The question is:
Let $Y_1 , Y_2, \dots$ be nonnegative i.i.d. random variables with $\mathbb{E}Y_m = 1$ and
$\mathbb{P} (Y_m = 1) < 1$. (i) Show that $X_n = \prod_{m \le n} Y_m$ defines a martingale. (ii) Use an argument by contradiction to show $X_n \to 0$ a.s.
(i) is easy to check. For (ii), by Martingale Convergence Theorem, we can show that $X_n$ converges to almost surely to some $X$ with $\mathbb{E}X \le \mathbb{E}X_0 = 1.$ ($X_0$ is not explicitly defined in the question, but to make $X_n$ a martingale, we need $X_0 = 1$.)
My guess is that $X = 0$ almost surely must comes from the fact $\mathbb{P} (Y_m = 1) < 1$. But I can't see how to continue from here.
Best Answer
Why an argument by contradiction? Note that $\log X_n$ is the sum of $n$ i.i.d. random variables with mean $m=\mathbb E(\log Y_1)$, hence, if $m\lt0$, by the strong law of large numbers, $\log X_n\to-\infty$.
But $m\leqslant\log\mathbb E(Y_1)=0$ by Jensen inequality and one knows that this convexity inequality is strict as soon as the random variable $Y_1$ is not almost surely constant. This is what the hypothesis $\mathbb P(Y_1=1)\lt1$ ensures. Hence, $m\lt0$, $\log X_n\to-\infty$ almost surely, and $X_n\to0$ almost surely.