or \(y = 0\). So the difference quotient \((f(hv) - f(0,0))/h = f(v)\) is independent of \(h \neq 0\), and

\begin{equation*} D_v f(0,0) = f(v) = \frac{v_1 v_2^2}{v_1^2 + v_2^2} \quad (v \neq 0), \qquad D_0 f(0,0) = 0 . \end{equation*}

Thus \(D_v f(x)\) exists for all \(x, v \in \mathbb{R}^2\); \(v = e_1, e_2\) gives the partials everywhere.

Total differentiability fails at the origin, because \(v \mapsto D_v f(0,0) = f(v)\) is not linear. Concretely, \(\frac{\partial f}{\partial x_1}(0,0) = f(1,0) = 0\) and \(\frac{\partial f}{\partial x_2}(0,0) = f(0,1) = 0\), so were \(f\) totally differentiable at \((0,0)\), Exercise 2.2.2 would force \(\nabla f(0,0) = (0,0)\) by (2.3) and hence \(D_v f(0,0) = v \cdot (0,0) = 0\) for every \(v\) by (2.2). But

\begin{equation*} D_{(1,1)} f(0,0) = f(1,1) = \frac{1 \cdot 1}{1 + 1} = \frac{1}{2} \neq 0 , \end{equation*}

a contradiction.

Suppose not, so that \(A := \{F \neq 0\}\) has \(m(A) > 0\) (local integrability includes measurability on each \(B(0,n)\), so \(A\) is measurable). Almost every point is a Lebesgue point of \(F\) by the Lebesgue differentiation theorem (Theorem 1.6.19, extended to locally integrable \(F\) by Exercise 1.6.14(ii)), so we may pick a Lebesgue point \(x_0 \in A\); replacing \(F\) by \(-F\) if necessary (which alters neither hypothesis nor conclusion) assume \(c := F(x_0) > 0\), and fix \(r > 0\) with

\begin{equation*} \int_{B(x_0,r)} |F(y) - F(x_0)| \, dy < \frac{c}{2^{d+1}} \, m\bigl( B(x_0,r) \bigr) . \qquad (\ast) \end{equation*}

Take \(g(x) := \varphi\bigl( |x-x_0|^2 / r^2 \bigr)\), where \(\varphi : \mathbb{R} \to [0,1]\) equals \(1\) for \(s \leq \tfrac14\), equals \(0\) for \(s \geq 1\), and equals \(\cos^2\!\bigl( \tfrac{2\pi}{3}(s - \tfrac14) \bigr)\) in between. Then \(\varphi\) is continuously differentiable, since on the middle interval \(\varphi^{\prime}(s) = -\tfrac{2\pi}{3}\sin\bigl( \tfrac{4\pi}{3}(s-\tfrac14) \bigr)\) matches both the values and the vanishing derivatives of the constant pieces at \(s = \tfrac14\) and \(s = 1\) (Check!). Hence \(g\) is continuously differentiable by the chain rule, with \(0 \leq g \leq 1\), \(g = 1\) on \(B(x_0,r/2)\), and \(g = 0\) off \(B(x_0,r)\) – so \(g\) has compact support and is an admissible test function, giving \(\int F g = 0\). Splitting that integral (both pieces absolutely convergent, \(g\) being bounded with compact support and \(F\) locally integrable) and writing \(V := m(B(x_0,r)) \in (0,\infty)\),

\begin{equation*} \begin{aligned} 0 = \int_{\mathbb{R}^d} F g &= F(x_0) \int_{\mathbb{R}^d} g + \int_{\mathbb{R}^d} \bigl( F - F(x_0) \bigr) g \\ &\geq c \, m\bigl( B(x_0,r/2) \bigr) - \int_{B(x_0,r)} |F - F(x_0)| \\ &> c \, 2^{-d} V - c \, 2^{-d-1} V = c \, 2^{-d-1} V > 0 , \end{aligned} \end{equation*}

using \(g \geq 0\) with \(g = 1\) on \(B(x_0,r/2)\) for the first term, \(|g| \leq 1\) supported in \(B(x_0,r)\) together with \((\ast)\) for the second, and the scaling \(m(B(x_0,r/2)) = 2^{-d} V\). This is absurd, so \(m(A) = 0\).

