Why does $\mathrm{E}[e^{-X}] = 0$ imply $\mathrm{P}(X = \infty)=1$?

Question

Came across the following problem:

For independent $(Y_n)_{n\geq1}$ with $Y_n \sim Exp(\lambda_n)$, let $X = \sum_{n\geq1} Y_n$. Show that if $\sum_{n\geq1} (\lambda_n)^{-1} = \infty$, then $\mathrm{P}(X = \infty)=1.$

With the following solution:

Notice that for $Y_n \sim Exp(\lambda_n)$, we have $\mathrm{E}[e^{-Y_n}] = \frac{\lambda_n}{1 + \lambda_n}.$ Consequently, $X_n := \sum_{k=1}^{n} Y_k$ satisfies $$\mathrm{E}[e^{-X_n}] \equiv (\prod_{k=1}^{n}(1 + \frac{1}{\lambda_k}) )^{-1} \leq (1 + \sum_{k=1}^{n} \frac{1}{\lambda_k})^{-1} \to 0$$ as $n \to \infty$. So that (MON) implies $\mathrm{E}[e^{-X}] = 0,$ equivalently, $\mathrm{P}(X = \infty)=1.$

I can follow the whole solution, except for the last statement: $\mathrm{E}[e^{-X}] = 0,$ equivalently, $\mathrm{P}(X = \infty)=1.$ Can someone walk me through the steps of this conclusion?

EDIT: Thanks for the help, and sorry for cross-posting. Didn’t realise that wasn’t allowed.

Answer 1

If $P(X=\infty)=1$ then $E e^{-X} = E[1\{X=\infty\}e^{-X}] = 0$. Conversely, suppose $P(X<\infty)>0$. Then $E[e^{-X}] \geq E[1\{X<\infty\}e^{-X}] > 0$ since $e^{-x}>0$ for all $x$ contradicting $E e^{-X} = 0$.

Answer 2

Hint: for (finite) real numbers $x$, $e^{-x} > 0$.