$L^{1}$ bounded but non uniformly integrable family Let $X_{k}$ be a collection of independent R.V.'s with $E(X_{k})=1$ and $X_{k}>0$ for all $k$. Define $M_{n}=\Pi_{k=1}^{n} X_{k}$ and $a_{k}=E(\sqrt{X_{k}})$. Then
1) If $\Pi_{k=1}^{n}a_{k}\rightarrow 0$, then $M_{n}\rightarrow0$ a.s.
2) If $\Pi_{k=1}^{n}a_{k}\nrightarrow 0$, then $M_{n}$ converges a.s. and in $L^{1}$ to some $M_{\infty}$ where $M_{\infty}>0$ a.s.
Clearly, $M_{n}$ defines a martingale, So martingale convergence theorem guarantees that there always exists an limit $M_{\infty}$. My attempt for the first question is to show the convergence in probability to $0$. That is, for $\delta,\epsilon >0$
$$P(|M_{n}-M_{n+1}|>\delta \epsilon)\geq P(|M_{n}|>\delta) \ P(|X_{n+1}-1|>\epsilon)
$$
If we can show that, for some $\epsilon>0$
$$
\limsup P(|X_{n+1}-1|>\epsilon) > 0 
$$
Then we are done. Jensen's inequality gives $E\log X_{k} \leq \log E(X_{k}) = 0$. So,the condition $\Pi_{k=1}^{n}a_{k}\rightarrow 0$, which means $\frac{1}{2}\sum_{k=1}^{n} E(\log X_{k})\leq \sum_{k=1}^{n}\log E(\sqrt{X_{k}}) \rightarrow -\infty$, only implies 
$$
P(X_{k}\neq 1)>0 
$$
for infinitely many k. But this is not sufficent for $\limsup P(|X_{n+1}-1|>\epsilon)$ to be positive. Can someone give me a hint on solving this.
My attempt for second question is to show $M_{n}$ is a uniform integrable family. I trying to use the test function of uniform integrability to show $E(\varphi(X_{k}))$ is uniformly bounded, where $\varphi(x)/x \rightarrow \infty$ as $x\rightarrow \infty$. However, I cannot see how the condition $\Pi_{k=1}^{n}a_{k}\nrightarrow 0$ can help me to achieve this. Can someone also give me a hint on this.
 A: For the first question, let $Y_n:=\sqrt{M_n}$. Then 
$$
\mathbb E\left[Y_n\right]=\mathbb E\left[\prod_{k=1}^n\sqrt{X_k}\right]=\prod_{k=1}^n\mathbb E\left[\sqrt{X_k}\right]=\prod_{k=1}^na_k
$$
hence $Y_n\to 0$ in probability. This implies that $M_n\to 0$ in probability. By the martingale convergence theorem, $(M_n)$ converges to some $Y$ almost surely hence $Y=0$. 
Notice that $0\leqslant a_k\leqslant 1$ hence the second condition $\prod_{k=1}^{n}a_{k}\nrightarrow 0$ means that the product converges to a non-zero value.
For the second question, consider the martingale 
$$
M'_n:=\sqrt{M_n}\left(\prod_{k=1}^na_k\right)^{-1}.
$$
Since $\mathbb E\left[{M'_n}^2\right]=\left(\prod_{k=1}^na_k\right)^{-2}$, the sequence $\left(M'_n\right)_{n\geqslant 1}$ is bounded in $\mathbb L^2$ and therefore, it converges in $\mathbb L^2$ and almost surely to some random variable $Z'$. Exploiting the convergence of $\prod_{k=1}^na_k$ to a positive value, it follows that $\sqrt{M_n}\to$ some $Z$ almost surely and in $\mathbb L^2$ hence we can take $M_\infty=Z^2$.
