Almost sure convergence of means of arbitrary sample sets?

Question

Suppose $\def\N{\mathbb{N}}(X_n)_{n\in\N}$ are i.i.d. integrable random variables with mean $\mu$. Let us write, for finite $I \subset \N$ $$\overline{X}_I = \frac1{|I|}\sum_{i\in I} X_i$$ (and $\overline{X}_\emptyset = 0$.) Suppose $(I_n)_{n\in\N}$ are finite subsets of $\N$ with $\lim_n|I_n| = +\infty$. Does it hold that $$\overline{X}_{I_n}\xrightarrow[n\to\infty]{~\text{a.s.}~}\mu ~ ?$$ The strong law of large numbers is the case $I_n = \{1,\dots,n\}$. I am mostly interested in the case where the $I_n$ are pairwise disjoint.

Answer 1

If you look carefully at the characteristic function proof of the weak law of large numbers, and the proof using finite fourth moments of the strong law of numbers in https://en.wikipedia.org/wiki/Law_of_large_numbers, the ordering of the variables plays no role in either proof, only the moments of $X_1$. Your statement at the very least holds for these assumptions and should likely hold for the full statement of the strong law of large numbers, but this isn't a guarantee.

Intuitively, your answer should be true, but I don't see why your statement immediately follows from the SLLN.