If we have that $\mid X_n \mid$ is bounded, why is it true that $X_n \overset{a.s.}\to c$ implies that $E(X_n) \to c$ with $c$ constant?

Question

I read in a probability book that said, without proof, that if $\mid X_n \mid$ is bounded, $X_n \overset{a.s.}\to c$ implies that $E(X_n) \to c$ with $c$ constant. Here, $X_n$ is the nth random variable of $X_1, \ldots, X_n$. I know that in general this is not true without the bounded part, ie, take $X_n = n^2 Bern(\frac{1}{n^2})$, but cant see what the boundedness changes. Does anyone have any hints? Thanks!

Answer 1

If we have

$$\lim X_n = c \ \text{a.s.}$$

let us take the expectation of both sides to get:

$$E[\lim X_n] = E[c] = c$$

Now do we have

$$E[\lim X_n] = \lim E[X_n]?$$

Note that boundedness of $|X_n|$'s means that $\exists M > 0$ s.t.

$$|X_n| \le M \ \forall n \ge 1$$

Note that $M$ is integrable $\because \ E[|M|] = E[M] = M < \infty$

By the dominated convergence theorem, we indeed have

$$E[\lim X_n] = \lim E[X_n]$$

$$\therefore c = \lim E[X_n] \ QED$$