:PROPERTIES: :ROAM_ALIASES: "Measure Theory Scratch" :ID: E64912FE-D60A-43D7-A891-93F4981B6C62 :END: #+TITLE: Informal Measure Theory Notes #+LATEX_COMPILER: lualatex #+LATEX_CLASS: article #+LATEX_CLASS_OPTIONS: [a4paper, 11pt] #+OPTIONS: title:nil toc:nil num:t author:nil date:nil ^:{} #+LATEX_HEADER: \setlength{\headheight}{13.59999pt} #+LATEX_HEADER: \definecolor{linkblue}{HTML}{00007B} #+LATEX_HEADER: \definecolor{exampleblue}{HTML}{4169E1} #+LATEX_HEADER: \definecolor{highlightred}{HTML}{B22222} #+LATEX_HEADER: \definecolor{highlightblue}{HTML}{0000FF} #+LATEX_HEADER: \definecolor{notegray}{HTML}{7F807F} #+LATEX_HEADER: \definecolor{highlightsalmon}{HTML}{F08080} #+LATEX_HEADER: \definecolor{highlightgreen}{HTML}{006400} #+LATEX_HEADER: \definecolor{examplebarbg}{rgb}{0.255, 0.41, 0.884} #+LATEX_HEADER: \usepackage{amsthm} #+LATEX_HEADER: \usepackage{mathtools} #+LATEX_HEADER: \usepackage{thmtools} #+LATEX_HEADER: \usepackage{mathrsfs} #+LATEX_HEADER: \usepackage{bm} 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\renewcommand{\qedsymbol}{$\blacksquare$} #+LATEX_HEADER: \pagestyle{empty} #+LATEX_HEADER: \geometry{a4paper, top=2cm, left=1cm, right=1cm} #+LATEX_LINKCOLOR: deep-navy #+BEGIN_EXPORT latex \pagestyle{fancy} \begin{center} {\bfseries \large Informal Measure Theory Notes} \\[0.5em] April 2026 \end{center} \tableofcontents \dotfill \newpage #+END_EXPORT o * Warm-up To start I think it might be worth-while to prove some basic limits: ** Epsilons and Limits #+begin_src latex \begin{definition} The statement $\lim_{x\to\infty} x_n = x$ means that for any real $\epsilon > 0 $, there is a number $N>0$ for which $n\ge N$ implies $|x_n-x|<\epsilon$. \end{definition} \begin{theorem}[Archimedean Property] \label{thm:archimedean} If $x\in\R$, $y\in\R$, and $x>0$, then there is a positive integer $n$ such that \begin{equation}\label{eq:1}nx>y\end{equation} \end{theorem} \begin{proof} Let $A$ be the set of all $nx$, where $n$ runs through the positive integers. If \ref{eq:1} were false ($nx\le y$), then $y$ would be an upper-bound of $A$. But then $A$ has a least upper bound in $\R$. Put $\alpha = \sup A$. Since $x>0$, $\alpha - x < \alpha$, and $\alpha - x$ is not an upper bound of $A$. Hence $\alpha - x < mx$ for some positive integer $m$. But then $\alpha < (m+1)x\in A$, which is impossible, since $\alpha$ is an upper bound of $A$. \end{proof} \begin{proposition} \[ \lim_{n\to \infty} \frac{1}{n} = 0\] \end{proposition} \begin{proof} Let $N\in\mathbb{N}$ such that $\tfrac1N < \epsilon$ (By \hyperref[thm:archimedean]{the Archimedean Property}). Then let $n\ge N$ such that: \begin{align*} n \ge N &> \tfrac1\epsilon\\ \tfrac1n \le \tfrac1N &< \epsilon \qquad\text{taking reciprocals} \\ |\tfrac1n - 0| = |\tfrac1n| &= \tfrac1n\qquad\text{since $\tfrac1n$ is always positive for $n\in\mathbb{N}$}\\ &<\epsilon \end{align*} \end{proof} \begin{proposition} \begin{align*} \lim_{n\to\infty}\frac{1}{n^k} &= 0\qquad\text{for any $k>0$}.\\ &\iff |\frac{1}{n^k} - 0| < \epsilon, \epsilon > 0, n > N \in \mathbb{N} \\ &\iff \frac{1}{n^k} < \epsilon\qquad\text{since $n^k > 0$} \end{align*} \end{proposition} \begin{proof} Let $N > \epsilon^{-\frac{1}{k}}$ (by Archimedean Property of Natural Numbers) and $n \ge N$. Then \begin{align*} n&> \left( \frac{1}{\epsilon} \right)^{\frac{1}{k}}\\ \Rightarrow n^k &> \left( \frac{1}{\epsilon} \right) \\ \Rightarrow \frac{1}{n^k} &< \epsilon \end{align*} \end{proof} \begin{proposition} $$\lim_{n\to\infty}2^{\frac{1}{n}} = 1$$ \end{proposition} \begin{proof} Let $\epsilon > 0$. Since $2^{1/n} > 1$ for all $n \in \mathbb{N}$, we have $|2^{1/n} - 1| = 2^{1/n} - 1$. Since $\epsilon > 0$, we have $\epsilon + 1 > 1$, so $\log(\epsilon+1) > 0$. Let $n \ge N > \log_{\epsilon+1} 2 = \frac{\log 2}{\log(\epsilon+1)}$ (by \hyperref[thm:archimedean]{Archimedean Property}). Then \begin{align*} n &> \frac{\log 