User:Tkojar/Sandbox/Bounded convergence theorem for Riemann integrals

Statement
Suppose $f_{n}:[a,b]\to \mathbb{R}$ are Riemman integrable and satisfy $|f_{n}(x)|\leq K$ for all n. If $f_{n}(x)\to f(x)$ pointwise and f is also Riemman integrable, then


 * $\displaystyle\int_{a}^{b}f_{n}(x)dx\to \int_{a}^{b}f(x)dx.$

Lemma
We call $E\subset \mathbb{R}$ an elementary subset if $E=\bigcup_{k=1}^{M} [a_{k},b_{k}]$ and we define $m(E)$ as the total length of these intervals minus their overlaps.

Lemma: Suppose $(A_{n})$ is a contracting sequence of bounded sets in R with an empty intersection. Let $a_{n}:=\sup\{m(E): E\subset A_{n}\text{ is an elementary subset} \}$, then $a_{n}\to 0$.

Proof of lemma
The sequence $a_{n}$ is decreasing and assume that $a_{n}\geq \delta>0$ to obtain a contradiction.

By the epsilon definition of supremm, for $\epsilon:=\frac{\delta}{2^{n}}$ there exists elementary subset $E_{n}$ such that


 * $\displaystyle m(E_{n})\geq a_{n}-\frac{\delta}{2^{n}}.$

For $H_{n}=\bigcap_{k=1}^{n}E_{k}\subset \bigcap_{k=1}^{n}A_{k}$, we will show that $H_{n}\neq \varnothing$ and thus contradict that $A_{n}$ have an empty intersection.

For each n, take any elementary subset $E\subset A_{k}\setminus E_{k}$, then we find


 * $\displaystyle m(E)+m(E_{k})=m(E\cup E_{k})\leq a_{k}\Rightarrow m(E)\leq \frac{\delta}{2^{k}}.$

Now take an elementary subset $S\subset A_{n}\setminus H_{n}=\bigcap_{k=1}^{n}(A_{n}\setminus E_{k})$, then we find


 * $\displaystyle E=(E\setminus E_{1})\cup … \cup (E\setminus E_{n}).$

Therefore, we get the bound


 * $\displaystyle m(E)\leq \sum_{k=1}^{n}m(E\setminus E_{k})\leq \sum_{k=1}^{n}\frac{\delta}{2^{n}}=\delta.$

In words, any elementary subset $E\subset A_{n}\setminus H_{n}$ was shown to have measure $m(E)\leq \delta$.

However, the inequality $a_{n}>\delta$ requires the existence of at least one elementary subset $U_{n}\subset A_{n}$ s.t. $m(U_{n})>\delta$.

Since all the elementary subset $E\subset A_{n}\setminus H_{n}$ satisfy $m(E)\leq \delta$, we mst have that $U_{n}\subset H_{n}$ for $n\geq 1$.

This contradicts the non-emptiness because $\lim_{n\to \infty} m(U_{n})>\delta$.

$\square$

Proof of main result
WLOG assume that $f_{n}\geq 0$ and $f_{n}\to 0$, so we will show that given $\epsilon>0$ there exists N s.t. forall $n\geq N$ we have.


 * $\displaystyle \int_{a}^{b}f_{n}(x)dx\leq \epsilon.$

Let $A_{n}:=\{x\in [a,b]:\text{ there exists }k\geq n \text{ such that} f_{k}(x)\geq \frac{\epsilon}{2(b-a)} \}$.

These sets are decreasing as $n\to +\infty$ and have empty intersection and so the sup $a_{n}$ from above goes to zero $a_{n}\to 0$.

So let $E_{n}\subset A_{n}$ be an elementary subset with $m(E_{n})\leq \frac{\epsilon}{2K}$ for all $n\geq N$, and consider the following subsets


 * $\displaystyle E:=\{x\in E_{n}:\text{ there exists }k\geq n \text{ such that} f_{k}(x)\geq \frac{\epsilon}{2(b-a)} \}\text{ and } F:=[a,b]\setminus E.$

Therefore, we find


 * $\displaystyle \int_{a}^{b}f_{n}(x)dx=\int_{E}f_{n}(x)dx+\int_{F}f_{n}(x)dx\leq K m(E_{n})+\frac{\epsilon}{2(b-a)} (b-a)\leq \epsilon.$

$\square$