= Solution
The <strong law of large numbers> states that if $(X_i)_{i\geq1}$ are <independent and identically distributed random variables> with $\mathbb E|X_1|<\infty$, then
$$
\frac1n\sum_{i=1}^nX_i\longrightarrow\mathbb E X_1
$$
<almost surely>. Under the stronger fourth-moment hypothesis, it has a short proof. Put $\mu=\mathbb EX_1$, $Y_i=X_i-\mu$, and $S_n=\sum_{i=1}^nY_i$. Independence and centering make every term in the expansion of $S_n^4$ vanish unless each index occurs at least twice, so
$$
\mathbb ES_n^4=n\mathbb EY_1^4+3n(n-1)(\mathbb EY_1^2)^2\leq Cn^2.
$$
For every $\varepsilon>0$, the <Markov inequality> gives
$$
\mathbb P(|S_n|>\varepsilon n)\leq\frac{C}{\varepsilon^4n^2}.
$$
The probabilities are summable, so the first <Borel-Cantelli lemma> says that $|S_n|>\varepsilon n$ occurs only finitely often almost surely. Apply this simultaneously to $\varepsilon=1,1/2,1/3,\ldots$ to obtain $S_n/n\to0$ almost surely, which is the claimed law.
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