My question is similar to Random index central limit theorem. I have a different proof of a similar theorem and would like to check if it is correct. I am suspicious that I have made an oversight because my proof seems to be considerably simpler.
Let $X_1, X_2, ...$ be i.i.d. with $EX_i = 0$ and $EX_i^2 = 1$. Let $a_n$ be an increasing sequence of natural numbers (I could just as well take $a_n = n$) such that $a_n \rightarrow \infty$ and $N_n$ be a random sequence of natural numbers with $\frac{a_n}{N_n} \rightarrow 1$ in probability. I wish to show that $\frac{S_{N_n}}{\sqrt{N_n}} \implies \mathcal{N}(0, 1)$, i.e., converges in distribution.
proof: First, $\frac{S_{a_n}}{\sqrt{a_n}} \implies \mathcal{N}(0, 1)$ by the CLT and since a subsequence of a weakly converging sequence converges weakly. Then, fix some $\epsilon > 0$ and let $A_n = \{|\frac{S_{N_n}}{\sqrt{N_n}} - \frac{S_{a_n}}{\sqrt{a_n}}| > \epsilon\}$. Conditioning on the event $|N_n - a_n| \ge 1$:
\begin{align*} P(A_n) &= P(A_n\ \big|\ |N_n - a_n| \ge 1)P(|N_n - a_n| \ge 1) + P(A_n\ \big|\ |N_n - a_n| < 1) P(|N_n - a_n| < 1)\\ &= P(A_n\ \big|\ |N_n - a_n| \ge 1)P(|N_n - a_n| \ge 1)\\ &\le P(|N_n - a_n| \ge 1)\\ &\rightarrow 0 \end{align*}
where I've used $|N_n - a_n| < 1 \implies N_n = a_n$ for the second inequality and $P(\cdot) \le 1$ in the third, and that $N_n/a_n \rightarrow 1$ in probability in the final step.
Since $\epsilon$ is arbitrary, it follows that $|\frac{S_{N_n}}{\sqrt{N_n}} - \frac{S_{a_n}}{\sqrt{a_n}}| \rightarrow 0$ in probability and therefore $\frac{S_{N_n}}{\sqrt{N_n}} \implies \mathcal{N}(0, 1)$ by Slutsky's theorem. $\square$
Is this proof valid?
Normally, you don't have $\lim_{n \rightarrow +\infty} \mathbb{P}( |N_n-a_n| \ge 1 ) =0$ unless more conditions are specified.