Proving a variant of the Kolmogorov SLLN Let $X_1, X_2, ...$ be a sequence of independent r.v.'s (not necessarily identical). Now for all $i$, we have $E(X_i) = 0$ and $E|X_i|^{1 + \delta} \leq C$ for some $\delta > 0$ and $C < \infty$. Prove
$$
n^{-1}\sum_{i = 1}^nX_i \to 0 \ a.s.
$$

My angle of attack:
If $\delta \geq 1$:
there is nothing to prove, it is just the simpliest case of Kolmogorov SLLN. So the challenging part is when $\delta \in (0,1)$. So Kolmogorov SLLN says that if
$$
\sum_{i=1}^\infty\frac{Var(X_i)}{i^2} < \infty,
$$
Then we will have convergence. So my thinking is that if I can show $E|X_n|^2 \sim \mathcal{O}(n^p)$ with $p \in (0,1)$, then I am done. This should obviously be connected with the condition that $E|X_i|^{1+\delta} \leq C$.
 A: If $\delta \in (0,1)$:
Truncate $X_n$. Let $Y_k = X_k\mathbf{1}\{|X_k| \leq k\}$. Since $E|X_i| < \infty$ for all $i$, then $P(X_k \neq Y_k \ i.o.) = 0$. So it is sufficient to show that
$$
\frac{\sum_i^n Y_i}{n} \to 0 \ a.s.
$$
Now
\begin{align}E|Y_i|^2 &\leq 1 + \sum_n^\infty P(|Y_i|^2 \geq n) \\
&\leq 1 + \sum_n^{i^2} P(|X_i|^2 \geq n) = 1 + \sum_n^{i^2} P(|X_i| \geq \sqrt{n})\\
&\leq 1 + \sum_{n=1}^{i^2}\frac{E|X_i|^{1+\delta}}{n^{(1+\delta)/2}} \leq 1 + \sum_{n=1}^{i^2}\frac{C}{n^{(1+\delta)/2}}\\
&< 1 + C'i^{1-\delta} \ (\text{integral approximation}).\end{align}
Where $C'$ is some constant. Then
\begin{equation}
\sum_{i = 1}^\infty Var(Y_i)/i^2 = \sum_{i = 1}^\infty E|Y_i|^2/i^2 < \sum_i^\infty\frac{1}{i^2} + \frac{C'}{i^{1+\delta}} < \infty.
\end{equation}
Therefore, the condition for Kolmogorv SLLN has met and we have
$$
\frac{\sum_i^n (Y_i - E(Y_i))}{n} \to 0 \ a.s.
$$
We now need to show $\sum_{i}^n\mu_i/n \to 0$ as $n \to \infty$ where $\mu_i = E(Y_i)$. To see this, we have $\mu_i = -E(X_i\mathbf{1}\{|X_i| > i\})$ since $E(X_i) = 0$. Therefore,
\begin{align}
|\mu_i| &= E(|X_i|\mathbf{1}\{|X_i| > i\})\\
        &= iE(|X_i|/i\mathbf{1}\{|X_i| > i\})\\
        &\leq iE(|X_i|^{1+\delta}/i^{1+\delta}\mathbf{1}\{|X_i| > i\})\\
        &\leq \frac{C}{i^\delta}.
\end{align}
Therefore, $\sum_{i}^n|\mu_i|/n \leq C\frac{n^{1-\delta}}{n} = Cn^{-\delta} \to 0$ as $n \to \infty$ and $\sum_{i}^n\mu_i/n \to 0$ as $n \to \infty$.
