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

Theorem
Let $f_n: \closedint a b \to \R$ be Riemman integrable real functions which satisfy $\size {\map {f_n} x} \le K$ for all $n \in \N$.

Let $\map {f_n} x \to \map f x$ pointwise and $f$ also be a Riemman integrable real function.

Then:


 * $\ds \int_a^b \map {f_n} x \rd x \to \int_a^b \map f x \rd x$

Lemma
Let $\sequence {A_n}$ be a contracting sequence of bounded sets in $\R$ with an empty intersection.

We call $E \subset \R$ an elementary subset $E = \bigcup_{k \mathop = 1}^M \closedint {a_k} {b_k}$.

We define $\map m E$ as the total length of these intervals minus their overlaps.

Let $a_n := \sup \set {\map m E: E \subset A_n \text{ is an elementary subset} }$.

Then:
 * $a_n \to 0$

Proof of Lemma
The sequence $a_n$ is decreasing.

$a_n \ge \delta > 0$.

By the epsilon definition of supremum, for $\epsilon := \dfrac \delta {2^n}$ there exists elementary subset $E_n$ such that:


 * $\map m {E_n} \ge a_n - \dfrac \delta {2^n}$

For $H_n = \bigcap_{k \mathop = 1}^n E_k \subset \bigcap_{k \mathop = 1}^n A_k$, we will show that $H_n \ne \O$

Thus we will contradict the assumption that the $A_n$ have an empty intersection.

For each $n$, take any elementary subset $E \subset A_k \setminus E_k$.

Then we find:


 * $\map m E + \map m {E_k} = \map m {E \cup E_k} \le a_k \implies \map m E \le \dfrac \delta {2^k}$

Take an elementary subset $S \subset A_n \setminus H_n = \bigcap_{k \mathop = 1}^n \paren {A_n \setminus E_k}$ (this equality is by De Morgan's laws).

Then we find:


 * $E = \paren {E \setminus E_1} \cup \dotsb \cup \paren {E \setminus E_n}$

Therefore, we get the bound:


 * $\ds \map m E \le \sum_{k \mathop = 1}^n \map m {E \setminus E_k} \le \sum_{k \mathop = 1}^n \dfrac \delta {2^n} = \delta$

In words, any elementary subset $E \subset A_n \setminus H_n$ was shown to have measure $\map m E \le \delta$.

However, the inequality $a_n > \delta$ requires the existence of at least one elementary subset $U_n \subset A_n$ such that $\map m {U_n} > \delta$.

Since all the elementary subset $E \subset A_n \setminus H_n$ satisfy $\map m E \le \delta$, we must have:
 * $U_n \subset H_n$

for $n \ge 1$.

This contradicts the non-emptiness because $\ds \lim_{n \mathop \to \infty} \map m {U_n} > \delta$.

Proof of main result
, assume that $f_n \ge 0$ and $f_n \to 0$.

We will show that given $\epsilon \in \R_{>0}$ there exists $N$ such that $\forall n \ge N$ we have:


 * $\ds \int_a^b \map {f_n} x \rd x \le \epsilon$

Let $A_n := \set {x \in \closedint a b: \exists k \ge n: \map {f_k} x \ge \dfrac \epsilon {2 \paren {b - a} } }$.

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 $\map m {E_n} \le \dfrac \epsilon {2 K}$ for all $n \ge N$, and consider the following subsets:


 * $E := \set {x \in E_n: \exists k \ge n: \map {f_k} x \ge \dfrac \epsilon {2 \paren {b - a} } }$ $F := \closedint a b \setminus E$

Therefore, we find


 * $\ds \int_a^b \map {f_n} x \rd x \le \int_{E_n} \map {f_n} x \rd x + \int_F \map {f_n} x \rd x \le K \map m {E_n} + \dfrac \epsilon {2 \paren {b - a} } \paren {b - a} \le \epsilon$