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\date{}
\title{Informal Measure Theory Notes}
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\hypersetup{
 pdfauthor={Aayush Bajaj},
 pdftitle={Informal Measure Theory Notes},
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}\begin{document}

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\begin{center}
{\bfseries \large Informal Measure Theory Notes} \\[0.5em]
April 2026
\end{center}
\tableofcontents
\dotfill
\newpage
\section{Warm-up}
\label{sec:org61cc80f}

To start I think it might be worth-while to prove some basic limits:
\subsection{Epsilons and Limits}
\label{sec:org4e4d6e8}

\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{document}
