\begin{proof} Sufficient to prove for $\mu=0$ (we can define $X_{i}=X_{i}'-\mu$) For $\mu=0$, $$\begin{align*} \mathbb{P}\left( \left| \frac{1}{n}\sum_{i=1}^{n}X_{i} \right| \ge\epsilon\right)&\le \frac{1}{\epsilon^{2}}\mathbb{E}\left[ \left| \frac{1}{n}\sum_{i=1}^{n}X_{i} \right| ^{2} \right]&\text{Chebyshev}\\ &= \frac{1}{(\epsilon n)^{2}}\sum_{i=1}^{n}\mathbb{E}[X_{i}^{2}]&\text{independence of variance}\\ &\le \frac{Vn}{\epsilon^{2}n^{2}}= \frac{V}{\epsilon^{2}n}=O_{\epsilon}\left( \frac{1}{n} \right)\to 0 \end{align*}$$ where the first inequality is due to Chebyshev Inequality, the second is by the fact that Independence of Variance and the third is by our upper bound on the Variance. \end{proof}

\begin{proof} Again we reduce to the case where $\mu=0$. Let $\sigma^{2}=\mathbb{E}[X_{i}^{2}]<\infty$. We first note that $$\mathbb{E}\left[ \left| \frac{X_{1}+\dots+X_{n}}{n} \right| ^{2} \right]=\frac{1}{n^{2}}(\sigma^{2}+\dots+\sigma^{2})=\frac{\sigma^{2}}{n}$$Let $Z_{n}=\left| \frac{X_{1}+\dots+X_{n}}{n} \right|^{2}$, we then note that $$\sum_{m\ge 1}\mathbb{E}[Z_{m^{2}}]=\sum_{m\ge 1} \frac{\sigma^{2}}{m^{2}}<\infty.$$So by Beppo-Levi (Probability) we have that:

• $\sum_{m\ge 1}Z_{m^{2}}<\infty$ a.s. and hence
• $Z_{m^{2}}\to0$ a.s. But, now suppose $n\to \infty$ and find $m=m(n)$ with $m^{2}\le n<(m+1)^{2}$. Have that $$n-m^{2}\le (m+1)^{2}-m^{2}=2m+1.$$Then, $$\left| \frac{X_{1}+\dots+X_{n}}{n} \right| =\left| \frac{X_{1}+\dots+X_{m^{2}}+O(2m+1)}{n} \right| \le \left| \frac{X_{1}+\dots+X_{m^{2}}}{m^{2}}+O\left( \frac{1}{m} \right) \right| \to0$$a.s..

\end{proof}

Intuition

You can think of this as a sequence of i.i.d. RVs with finite expectation converging to the mean almost surely (or w.p.1.).