Do random subsequences of converging random variables converge as well? I am trying to understand the convergence of the sequence of random variables $(X_{N_n})_n$ where $(X_n)_n$ is a sequence of random variables and $(N_n)_n$ is a sequence of random positive integers that "increase to $\infty$" (since this is a sequence of random variables there is different notions of how this $N_n \to \infty$). There are two scenarios that I am trying to analyze.
Suppose $X_n\xrightarrow{P}0$ and $N_n\xrightarrow{a.s.}\infty$. Is it neccesarily true that $X_{N_n}\xrightarrow{P}0$?. Similarly if $X_n\xrightarrow{a.s.} 0$ and $N_n\xrightarrow{P}\infty$ (This just means $P(|N_n|\le K)\to 0$ for all $K>0$) is it necessary that $X_{N_n}\xrightarrow{a.s.} 0$?
 A: A counterexample where $X_n \overset{p}{\to}0$ and $N_n \overset{\text{a.s.}}{\to}\infty$ does not imply $X_{N_n} \overset{p}{\to}0$.
Consider a sequence of independent events $A_1,\ldots$ such that $\forall n\geq 1$, $P(A_n)\geq \frac 1n$ and let $X_n = 1_{A_n}$, so that $X_n \overset{p}{\to}0$.
By Borel-Cantelli lemma, $P(\limsup_n A_n) = 1$ and for each $\omega\in \limsup_n A_n$, you can define a subsequence $N_n(\omega)$ such that $\forall n$, $\omega\in A_{N_n(\omega)}$ and $N_n(\omega)\to \infty$. This implies $X_{N_n(\omega)}(\omega)=1$, thus a.s. $X_{N_n}=1$ and $X_{N_n}$ does not converge in probability to $0$.

A proof that $X_n \overset{a.s.}{\to}0$ and $N_n \overset{P}{\to}\infty$ implies $X_{N_n} \overset{a.s.}{\to}0$.
It suffices to show that $N_n \overset{a.s.}{\to}\infty$.
Since $N_n$ is increasing,the following inclusion of events holds: $$(N_n \to \infty)^c \subset \Big(\bigcup_{m\geq 1} \liminf_n(N_n\leq m)\Big)$$
so it suffices to show that for fixed $m\geq 1$, $P(\liminf_n(N_n\leq m)) = 0$.
The next step is to write $$P(\liminf_n(N_n\leq m)) = \lim_{n_0}P\Big(\bigcap_{n\geq n_0} (N_n\leq m)\Big)$$
and note that
$$P\Big(\bigcap_{n\geq n_0} (N_n\leq m)\Big)\leq P(N_{n_0}\leq m).$$ By the hypothesis $N_n \overset{P}{\to}\infty$, $$\lim_{n_0\to \infty} P(N_{n_0}\leq m) = 0$$
hence by squeezing, $\lim_{n_0}P\Big(\bigcap_{n\geq n_0} (N_n\leq m)\Big) = 0$, thus $P(\liminf_n(N_n\leq m)) = 0$ and finally
$$P(N_n \to \infty) = 1.$$
A: Below I attempt to prove the first of your two questions. [There may be errors...]
I assume the sequences $(X_n)$ and $(N_n)$ are independent of each other.

Suppose $X_n \overset{p}{\to}0$ and $N_n \overset{\text{a.s.}}{\to}\infty$.
Let $E=\{N_n \to \infty\}$
\begin{align}
&\lim_{n \to \infty} P(|X_{N_n}| > \epsilon)
\\
&\le
\lim_{n \to \infty} P(\{|X_{N_n}| > \epsilon\} \cap E)
+ \underbrace{\lim_{n \to \infty} P(\{|X_{N_n}| > \epsilon\} \cap E^c)}
_{\le P(E^c) = 0}
\\
&= \lim_{n \to \infty}
E\left[
P\left(
\{|X_{N_n}| > \epsilon\} \cap E \mid (N_n)_n
\right)
\right]
\\
&=
E\left[
\lim_{n \to \infty}
P\left(
\{|X_{N_n}| > \epsilon\} \cap E \mid (N_n)_n
\right)
\right]
& \text{dom. convergence thm.}
\\
&=0.
\end{align}