Suppose such a \(\mathbf{P}\) existed. Since \(\{n\} = \{0\} + n\), translation-invariance forces every singleton to carry the same mass \(p := \mathbf{P}(\{0\}) \in [0,1]\); and the singletons are disjoint events (the \(\sigma\)-algebra being discrete) whose union is \(\mathbb{Z}\), so for any enumeration \(n_1, n_2, \dots\) of \(\mathbb{Z}\), countable additivity gives

\begin{equation*} 1 = \mathbf{P}(\mathbb{Z}) = \sum_{k=1}^{\infty} \mathbf{P}(\{n_k\}) = \sum_{k=1}^{\infty} p , \end{equation*}

the series of unsigned terms having a value in \([0,+\infty]\) independent of the enumeration. But its partial sums are \(Kp\), so the sum is \(0\) when \(p = 0\) and \(+\infty\) when \(p > 0\), never \(1\) – a contradiction.

Suppose such a \(\mathbf{P}\) existed, and chop the line into the half-open unit intervals \(I_n := [n,n+1)\), \(n \in \mathbb{Z}\). These are Borel, hence events (and \(E+x\) stays Lebesgue measurable by Exercise 1.2.20, so the hypothesis makes sense); they are pairwise disjoint and cover \(\mathbb{R}\), since \(x \in I_{\lfloor x \rfloor}\); and \(I_n = I_0 + n\), so translation-invariance gives \(\mathbf{P}(I_n) = \mathbf{P}(I_0) =: c\) for every \(n\). Countable additivity over any enumeration \(n_1, n_2, \dots\) of \(\mathbb{Z}\) then gives

\begin{equation*} 1 = \mathbf{P}(\mathbb{R}) = \sum_{k=1}^{\infty} \mathbf{P}(I_{n_k}) = \sum_{k=1}^{\infty} c , \end{equation*}

the unsigned series having a value in \([0,+\infty]\) independent of the enumeration. But its partial sums are \(Kc\), so the sum is \(0\) when \(c = 0\) and \(+\infty\) when \(c > 0\), never \(1\) – a contradiction.

Everything follows from two remarks about preimages. Write \(\pi_{C \leftarrow B} : X_B \to X_C\) for the subprojection \((x_\alpha)_{\alpha \in B} \mapsto (x_\alpha)_{\alpha \in C}\) when \(C \subset B \subset A\), so \(\pi_B = \pi_{B \leftarrow A}\), \(\pi_\beta = \pi_{\{\beta\} \leftarrow A}\), and \(\pi_{D \leftarrow C} \circ \pi_{C \leftarrow B} = \pi_{D \leftarrow B}\); write \(\varphi^*(\mathcal{C}) := \{\varphi^{-1}(F) : F \in \mathcal{C}\}\) and \(\langle \mathcal{F} \rangle\) for the generated \(\sigma\)-algebra (Exercise 1.4.13), so that by definition

\begin{equation*} \mathcal{B}_B = \Big\langle \bigcup_{\beta \in B} \pi_{\beta \leftarrow B}^*(\mathcal{B}_\beta) \Big\rangle . \end{equation*}

Since \(\varphi^{-1}\) commutes with complements and countable unions and sends \(\emptyset\) to \(\emptyset\):

• Lemma A. \(\varphi^*(\mathcal{C})\) is a \(\sigma\)-algebra on \(X\) whenever \(\mathcal{C}\) is one on \(Y\).
• Lemma B. \(\{F \subset Y : \varphi^{-1}(F) \in \mathcal{B}\}\) is a \(\sigma\)-algebra, so if \(\varphi^{-1}(F) \in \mathcal{B}\) for every \(F\) in a family \(\mathcal{F}\), then \(\varphi : (X,\mathcal{B}) \to (Y, \langle \mathcal{F} \rangle)\) is a measurable morphism.

Note \(\mathcal{B}_B\) depends only on the subfamily indexed by \(B\), so each part below holds verbatim with \(A\) replaced by any \(B \subset A\).

(1) \(\mathcal{B}_A\) makes each \(\pi_\beta\) measurable, since \(\pi_\beta^*(\mathcal{B}_\beta) \subset \mathcal{B}_A\) by construction. Conversely any \(\sigma\)-algebra \(\mathcal{C}\) doing so contains \(\bigcup_{\beta \in A} \pi_\beta^*(\mathcal{B}_\beta)\), hence contains the \(\sigma\)-algebra that union generates: \(\mathcal{B}_A \subset \mathcal{C}\), i.e. \(\mathcal{B}_A\) is coarser than \(\mathcal{C}\) (Remark 1.4.33). So \(\mathcal{B}_A\) is the coarsest such.

