<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Real-Analysis on Aayush Bajaj's Augmenting Infrastructure</title><link>https://abaj.ai/tags/real-analysis/</link><description>Recent content in Real-Analysis on Aayush Bajaj's Augmenting Infrastructure</description><generator>Hugo</generator><language>en</language><copyright>© 2026 Aayush Bajaj</copyright><lastBuildDate>Mon, 24 Aug 2026 06:00:15 +1000</lastBuildDate><atom:link href="https://abaj.ai/tags/real-analysis/index.xml" rel="self" type="application/rss+xml"/><item><title>Solutions to Bogachev's Measure Theory, Volume 1</title><link>https://abaj.ai/words/library/books/bogachev-measure-theory/</link><pubDate>Sun, 23 Aug 2026 20:10:00 +1000</pubDate><guid>https://abaj.ai/words/library/books/bogachev-measure-theory/</guid><description>&lt;p>Solutions to the exercises of V.I. Bogachev&amp;rsquo;s &lt;em>Measure Theory&lt;/em>, Volume 1 (Springer, 2007) — exercises 1.12.47–160, 2.12.25–117, 3.10.29–125, 4.7.42–154, and 5.8.37–142. Exercises marked \(\circ\) in the book are the basic ones. The book itself is filed at &lt;a
 href="https://abaj.ai/roam/measure_theory_bogachev/"
 
 
>Measure Theory (Bogachev)&lt;/a>.&lt;/p>


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&lt;h2 id="constructions-and-extensions-of-measures">Constructions and Extensions of Measures&lt;a href="#constructions-and-extensions-of-measures" class="post-heading__anchor" aria-hidden="true">#&lt;/a>
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&lt;h2 id="the-lebesgue-integral">The Lebesgue Integral&lt;a href="#the-lebesgue-integral" class="post-heading__anchor" aria-hidden="true">#&lt;/a>
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&lt;h2 id="operations-on-measures-and-functions">Operations on Measures and Functions&lt;a href="#operations-on-measures-and-functions" class="post-heading__anchor" aria-hidden="true">#&lt;/a>
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&lt;h2 id="the-spaces-l-p-and-spaces-of-measures">The Spaces \(L^p\) and Spaces of Measures&lt;a href="#the-spaces-l-p-and-spaces-of-measures" class="post-heading__anchor" aria-hidden="true">#&lt;/a>
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&lt;h2 id="connections-between-the-integral-and-derivative">Connections between the Integral and Derivative&lt;a href="#connections-between-the-integral-and-derivative" class="post-heading__anchor" aria-hidden="true">#&lt;/a>
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&lt;/div></description></item><item><title>Solutions to Kolmogorov and Fomin's Introductory Real Analysis</title><link>https://abaj.ai/words/library/books/ira-fomin/</link><pubDate>Sun, 20 Jul 2025 02:42:32 +1100</pubDate><guid>https://abaj.ai/words/library/books/ira-fomin/</guid><description>&lt;div class="collapse-marker" data-target-path="/words/library/books/ira-fomin/" data-folded="true" data-lvl="0">&lt;/div>
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&lt;h2 id="set-theory">Set Theory&lt;a href="#set-theory" class="post-heading__anchor" aria-hidden="true">#&lt;/a>
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&lt;h3 id="sets-and-functions">Sets and Functions&lt;a href="#sets-and-functions" class="post-heading__anchor" aria-hidden="true">#&lt;/a>
&lt;/h3>
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 &lt;div class="math-problem-header">
 &lt;strong>Problem&lt;span class="problem-counter">&lt;/span>&lt;/strong>
 &lt;/div>
 &lt;div class="math-problem-content">
 Prove that if \(A\cup B = A\) and \(A \cap B = A\), then \(A = B\).
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&lt;div class="math-solution">
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 &lt;strong>Solution&lt;/strong>
 &lt;/div>
 &lt;div class="math-solution-content">
 To show \(A=B\), show \(A\subseteq B\) and \(B\subseteq A\). Suppose \(x\in B\), then we know by definition that \(x\in (A\cup B)\) if \(x\in B\) or \(x\in A\). Which then implies that \(x\in A\) from rule 1. Thus \(B\subseteq A\).
Now suppose \(x\in A\) which implies \(x\in (A\cap B)\) (rule 2). The definition of this &lt;strong>means&lt;/strong> that \(x\in A\) and \(x\in B\). \(\therefore x\in A \implies x\in B\), i.e. \(A\subseteq B\), so \(A=B\).
 &lt;/div>
&lt;/div>
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 &lt;div class="math-problem-header">
 &lt;strong>Problem&lt;span class="problem-counter">&lt;/span>&lt;/strong>
 &lt;/div>
 &lt;div class="math-problem-content">
 Show that in general \((A-B)\cup B \neq A\).
 &lt;/div>
&lt;/div>
&lt;div class="math-solution">
 &lt;div class="math-solution-header">
 &lt;strong>Solution&lt;/strong>
 &lt;/div>
 &lt;div class="math-solution-content">
 This only holds for \(B\subseteq A\). We proceed by counterexample.
Let \(A={1,2}, B={3,4}\). Then \((A-B) = {1,2}\) and \((A-B)\cup B = {1,2,3,4} \neq {1,2}\).
 &lt;/div>
&lt;/div>
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 &lt;strong>Problem&lt;span class="problem-counter">&lt;/span>&lt;/strong>
 &lt;/div>
 &lt;div class="math-problem-content">
 Let \(A = {2,4,&amp;hellip;,2n,&amp;hellip;}\) and \(B = {3,6,&amp;hellip;,3n,&amp;hellip;}\). Find \(A\cap B\) and \(A-B\)
 &lt;/div>
&lt;/div>
&lt;div class="math-solution">
 &lt;div class="math-solution-header">
 &lt;strong>Solution&lt;/strong>
 &lt;/div>
 &lt;div class="math-solution-content">
 \[
A\cap B = {6n \mid n\in \mathbb{N}}
\]
\[
A - B = {2n \mid n\in \mathbb{N}, 2n \not\in {6m \mid m\in \mathbb{N}}}
\]
 &lt;/div>
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 &lt;strong>Problem&lt;span class="problem-counter">&lt;/span>&lt;/strong>
 &lt;/div>
 &lt;div class="math-problem-content">
 Prove that:
 &lt;/div>
&lt;/div>
&lt;div class="math-subproblem">
 &lt;div class="math-subproblem-header">
 &lt;strong>(&lt;span class="subproblem-counter">&lt;/span>)&lt;/strong>
 &lt;/div>
 &lt;div class="math-subproblem-content">
 \((A-B)\cap C = (A\cap C) - (B\cap C)\)
 &lt;/div>
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&lt;div class="math-subsolution">
 &lt;div class="math-subsolution-header">
 &lt;strong>Solution:&lt;/strong>
 &lt;/div>
 &lt;div class="math-subsolution-content">
 &lt;p>Let \(x\in (A-B)\cap C\).
Then:&lt;/p></description></item></channel></rss>