% This is lecture no. 9 of MH3100/: Real Analysis I.

\chapter{Monotone Convergence Theorem}%
\label{cha:monotone_convergence_theorem}

\section{The Monotone Convergence Theorem}%
\label{sec:the_monotone_convergence_theorem}

\begin{definition}
	Monotone Sequence.
	A sequence $\left\{a_n\right\}$ is said to be \textbf{increasing} if
	\[
	\forall n\in\mathbb{N},\ a_n\leqslant a_{n+1}
	,\]
	and is said to be \textbf{decreasing} if
	\[
	\forall n\in\mathbb{N},\ a_n\geqslant a_{n+1}
	.\]
	In either case, the sequence is said to be \textbf{monotone}.
\end{definition}

\begin{thrm}
	Monotone Convergence Theorem.
	If a sequence $\left\{a_n\right\}$ is \textbf{eventually monotone} and \textbf{bounded}, then it is \textbf{convergent}.
\end{thrm}

\begin{proof}
	The Bolzano-Weierstrass Theorem asserts that the bounded sequence $\left\{a_n\right\}$ has a convergent subsequence $\left\{a_{n_{k}}\right\}$.
	Let $a$ denote the limit of the subsequence $\left\{a_{n_{k}}\right\}$.
	The definition of $a=\lim a_{n_{k}}$ is
	\[
	\forall\varepsilon>0,\exists K\in\mathbb{N}\suchthat\forall k\in\mathbb{N},\ k>K\implies\abs{a_{n_{k}}-a}<\varepsilon
	.\]
	Since $\left\{a_n\right\}$ is monotone, for any three natural numbers $m\geqslant m^{\prime}\geqslant m^{\prime\prime}$,
	\[
	\abs{a_{m^{\prime}}-a}\leqslant\max\left\{\abs{a_m-a},\abs{a_{m^{\prime\prime}}-a}\right\}
	.\]
	Therefore, $\forall\varepsilon>0$, $\exists N=n_{K+1}\in\mathbb{N}\suchthat\forall n\in\mathbb{N}$ 
	\begin{align*}
		n>N & \implies n_n\geqslant n>N = n_{K+1}\geqslant K+1>K\\
		    & \implies \abs{a_n-a}\leqslant\max\left\{\abs{a_{n_n}-a},\abs{a_{n_{K+1}}-a}\right\}<\varepsilon.
	\end{align*}
	Which is the definition for $0=\lim a_n$.
\end{proof}
