Prove $0 = E[e^{-M_{\infty}}] \to M_{\infty} = \infty$ a.s. w/ and then w/o Kolmogorov

Question

In Independent events and Kolmogorov, it seems Petite Etincelle is trying to answer without Kolmogorov 0-1 Law.

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Without using Kolmogorov 0-1 Law how do you prove the last step? I tried:

$$0 = E[e^{-M_{\infty}}] = E[e^{-M_{\infty}}1_{M_{\infty}=\infty}+e^{-M_{\infty}}1_{M_{\infty}<\infty}]$$

$$=E[0 \times 1_{M_{\infty}=\infty}+e^{-M_{\infty}}1_{M_{\infty}<\infty}]$$

Then what?

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With Kolmogorov 0-1 Law, how do you prove the last step?

I tried to suppose on the contrary that $P(e^{-\sum_{k=1}^\infty I_{E_k}} = 0)<1$. Then by Kolmogorov 0-1 Law, $$P(e^{-\sum_{k=1}^\infty I_{E_k}} = 0)=0$$

$$\to P(\sum_{k=1}^\infty I_{E_k} < \infty) = 1$$

$$\to E[\sum_{k=1}^\infty I_{E_k}] < \infty$$

$$\to \sum_{k=1}^\infty E[I_{E_k}] < \infty$$

$$\to \sum_{k=1}^\infty P(E_k) < \infty$$

Answer 1

(Expanded from comments)

Recall the fact that if $X$ is a non-negative random variable satisfying $\mathbf{E}[X] = 0$, then $X = 0$ $\mathbf{P}$-a.s. Indeed, by the Markov inequality

$$ \mathbf{P}[X \geq \epsilon] \leq \frac{1}{\epsilon}\mathbf{E}[X] = 0 $$

and hence $\mathbf{P}[X > 0] = \mathbf{P}[\cup_{n\geq 1}\{X \geq 1/n\}] \leq \sum_{n\geq1} \mathbf{P}[X\geq1/n] = 0 $.

Now you can apply this claim to $X = e^{-M_{\infty}}$.

Answer 2

Use Markov's inequality with $X = e^{-M_{\infty}}$:

$$ \mathbf{P}[X \geq \epsilon] \leq \frac{1}{\epsilon}\mathbf{E}[X] = 0 $$

$$ \to \mathbf{P}[X \leq \epsilon] = 1 $$

$$ \to \mathbf{P}[X = 0] = 1 $$