Proof of Strong Law of Large Numbers under a Fourth Moment Bound

Step 0 (reduction to mean zero). Set $Y_m=X_m-\mu$. The sequence $(Y_m)$ is independent and identically distributed: translation by $-\mu$ is a continuous, hence Borel measurable, map (generator criterion of Measurable Function and Real-Valued Measurable Function), so the defining events $\{Y_m\in B\}$ are events of the form $\{X_m\in B'\}$ for Borel $B'$, and both the independence product formula and equality of distributions transfer. Moreover $\mathbb{E}[Y_1]=0$, and $Y_1^{4}$ has finite expectation since $(x-\mu)^{4}$ expands binomially into powers $x^{k}$, $k\le 4$, each with finite expectation by hypothesis and linearity (claim 2 of Linearity and Monotonicity of the Lebesgue Integral). Writing $T_n=Y_1+\cdots+Y_n=S_n-n\mu$, it suffices to prove $T_n/n\to 0$ almost surely.

Step 1 (fourth moment bound). Expanding $T_n^{4}$ multinomially and using linearity,

For a term whose index multiset contains an index $i$ appearing exactly once, group the remaining three factors into the single random variable $Z$ (a product of powers of the $Y$'s with indices different from $i$); the pair $(Y_i,Z)$ satisfies the independence hypothesis of Expectation of a Product of Independent Random Variables, since $Z$ is obtained from the finitely many $Y_r$, $r\ne i$, by products and powers, and the events $\{Z\in B\}$ lie in the collection generated by events $\{Y_r\in B_r\}$, $r\ne i$, for which the product formula with events of $Y_i$ holds by independence of finite subfamilies; integrability of the factors holds by the moment bounds of Step 0. Hence such terms equal $\mathbb{E}[Y_i]\,\mathbb{E}[Z]=0$. The surviving terms are the $n$ diagonal terms $\mathbb{E}[Y_i^{4}]$ and the $3n(n-1)$ terms $\mathbb{E}[Y_i^{2}Y_j^{2}]$ with $i\ne j$, each equal to $\mathbb{E}[Y_1^{2}]^{2}$ by Expectation of a Product of Independent Random Variables and Step 0 of the proof of Weak Law of Large Numbers (identically distributed variables share moments). Therefore, with $C=\mathbb{E}[Y_1^{4}]+3\,\mathbb{E}[Y_1^{2}]^{2}$,

Step 2 (Borel–Cantelli). Fix $\varepsilon>0$. By Markov's inequality (Markov's and Chebyshev's Inequalities) applied to the nonnegative random variable $T_n^{4}$ with threshold $n^{4}\varepsilon^{4}$,

and $\sum_n C/(n^{2}\varepsilon^{4})<\infty$ (the partial sums of $\sum 1/n^{2}$ are bounded, e.g. by $2$, via $1/n^{2}\le 1/(n(n-1))=1/(n-1)-1/n$ for $n\ge 2$). By the first Borel–Cantelli lemma, for each $j\in\mathbb{N}$ the event $N_j=\limsup_n\{|T_n/n|\ge 1/j\}$ has $P(N_j)=0$.

Step 3 (conclusion). Let $N=\bigcup_j N_j$; by countable subadditivity of $P$ (disjointify and use countable additivity with monotonicity, as in claim 1 of the proof of Existence of Lebesgue Measure on the Real Line for measures generally), $P(N)=0$. For $\omega\notin N$: for every $j$ there are only finitely many $n$ with $|T_n(\omega)/n|\ge 1/j$, which is precisely the statement that $T_n(\omega)/n\to 0$ in the sense of Limit of a Sequence of Real Numbers. Hence the convergence set contains the complement of $N$ and has probability $1$: $T_n/n\to 0$, i.e. $S_n/n\to\mu$, almost surely. $\blacksquare$