Dominated Convergence Theorem.

Lebesgue Dominated Convergence Theorem ( Real analysis)
Lebesgue Dominated Convergence Theorem ( Real analysis)

  1. 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.

  1. 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).

  1. $N_{\epsilon}$ 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. 😛 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.

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