(2) By Lemma B it suffices to pull back the generators of \(\mathcal{B}_B\): for \(\beta \in B\) and \(E_\beta \in \mathcal{B}_\beta\), the composition law \(\pi_{\beta \leftarrow B} \circ \pi_B = \pi_\beta\) gives

\begin{equation*} \pi_B^{-1}\big( \pi_{\beta \leftarrow B}^{-1}(E_\beta) \big) = (\pi_{\beta \leftarrow B} \circ \pi_B)^{-1}(E_\beta) = \pi_\beta^{-1}(E_\beta) \in \mathcal{B}_A \end{equation*}

by (1). Applied with \(A\) replaced by \(B\), this also gives the transfer property: \(\pi_{C \leftarrow B} : (X_B, \mathcal{B}_B) \to (X_C, \mathcal{B}_C)\) is a measurable morphism for \(C \subset B \subset A\).

(3) The family \(\mathcal{C}\) of all sets \(\pi_B^{-1}(E_B)\), with \(B \subset A\) at most countable and \(E_B \in \mathcal{B}_B\), is a \(\sigma\)-algebra containing every generator \(\pi_\beta^{-1}(E_\beta)\) (take \(B := \{\beta\}\), for which \(\mathcal{B}_{\{\beta\}} = \mathcal{B}_\beta\)). Indeed it contains \(\emptyset = \pi_\emptyset^{-1}(\emptyset)\); it is closed under complements, as \(X_A \setminus \pi_B^{-1}(E_B) = \pi_B^{-1}(X_B \setminus E_B)\); and it is closed under countable unions, since for \(E_n = \pi_{B_n}^{-1}(F_n)\) and \(B := \bigcup_n B_n\) (at most countable) the transfer property puts \(G_n := \pi_{B_n \leftarrow B}^{-1}(F_n)\) in \(\mathcal{B}_B\) with \(\pi_B^{-1}(G_n) = \pi_{B_n}^{-1}(F_n) = E_n\), whence

\begin{equation*} \bigcup_{n} E_n = \pi_B^{-1}\Big( \bigcup_n G_n \Big) \in \mathcal{C} . \end{equation*}

By minimality \(\mathcal{B}_A \subset \mathcal{C}\), which is the assertion.

(4) We may assume \(X_A \neq \emptyset\) (otherwise take \(B := \emptyset\) and \(f_B :\equiv 0\), both sides being the empty function); then every \(X_\alpha \neq \emptyset\), so by the axiom of choice every \(\pi_B\) is surjective. For each rational \(q \geq 0\), (3) supplies an at most countable \(B_q\) and \(F_q \in \mathcal{B}_{B_q}\) with \(\{f>q\} = \pi_{B_q}^{-1}(F_q)\); put \(B := \bigcup_{q} B_q\), at most countable, and transfer as in (3) to get \(\{f>q\} \in \mathcal{D} := \pi_B^*(\mathcal{B}_B)\), a \(\sigma\)-algebra by Lemma A. Since \(\{f > \lambda\} = \bigcup_{q \in \mathbb{Q},\, q > \lambda} \{f > q\}\) for \(\lambda \geq 0\) and \(\{f>\lambda\} = X_A\) for \(\lambda < 0\), \(f\) is \(\mathcal{D}\)-measurable by Exercise 1.4.29(i): say \(\{f>\lambda\} = \pi_B^{-1}(H_\lambda)\) with \(H_\lambda \in \mathcal{B}_B\).

Hence \(f\) is constant on the fibres of \(\pi_B\): if \(\pi_B(x_A) = \pi_B(x_A^{\prime})\) then \(x_A \in \{f>\lambda\} \iff \pi_B(x_A) \in H_\lambda \iff x_A^{\prime} \in \{f>\lambda\}\) for every \(\lambda\), forcing \(f(x_A) = f(x_A^{\prime})\). Surjectivity of \(\pi_B\) therefore lets us define \(f_B(x_B) := f(x_A)\) for any \(x_A \in \pi_B^{-1}(\{x_B\})\), so that \(f = f_B \circ \pi_B\). Finally \(\pi_B^{-1}(\{f_B > \lambda\}) = \{f>\lambda\} = \pi_B^{-1}(H_\lambda)\), and \(S \mapsto \pi_B^{-1}(S)\) is injective for surjective \(\pi_B\), so \(\{f_B > \lambda\} = H_\lambda \in \mathcal{B}_B\) and \(f_B\) is measurable (Exercise 1.4.29(i)).

