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

Theorem
Suppose $f_n:[a,b]\to \R$ 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 $E\subset \R$ 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 supremum, 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 $S\subset A_n\setminus H_n=\bigcap_{k=1}^n (A_n\setminus E_k)$ (this equality is by DeMorgan's laws), 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 $\lim_{n\to \infty} m(U_n)>\delta$.

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


 * $\displaystyle \int_a^b f_n(x)dx\leq \int_{E_n}f_n(x)dx+\int_{F}f_n(x)dx\leq K m(E_n)+\frac{\epsilon}{2(b-a)} (b-a)\leq \epsilon.$