Proof of Weak Law of Large Numbers

Step 0 (identically distributed variables share moments). Let $Z$ and $W$ be identically distributed random variables with $Z,W\ge 0$. With the dyadic functions $\varphi_m$ of Step 2 of the proof of Expectation of a Product of Independent Random Variables, $\varphi_m(Z)=\sum_i c_i\mathbf{1}_{\{Z\in B_i\}}$ with Borel sets $B_i$, so $\mathbb{E}[\varphi_m(Z)]=\sum_i c_i P_Z(B_i)$ depends only on the distribution $P_Z$; by Monotone Convergence Theorem, $\mathbb{E}[Z]=\sup_m\mathbb{E}[\varphi_m(Z)]$ likewise. Applying this to positive and negative parts (whose distributions are determined by the original distribution, as their defining preimages are preimages of Borel sets under continuous maps, cf. Step 3 of the same proof) shows that identically distributed variables have equal expectations, and, applying it to squares, equal second moments and equal variances. In particular $\mathbb{E}[X_m]=\mu$ and $\operatorname{Var}(X_m)=\operatorname{Var}(X_1)$ for every $m$.

Step 1. Fix $n$. By linearity (claim 2 of Linearity and Monotonicity of the Lebesgue Integral) and Step 0, $\mathbb{E}[S_n]=n\mu$, so $\mathbb{E}[S_n/n]=\mu$. The variables $X_1,\dots,X_n$ are independent with finite second moments, so by Expectation of a Product of Independent Random Variables and Step 0,

$$\operatorname{Var}(S_n)=\sum_{m=1}^{n}\operatorname{Var}(X_m)=n\operatorname{Var}(X_1),$$

and from the definition of the variance together with linearity, $\operatorname{Var}(S_n/n)=\operatorname{Var}(S_n)/n^{2}=\operatorname{Var}(X_1)/n$.

Step 2. By Chebyshev's inequality (Markov's and Chebyshev's Inequalities) applied to $S_n/n$, for every $\varepsilon>0$,

$$P\Bigl(\Bigl|\frac{S_n}{n}-\mu\Bigr|\ge\varepsilon\Bigr)\ \le\ \frac{\operatorname{Var}(S_n/n)}{\varepsilon^{2}}=\frac{\operatorname{Var}(X_1)}{n\,\varepsilon^{2}}\longrightarrow 0\qquad(n\to\infty).$$

This is exactly convergence of $S_n/n$ to the constant $\mu$ in probability, in the sense of Almost Sure Convergence, Convergence in Probability, and Convergence in Distribution. $\blacksquare$