(5) Let \(\mathcal{P}\) be the collection of products \(\prod_{\beta \in A} E_\beta\) with \(E_\beta \in \mathcal{B}_\beta\), and \(\mathcal{G} := \langle \mathcal{P} \rangle\). Always \(\mathcal{B}_A \subset \mathcal{G}\), since \(\pi_\beta^{-1}(E_\beta)\) is the product with \(E_\gamma := X_\gamma\) for \(\gamma \neq \beta\). Conversely

\begin{equation*} \prod_{\beta \in A} E_\beta = \bigcap_{\beta \in A} \pi_\beta^{-1}(E_\beta) , \end{equation*}

which for \(A\) at most countable is a countable intersection of members of \(\mathcal{B}_A\), hence lies in \(\mathcal{B}_A\) (De Morgan). So \(\mathcal{P} \subset \mathcal{B}_A\), \(\mathcal{G} \subset \mathcal{B}_A\), and the two agree.

(6) Let \(A\) be uncountable and fix, by non-triviality, \(E_\alpha \in \mathcal{B}_\alpha\) with \(\emptyset \neq E_\alpha \neq X_\alpha\) for each \(\alpha\); put \(E := \prod_{\alpha \in A} E_\alpha \in \mathcal{P} \subset \mathcal{G}\) and choose \(x = (x_\alpha) \in E\) (axiom of choice). If \(E\) lay in \(\mathcal{B}_A\), then \(E = \pi_B^{-1}(E_B)\) for some at most countable \(B\) by (3); pick \(\gamma \in A \setminus B\) and \(y_\gamma \in X_\gamma \setminus E_\gamma\), and let \(x^{\prime}\) agree with \(x\) off \(\gamma\) with \(x^{\prime}_\gamma := y_\gamma\). Then \(\pi_B(x^{\prime}) = \pi_B(x) \in E_B\) gives \(x^{\prime} \in E\), while \(x^{\prime}_\gamma \notin E_\gamma\) gives \(x^{\prime} \notin E\) – absurd. Hence \(E \in \mathcal{G} \setminus \mathcal{B}_A\) and \(\mathcal{B}_A \subsetneq \mathcal{G}\).

(7) \(E_{x_{A \setminus B}, B} = \iota^{-1}(E)\) for the insertion map \(\iota : X_B \to X_A\), \(\iota(x_B) := (x_B, x_{A\setminus B})\), so it suffices that \(\iota\) be a measurable morphism, and by Lemma B only \(\iota^{-1}(\pi_\beta^{-1}(E_\beta)) = (\pi_\beta \circ \iota)^{-1}(E_\beta) \in \mathcal{B}_B\) need be checked:

• (i) \(\beta \in B\): \(\pi_\beta \circ \iota = \pi_{\beta \leftarrow B}\), so the preimage lies in \(\pi_{\beta \leftarrow B}^*(\mathcal{B}_\beta) \subset \mathcal{B}_B\).
• (ii) \(\beta \in A \setminus B\): \(\pi_\beta \circ \iota\) is constant with value \(x_\beta\), so the preimage is \(X_B\) or \(\emptyset\).

(8) \(f_{x_{A \setminus B}, B} = f \circ \iota\) with \(\iota\) as in (7), and compositions of measurable morphisms are measurable: for every real \(\lambda\),

\begin{equation*} \{ x_B \in X_B : f_{x_{A \setminus B}, B}(x_B) > \lambda \} = \iota^{-1}\big( \{ f > \lambda \} \big) \in \mathcal{B}_B , \end{equation*}

since \(\{f>\lambda\} \in \mathcal{B}_A\) (Exercise 1.4.29(i)) and \(\iota\) is measurable by (7). So \(f_{x_{A \setminus B}, B}\) is \(\mathcal{B}_B\)-measurable, again by Exercise 1.4.29(i).