Show by contradiction that a martingale converge to zero almost surely. 
Theorem 5.2.9. If $X_n \geq 0$ is a supermartingale then as $n \rightarrow \infty, X_n \rightarrow X$ a.s. and $E X \leq E X_0$


5.2.9. Let $Y_1, Y_2, \ldots$ be nonnegative i.i.d. random variables with $E Y_m=1$ and $P\left(Y_m=1\right)<1$.

*

*(i) Show that $X_n=\prod_{m \leq n} Y_m$ defines a martingale.

*(ii) Use Theorem $5.2 .9$ and an argument by contradiction to show $X_n
   \rightarrow 0$ a.s.


I am trying to show that $X_n \to 0$ a.s., and I want to try to contradict the fact that $X_n$ is a martingale and must not be a supermartingale by showing the inequality is actually strict. To contradict the statement, my idea is to assume there exists a set $A$ of positive probability on which $\lim X_n $ does not converge to zero, and derive the contradiction from this. However, all I would obtain is $0<E[X]\leq E[X_0]$ however $E[X_0]=E[Y_1]=1$ which I am having trouble realizing as a contradcition.
Thank you
 A: The idea given by Michael in the comment is very good : $\mathbb P(Y_1>1)$ is positive. Otherwise, since $\mathbb E[Y_1]=1$ and $Y_1\leqslant 1$, this would force $Y_1=1$ almost surely which contradicts $\mathbb P(Y_m=1)<1$. Since $\{Y_1>1\}=\bigcup_{k\geqslant 1}\{Y_1>1+1/k\}$ and $(\{Y_1>1+1/k\})_{k\geqslant 1}$ is non-decreasing, one get that for some $k_0\geqslant 1$,
$
\mathbb P(Y_1>1+1/k_0)>0.
$
By the second Borel-Cantelli lemma, we infer that for almost every $\omega$, there exists an increasing sequence of integers $\left(m_{j}(\omega)\right)_{j\geqslant 1}$ such that for each $j\geqslant 1$, $ Y_{m_{j}(\omega)}(\omega)>1+1/k_0$.
As a consequence, one has
$$
X_{m_{j}(\omega) }(\omega)=X_{m_{j}(\omega)-1 }(\omega)Y_{m_{j}(\omega)}(\omega)
\geqslant X_{m_{j}(\omega)-1 }(\omega)\left(1+\frac 1{k_0}\right)
$$
By Theorem 5.2.9., we know that $X_{m_{j}(\omega) }(\omega)\to X(\omega)$ and
$X_{m_{j}(\omega)-1 }(\omega)\to X(\omega)$ hence taking the limit in (1) gives $X(\omega)\geqslant X(\omega)\left(1+\frac 1{k_0}\right)$ hence $X(\omega)=0$.
