how that $P(G)=1$ iff $\sum_n \Bbb P(A \cap E_n )=\infty$ for all events $A$ having $\Bbb P(A)>0$. Two probability problems:
1. Let $a>0$ and let $X_n$, $n \geq 1$, be iid r.v. that are uniform on $(0,a)$ and let $Y_n = \prod_{k=1}^{n} X_k$. Determine all values of $a$ for which $\lim_{n \rightarrow \infty} Y_n =0$ with probability one.


*Let $E_1, E_2,\ldots$ be arbitrary events. Let $G= \limsup_n E_n$.
Show that $P(G)=1$ iff $\sum_n \Bbb P(A \cap E_n )=\infty$ for all events $A$ having $\Bbb P(A)>0$.


For problem 1, I guess I need to discuss the case when $a>1$ and $a<1$ but not sure whether it's the right track. I don't know how to deal with the almost surely convergence here.
For problem 2, we know $G=\cap \bigcup E_n$. It seems the result is quite direct but I don't know how to argue.
 A: Suppose there exists $A$ such that $\sum_{n=1}^\infty \mathbb{P}(A\cap E_n)<\infty$. By the Borel-Cantelli Lemma, $$\mathbb{P}\left(\limsup_n (A \cap E_n)\right)=0.$$
Note that $\limsup_n (A \cap E_n) = A \cap G$.
Thus, $\mathbb{P}(A \cap G)=0$, so $\mathbb{P}(G)=\mathbb{P}(G \setminus A) \le 1-\mathbb{P}(A)<1$.

Conversely, if $P(G)<1$, then let $A$ be the complement of $G$; note that $\mathbb{P}(A)>0$. By definition, $A$ consists of the points that appear in finitely many $E_n$.
Define $$A_k:= A \setminus \bigcup_{n=k+1}^\infty E_n,$$ i.e., the set of points of $A$ that don't appear in $E_n$ for all $n>k$.
Note that $A_1 \subset A_2 \subset \cdots,$ and $\bigcup_{k=1}^\infty A_k=A$ by the definition of $A$. Therefore $$\lim_{k \to \infty} \mathbb{P}(A_k) = \mathbb{P}(A)>0,$$ so there exists some $k^*$ such that $\mathbb{P}(A_{k^*})>0$.
Then by definition, $\mathbb{P}(A_{k^*} \cap E_n)=0$ for all $n>k^*$, so $$\sum_{n=1}^\infty \mathbb{P}(A_{k^*} \cap E_n)=\sum_{n=1}^{k^*}\mathbb{P}(A_{k^*} \cap E_n) <\infty.$$
