Let $(X_n)$ be a sequence of non-negative, integrable, independent random variables such that $\forall n, E(X_n)=1$. Let $Y_0=1$ and $Y_n=\prod_{k=1}^n X_k$. Suppose that $\prod_{k=1}^\infty E(\sqrt {X_k})>0$ (the limit of the product exists and is positive). Prove that $Y_n$ converges with respect to the $L^1$ norm.
It's easily seen that $Y_n$ is a martingale with respect to the filtration $\mathcal F_n =\sigma(X_0,\ldots X_n)$. Then $\sqrt{Y_n}$ is a positive supermartingale, hence converges almost surely to some $Z$. Thus $Y_n$ converges almost surely to $Z^2$. I haven't been able to prove that $Y_n$ converges in $L^1$ to $Z^2$... I don't see how I can take advantage of the assumption $\prod_{k=1}^\infty E(\sqrt {X_k})>0$.
Define $a_n:=E(\sqrt{X_n})$ and $$M_n:=\prod_{k=1}^n\frac{\sqrt{X_k}}{a_k}.$$ Then $\{M_n\}$ is a non-negative martingale with respect to $\{\mathcal F_n\}$. Moreover, we have $$E(M_n^2)=E\left(\prod_{k=1}^n\frac{X_k}{a_k^2}\right)=\frac{E(Y_n)}{\left(\prod_{k=1}^na_k\right)^2}\le\frac1{\left(\prod_{k=1}^\infty a_k\right)^2}<\infty,$$ where the inequality follows since $E(Y_n)=1$ and $a_k\le1$ by Jensen's inequality. Thus by the $L^2$-martingale convergence theorem, $M_n\to M_\infty$ almost surely and in $L^2$. Since $Y_n=\left(\prod_{k=1}^na_k\right)^2M_n^2$, it follows that $\{Y_n\}$ is uniformly integrable and thus $Y_n\to Z$ in $L^1$.
Note: the converse holds. That is, if $\prod_{k=1}^\infty a_k=0$, then $Y_n\to0$ almost surely. Together these two results are called Kakutani's theorem.