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Let $(X_n)_n$ be a sequence of random variables in $\mathbb{Z}_+$ such that $X_n\le n$ and $\frac{X_n}{n}\to c>0$ almost surely.

Can I conclude that $\frac{X_{N_k}}{N_k}\to c$ almost surely, where $(N_k)_k$ is a sequence of random variables such that $N_k\in\mathbb{Z}_+$, $N_k\le N_{k+1}$ and $N_k\to\infty$?

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Yes, that's correct.

For fixed $\omega$ and $\epsilon>0$ there exists (with probability 1) some number $M \in \mathbb{N}$ such that

$$\forall n \geq M: \left| \frac{X_n(\omega)}{n} - c \right| \leq \epsilon. \tag{1}$$

Now choose $K=K(\omega) \in \mathbb{N}$ sufficiently large such that $N_k(\omega) \geq M$ for all $k \geq K$. Then, by $(1)$,

$$\forall k \geq K: \left| \frac{X_{N_k}(\omega)}{N_k(\omega)}-c \right| \leq \epsilon.$$

Since $\epsilon>0$ is arbitrary, this proves $X_{N_k}/N_k \to c$ almost surely.

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