Weak law of large numbers

When the variance $\sigma^2 = \operatorname{Var}(X_1)$ is finite, linearity and independence give $\mathbb{E}[\bar{X}_n] = \mu$ and $\operatorname{Var}(\bar{X}_n) = \frac{\sigma^2}{n}$. Applying Chebyshev's inequality yields $\mathbb{P}(|\bar{X}_n - \mu| \ge \varepsilon) \le \frac{\operatorname{Var}(\bar{X}_n)}{\varepsilon^2} = \frac{\sigma^2}{n\varepsilon^2} \to 0$ as $n \to \infty$. When only $\mathbb{E}[|X_1|] < \infty$ is assumed (Khinchin's theorem), one truncates at level $n$ by setting $Y_{i} = X_i \mathbf{1}_{\{|X_i| \le n\}}$, uses $\mathbb{E}[|X_1|] < \infty$ to show $\mathbb{P}(\exists i \le n : X_i \ne Y_i) \le n\mathbb{P}(|X_1| > n) \to 0$, and applies Chebyshev's inequality to the truncated average.