If $\{X_n\}$ are independent random variables such that the fourth moments $\mathscr{E}(X_n^4)$ have a common bound and define $S_n=\sum_{j=1}^nX_j$, then $$\frac{S_n-\mathscr{E}(S_n)}{n}\rightarrow0\mbox{ a.e.}$$
$\bullet$ Proof.
WLOG, suppose $\mathscr{E}(X_n)=0$ for all $n$ and denote the common bound of $\mathscr{E}(X_n^4)$ to be $$\mathscr{E}(X_n^4)\leq M_4<\infty\mbox{ for all }n.$$Then by Lyapunov's inequality, we have the second moments $$\mathscr{E}|X_n|^2\leq\left[\mathscr{E}|X_n|^4\right]^\frac{2}{4}\leq \sqrt{M_4}<\infty.$$Consider the fourth moment of $S_n$, $$\begin{array}{rl}\mathscr{E}(S_n^4)
&=\mathscr{E}\left[\left(\sum_{j=1}^nX_j\right)^4\right]\\ &= \mathscr{E}\left[\sum_{j=1}^nX_j^4+{4\choose1}\sum_{i\neq j}X_iX_j^3+{4\choose2}\sum_{i\neq j}X_i^2X_j^2\right.\\ &\quad\left.+{4\choose1}{3\choose1}\sum_{i\neq j\neq k}X_iX_jX_k^2+{4\choose1}{3\choose1}{2\choose1}\sum_{i\neq j\neq k\neq l}X_iX_jX_kX_l\right]\\
&=\sum_{j=1}^n\mathscr{E}(X_j^4)+4\sum_{i\neq j}\mathscr{E}(X_i)\mathscr{E}(X_j^3)+6\sum_{i\neq j}\mathscr{E}(X_i^2)\mathscr{E}(X_j^2)\quad(\because\mbox{ indep.})\\ &\quad+12\sum_{i\neq j\neq k}\mathscr{E}(X_i)\mathscr{E}(X_j)\mathscr{E}(X_k^2)+24\sum_{i\neq j\neq k\neq l}\mathscr{E}(X_i)\mathscr{E}(X_j)\mathscr{E}(X_k)\mathscr{E}(X_l)\\ &=\sum_{j=1}^n\mathscr{E}(X_j^4)+6\sum_{i\neq j}\mathscr{E}(X_i^2)\mathscr{E}(X_j^2)\qquad\qquad(\because\mbox{ assuming }\mathscr{E}(X_n)=0.) \\ &\leq nM_4+3n(n-1)\sqrt{M_4}\sqrt{M_4}=n(3n-2)M_4.\end{array}$$By Markov's inequality, for $\varepsilon>0$, $$\mathscr{P}\{|S_n|>n\varepsilon\}\leq\frac{\mathscr{E}(S_n^4)}{n^4\varepsilon^4}\leq\frac{n(3n-2)M_4}{n^4\varepsilon^4}=\frac{3M_4}{n^2\varepsilon^4}+\frac{2M_4}{n^3\varepsilon^4}.$$Thus, $$\sum_n\mathscr{P}\{|S_n|>n\varepsilon\}\leq\sum_n\frac{3M_4}{n^2\varepsilon^4}+\frac{2M_4}{n^3\varepsilon^4}<\infty.$$By Borel-Cantelli Lemma I, we have $$\mathscr{P}\{|S_n|>n\varepsilon\mbox{ i.o.}\}=0\implies\frac{S_n}{n}\rightarrow0\mbox{ a.e.}$$
$\Box$
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