# Apostol's proof of Arzelà's Bounded Convergence Theorem

In the first edition of Apostol's Mathematical Analysis, there is a proof of Arzelà's bounded convergence theorem:

$\textbf{Theorem 13-17}\,(\textit{Arzel}\grave{a})\textbf{.}$ Assume that $\lbrace f_{n}\rbrace$ is a uniformly bounded, pointwise convergent on $[a,b]$ and suppose that each $f_{n}$ is Riemann-integrable on $[a,b]$. Suppose also that the limit function $f(x):=\lim_{n}f_{n}(x)$ is Riemann-integrable on $[a,b]$. Then, $$\lim_{n}\int_{[a,b]}f_{n}=\int_{[a,b]}\lim_{n}f_{n}=\int_{[a,b]}f.$$ $\textit{Proof}.$ Let $g_{n}(x)=\left|f_{n}(x)-f(x)\right|$. We will prove that $$\lim_{n}\int_{[a,b]}g_{n}=0.$$ For this purpose we define a new sequence of functions $\lbrace h_{n}\rbrace$ as follows: $$h_{n}(x)=\sup_{m\ge n}f_{m}(x),\qquad\text{ if }x\in[a,b]\text{ and }n\in\mathbf{N}.$$ Note that, for each fixed $x$, we have $$0\le g_{n}(x)\le h_{n}(x),\qquad h_{n+1}(x)\le h_{n}(x),\qquad \lim_{n} h_{n}(x)=0.$$ Hence, $0\le\int_{[a,b]}g_{n}=\mathop{\underline{\int}}_{[a,b]}g_{n}\le\mathop{\underline{\int}}_{[a,b]}h_{n}.$ (The lower integral $\mathop{\underline{\int}}_{[a,b]} h_{n}$ exists because each $h_{n}$ is bounded on $[a,b]$. However, $h_{n}$ need not to be Riemann-integrable. On the other hand, $g_{n}$ $\textit{is}$ Riemann-integrable, since both $f_{n}$ and $f$ are integrable.) Let $I_{n} = \mathop{\underline{\int}}_{[a,b]} h_{n}$. The proof will be complete if we can show that $I_{n}\to 0$ as $n\to\infty$. Observe that the sequence $\lbrace I_{n}\rbrace$ convergs to $\textit{some}$ non-negative limit. Let $I=\lim_{n}I_{n}$. We will show that the inequality $I>0$ leads to a contradiction. Assume that $I>0$. Since $I_{n}\ge I > I/2$ for each $n$, there is a partition $P_{n}$ of $[a,b]$ and a lower Riemann sum $L(P_{n},h_{n})$ such that $L(P_{n},h_{n})>I/2.$ Let $\varepsilon=\tfrac{1}{2}I/(M+b-a)$, where $M$ is a uniform bound for $\lbrace h_{n}\rbrace$ on $[a,b]$. The lower sum $L(P_{n},h_{n})$ can be split into two parts as follows: $$L(P_{n},h_{n})=\sum_{i\in A_{n}}m_{i}(h_{n})\Delta x_{i}+\sum_{i\in B_{n}}m_{i}(h_{n})\Delta x_{i},$$ where $m_{i}(h_{n})$ denotes the $\inf$ of $h_{n}$ in the $i$th subinterval of $P_{n}$ and $$A_{n}=\lbrace i: m_{i}(h_{n})>\varepsilon\rbrace,\quad B_{n}=\lbrace i:m_{i}(h_{n})\le\varepsilon\rbrace.$$ The inequality $L(P_{n},h_{n}) >I/2$ implies $$\tfrac{1}{2} I <\sum_{i\in A_{n}}M\Delta x_{i}+\varepsilon\sum_{i\in B_{n}}\Delta x_{i}\le M\sum_{i\in A_{n}}\Delta x_{i}+\varepsilon(b-a),$$ from which we obtain $\sum_{i\in A_{n}}>\varepsilon$. Since refinement of partitions increases lower sums, there is no loss in generality in assuming that the $P_{n}$ are such that $P_{n}\subset P_{n+1}.$ Now let $S_{n}$ denote the union of these subintervals $[x_{i-1},x_{i}]$ of $P_{n}$ for which $i\in A_{n}$. Then $S_{n}$ is a closed set and its Jordan content is $$c(S_{n})=\sum_{i\in A_{n}}\Delta x_{i}>\varepsilon.$$ This implies that there is ate least one $x$ belonging to infinitely many of the sets $S_{n}$. Hence, for this $x$ we have $h_{n}(x)>\varepsilon$ infinitely often, contradicting the fact that $h_{n}(x)\to 0$ as $n\to\infty$. Thus, $I=0$.

Why does $c(S_{n})>\varepsilon$ for all $n$ imply that there exists some $x$ such that $x\in S_{n}$ for infinitely many $n$? Apostol doesn't prove this, but this claim doesn't seem to be entirely obvious. I would appreciate any help.