I am working on an exercise as follows.
Given i.i.d non-negative random variables $(X_{i})_{i=1}^{\infty}$ with $\mathbb{P}(X_{1}=0)>0$. Show that $$\lim_{n\rightarrow\infty}\sqrt[n]{X_{1}\dots X_{n}}=0.$$
I can show the almost surely convergence but the exercise asks for "surely" convergence. For the almost surely convergence, it is easy to see that $\sum_{n=1}^{\infty}\mathbb{P}(X_{n}=0)=\infty$, and thus using independence and Borel-Cantelli lemma, we have $\mathbb{P}(X_{n}=0\ \text{i.o.})=1$. Hence, with probability $1$, for every $n\geq 1$, there exists $k\geq n$ such that $X_{k}=0$, which implies that $X_{1}\dots X_{k}\dots X_{n}=0$ for all $n\geq k$.
However, how to extend this result to all $\omega\in\Omega$ instead of just almost surely convergence? In other words, I also have to show that $Y_{n}\rightarrow 0$ for all $\omega$ such that $X_{n}(\omega)\neq 0$ eventually. I've been trying to show that in fact $\{X_{n}\neq 0\ e.v.\}=\varnothing$, but I do not see a doable way to achieve this.
Can this "surely convergence" be true? Is there an example where such a "weighted product" converges almost surely but not convergence surely?
Whether it's $\emptyset$ or not would depend on the probability space and how the random variables (as functions) are defined.
For example, take the countable product of $([0,1],\mathcal{B}([0,1]),\lambda)$.
If you define $X_{n}:[0,1]^{\mathbb{N}}\to\Bbb{R}$ by $X_{n}=\mathbf{1}_{[0,\frac{1}{2}]}\circ\pi_{n}$ . Then $X_{n}$ are iid. But see that $(0,0,...)\in\{X_{n}\neq 0\ e.v.\}$ (i.e. the $0$ sequence) is in this set and hence it's not empty.
Intuitively, in an infinite coin tossing scenario, you "CAN" end up with all tosses being heads but it will happen with $0$ probability measure.