The states that for independent identically distributed integrable random variables,
$$
\frac{S_n}{n}\longrightarrow\mu
\quad\text{almost surely}.
$$
To prove it, set $Y_n=X_n\mathbf1_{\{|X_n|\leq n\}}$. The tail-sum formula gives $\sum_n\mathbb P(X_n\ne Y_n)<\infty$, so the <Borel-Cantelli lemmas> make the two sequences eventually equal. Also
$$
\sum_{n=1}^\infty\frac{\operatorname{Var}(Y_n)}{n^2}
\leq\mathbb E\left[
X_1^2\sum_{n\geq|X_1|}\frac1{n^2}\right]
\leq C\mathbb E|X_1|<\infty.
$$
The <Kolmogorov convergence theorem> implies that $\sum_n(Y_n-\mathbb EY_n)/n$ converges almost surely, and <Kronecker lemma> yields
$$
\frac1n\sum_{j=1}^n(Y_j-\mathbb EY_j)\to0.
$$
Finally $\mathbb EY_n\to\mu$, so the <Cesaro mean> of these expectations tends to $\mu$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 201 / 2 / a / Solution

{c}

If independent centered random variables $(Y_n)$ satisfy $\sum_n\operatorname{Var}(Y_n)<\infty$, then $\sum_nY_n$ converges almost surely. It follows from Kolmogorov's maximal inequality and a Cauchy-tail argument.

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 Incoming links: Kolmogorov convergence theorem