Dominated Convergence Theorem. Dominated Convergence Theorem
"Suppose $X_{n}\rightarrow X$ a.s., and there is a random variable $Y$ with $E[Y]<\infty$ such that $|X_{n}|<Y$ for all $n$. 
Then 
$E[\lim_{n \to \infty}X_{n}]=\lim_{n\to \infty}E[X_{n}]$." 




Pardon the picture quality. Sorry guys i just have trouble grasping the red parts of this proof especially: 


*

*What is the role of the $\epsilon$ in the proof the red parts.?

*What is meant by and what is the relevance of the inequality $|X_{n}+Y|<2Y$?

*Why are we using the $N_{\epsilon}=\min \{|X_{i}-X|<\epsilon\}$  for all $i\ge n$? 

 A: *

*If we combine (a) and (b), we get


$\forall \epsilon > 0, \exists N > 0$ s.t. if $n > N$, then
$$E[X] - 3\epsilon \le E[X_n] \le E[X] + 3\epsilon$$
$$\iff - 3\epsilon \le E[X_n] - E[X] \le + 3\epsilon$$
$$\iff |E[X_n] - E[X]| \le 3\epsilon$$
$$\iff |E[X_n] - E[X]| \le \epsilon$$
which is the $\epsilon$-N definition of $\lim E[X_n] = E[X]$
The reason for the factor of 3 is related to the proof.


*If $X_n \ge 0$ and $|X_n| < Y$, then $$0 \le |X_n| = X_n < Y$$


$$\to |X_n| + Y  = X_n + Y\ge 0$$
Also, $|X_n + Y| \le |X_n| + |Y| = X_n + Y < Y + Y = 2Y$
The relevance of $|X_n + Y| < 2Y$ and $X_n + Y\ge 0$ is for proving the dominated convergence theorem for $X_n \in \mathbb{R}$ after proving it for the case $X_n \ge 0$ (as is done above).


*$N_{\epsilon}$ is our choice of $N$ (see 1), and it is the first time in the sequence $X_n, X_{n+1}, X_{n+2}, \dots$ where a random variable is w/in $\epsilon$-units of $X$


Look through the proof to see whenever $N_{\epsilon}$ is used and understand why that particular part is correct.
For instance, if $X_n$ converges almost surely to $X$, then $N_{\epsilon}$ is almost surely some finite number i.e. there is a 'first time in the sequence $X_n, X_{n+1}, X_{n+2}, \dots$ where a random variable is w/in $\epsilon$-units of $X$'
If you want to know how to come up w/ $N_{\epsilon}$, contact the authors. :P Why it's being used is simply that it works. It's not necessarily the best choice of $N$, but it's one that works.
