Let $(\Omega, \Sigma, \mathbb{P})$ be a probability space. If $A_1, A_2, \ldots$ are independent events and $\bar{p}_n$ and $N$ are defined as $$ \bar{p}_n=\frac{1}{n}\sum_{i=1}^n\mathbb{P}(A_i) \quad \text{and} \quad N_n=\sum_{i=1}^n \mathbf{1}_{A_i}$$ then $\frac{N_n}{n}-\bar{p}_n \to 0$ in probability.
This problem is from Billingsley (3rd edition, #6.5). I currently have the laws of large numbers at my disposal but since the sequence here is not i.i.d., I guess I need to invoke heavier machinery. Thank you for any help!
This is a corollary of Chebyshev's inequality. Fix $\delta > 0$. Then since $\mathbb E(N_n/n) = \overline p_n$, $$ \mathbb P \left( \left| \frac{N_n}n - \overline p_n \right| \ge \delta \right) \le \frac{\mathrm{Var}(N_n/n)}{\delta^2}. $$ It suffices to show now that the variance of $N_n/n$ goes to $0$. It is not hard to show that $f : [0,1] \to \mathbb R$ defined by $x \mapsto x(1-x)$ has a maximum at $x = 1/2$ which implies $f(x) \le 1/4$. Now $$ \mathrm{Var}(N_n) = \sum_{i=1}^n \mathrm{Var}(\mathbb 1_{A_i}) = \sum_{i=1}^n \mathbb P(A_i)(1-\mathbb P(A_i)) \le \frac n4, $$ hence $\mathrm{Var}(N_n/n) \le \frac 1{4n} \to 0$.
Hope that helps,