measure theory - Almost surely convergence to $0$ if and only if convergence to $0$ in probability

I am working on this question:

Prove that $X_{n}\rightarrow 0$ a.s. if and only if for every $\epsilon>0$, there exists $n$ such that the following holds: for every random variable $N:\Omega\rightarrow\{n,n+1,\cdots\}$, we have

$$P\Big(\{\omega:|X_{N(\omega)}(\omega)|>\epsilon\}\Big)<\epsilon.$$

Is this question equivalent to asking me to prove "almost surely convergence to $0$ if and only if convergence to $0$ almost surely"?

If so, the direction $(\Rightarrow)$ can be proved following this: Convergence in measure and almost everywhere

However, isn't the direction $(\Leftarrow)$ not generally true? I can surely prove that there exists a subsequence $X_{k_{n}}$ of $X_{n}$ converges to $0$ almost surely...

Could someone tell me what this question is really asking about? I don't really want to spend time proving a wrong thing..

Thank you!

Answer

First, $X_n\to 0$ a.s. iff for any $\epsilon>0$, there exists $n\ge 1$ s.t. $\mathsf{P}(\sup_{m\ge n}|X_m|>\epsilon)<\epsilon$. In the following we fix $\epsilon>0$.

$(\Rightarrow)$ Suppose that $X_n\to 0$ a.s. Then since for any r.v. $N$ on $\{n,n+1,\ldots\}$, $|X_{N}|\le \sup_{m\ge n}|X_m|$, the result follows from the above statement.

$(\Leftarrow)$ Let $N_n':=\inf\{m\ge n:|X_m|>\epsilon\}$. Define $N_n:=N_n'1\{N_n'<\infty\}+n1\{N_n'=\infty\}$. Then $\{\sup_{m\ge n}|X_m|>\epsilon\}=\{|X_{N_n}|>\epsilon\}$. However, there exists $n\ge 1$ s.t. $\mathsf{P}(|X_{N_n}|>\epsilon)<\epsilon$.