2}{\log(\epsilon+1)} \\ \Rightarrow \frac{1}{n} &< \frac{\log(\epsilon+1)}{\log 2} \\ \Rightarrow \log\left(2^{\frac{1}{n}}\right) &< \log(\epsilon+1) \\ \Rightarrow 2^{\frac{1}{n}} &< \epsilon + 1 \\ \Rightarrow \left|2^{\frac{1}{n}} - 1\right| &< \epsilon. \qedhere \end{align*} \end{proof} \begin{proposition} If $\lim_{n\to\infty} x_n = x$ and $a \in \mathbb{R}$, then $\lim_{n\to\infty} ax_n = ax$. \end{proposition} \begin{proof} If $a = 0$ then $ax_n = 0 = ax$ for all $n$ and the result is immediate. Now suppose $a \neq 0$ and let $\epsilon > 0$. Since $\lim_n x_n = x$, there exists $N \in \mathbb{N}$ such that $n \ge N$ implies, $|x_n -x | < \frac{\epsilon}{|a|}$. Then for $n\ge N$, \[ |ax_n - ax| = |a||x_n - x| < |a| \cdot \frac{\epsilon}{|a|} = \epsilon. \] \end{proof} \begin{proposition} If $\lim_{n\to\infty} x_n = x$ and $\lim_{n\to\infty} y_n = y$, then $\lim_{n\to\infty} (x_n + y_n) = x + y$. \end{proposition} \begin{proof} Let $\epsilon > 0$. Since $\lim_n x_n = x$, there exists $N_1 \in \mathbb{N}$ such that $n \ge N_1$ implies $|x_n - x| < \frac{\epsilon}{2}$. Since $\lim_n y_n = y$, there exists $N_2 \in \mathbb{N}$ such that $n \ge N_2$ implies $|y_n - y| < \frac{\epsilon}{2}$. Let $N = \max(N_1, N_2)$. Then for $n \ge N$, by the triangle inequality, \[ |(x_n + y_n) - (x+y)| \le |x_n - x| + |y_n - y| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \qedhere \] \end{proof} \begin{proposition} If $\lim_{n\to\infty} x_n = x$ and $\lim_{n\to\infty} y_n = y$, then $\lim_{n\to\infty} (x_n \cdot y_n) = x\cdot y$. \end{proposition} \begin{proof} Let $\epsilon > 0$. Consider \begin{align*} x_n y_n - xy &= x_n y_n - x_n y + x_n y - xy \\ &= x_n(y_n -y ) + y(x_n -x) \end{align*} Then by the Triangle inequality and by the boundedness of convergent sequences we obtain: \begin{align*} |x_n y_n -xy| &\le |x_n | | y_n -y | + |y| |x_n - x| \\ &< M \frac{\epsilon}{2M} + |y| \frac{\epsilon}{2|y|}\\ &=\epsilon \end{align*} where $M>0$. \end{proof} \begin{proof} Let $\epsilon > 0$. If $y = 0$ then $x_n y_n = 0 = xy$ for all $n$ and the result is immediate. Now suppose $y \neq 0$. Since $x_n \to x$, the sequence $(x_n)$ is bounded, so there exists $M > 0$ such that $|x_n| \le M$ for all $n$. Since $\lim_n x_n = x$, there exists $N_1 \in \mathbb{N}$ such that $n \ge N_1$ implies $|x_n - x| < \frac{\epsilon}{2|y|}$. Since $\lim_n y_n = y$, there exists $N_2 \in \mathbb{N}$ such that $n \ge N_2$ implies $|y_n - y| < \frac{\epsilon}{2M}$. Let $N = \max(N_1, N_2)$. Then for $n \ge N$, by the add-and-subtract identity and the triangle inequality, \begin{align*} |x_n y_n - xy| &= |x_n(y_n - y) + y(x_n - x)| \\ &\le |x_n||y_n - y| + |y||x_n - x| \\ &< M \cdot \frac{\epsilon}{2M} + |y| \cdot \frac{\epsilon}{2|y|} \\ &= \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \qedhere \end{align*} \end{proof} \begin{proposition} If $\lim_{n\to\infty} x_n = x$ and $x \neq 0$, then $\lim_{n\to\infty} \frac{1}{x_n} = \frac{1}{x}$. \end{proposition} \begin{proof} Let $\epsilon > 0$. Since $x_n \to x$ and $|x| > 0$, there exists $N_1 \in \mathbb{N}$ such that $n \ge N_1$ implies $|x_n - x| < \frac{|x|}{2}$. By the reverse triangle inequality this gives $|x_n| \ge |x| - |x_n - x| > \frac{|x|}{2}$, so $\frac{1}{|x_n|} < \frac{2}{|x|}$ for all $n \ge N_1$. There also exists $N_2 \in \mathbb{N}$ such that $n \ge N_2$ implies $|x_n - x| < \frac{|x|^2 \epsilon}{2}$. Let $N = \max(N_1, N_2)$. Then for $n \ge N$, \begin{align*} \left|\frac{1}{x_n} - \frac{1}{x}\right| &= \frac{|x_n - x|}{|x_n||x|} \\ &< \frac{2}{|x|^2} |x_n - x| \\ &< \frac{2}{|x|^2} \cdot \frac{|x|^2 \epsilon}{2} = \epsilon. \qedhere \end{align*} \end{proof} #+end_src #+LATEX:\newpage ** Infimums and Supremums #+begin_src latex \begin{proposition} Let $R,S \subseteq \R$ each be non-empty and bounded below. Prove that $\inf(R\cup S) = \min(\inf R, \inf S)$. \end{proposition} \begin{proof} \end{proof} #+end_src