The Weak one

\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 Variance and the third is by our upper bound on the variance. \end{proof}

The Strong one

\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 Theorem 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.).