Prove $X_{\infty} < \infty$

Question

From Williams' Probability with Martingales:

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How exactly do we prove $X_{\infty} < \infty$ a.s.?

$$E[|X_{\infty}|] = E[|\lim X_n|] = E[|\liminf X_n|] = E[\liminf |X_n|]$$

$$ \le \liminf E[|X_n|] \ \text{by Fatou's Lemma}$$

$$ \le \limsup E[|X_n|]$$

$$ \le \sup E[|X_n|]$$

$$< \infty \ \text{by assumption}$$

$$\to |X_{\infty}| < \infty \ a.s.$$

$$\to X_{\infty} < \infty \ a.s.$$

Answer 1

The proof is correct, but notice that you get something stronger than what you state in the last line of your question, indeed $X_{\infty}$ is finite almost surely, which is the same as saying $-\infty < X_{\infty} < \infty$ (and not only $X_{\infty} < \infty$).

To go from the fact that $E(|X_{\infty}|) < \infty$ to the fact that $X_{\infty} \in \mathbb{R}$ a.s., notice that since $X_{\infty} \in L^1$, then Lebesgue's DCT implies that $|X_{\infty}|1_{\{|X_{\infty}| > n\}} \to 0$. In turn, this shows that $\lambda\big(\{w : |X_{\infty}(w)| > n\}\big) \to 0$ as $n\to \infty$.