Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and ${B_i}$ are all independent events, $B_i \in \mathcal{F}$. Let $S_n = \sum_{m=1}^n 1_{B_m}$, where $1_{B_m}$ is the indicator function. Set up $X_n = \dfrac{S_n}{n}$ and $Y_n = \dfrac{\sum_{i=1}^n P(B_i)}{n}$. I need to prove that $X_n - Y_n \rightarrow 0$ in probability when $n \rightarrow \infty$.
My try: I started with proving that $\dfrac{\sum_{m=1}^n 1_{B_m}}{\sum_{i=1}^n P(B_i)} \rightarrow 1$ almost surely. However, after that I get stuck to find the way to continue.
Note that $P(B_m)=\mathbb E[\mathbf 1_{B_m}]$. So, actually the $\mathbf 1_{B_m}$, for $m\in \mathbb N$, are Bernoulli random variables with probability of success $p_m=P(B_m)$. Now, by linearity of expectation $$\mathbb E[S_n]=\mathbb E\left[\sum_{m=1}^n\mathbf1_{B_m}\right]=\sum_{m=1}^n\mathbb E\left[\mathbf1_{B_m}\right]=\sum_{m=1}^n\mathbb P(B_m)=nY_n$$ Moreover, $Var(\mathbf 1_{B_m})=P(B_m)(1-P(B_m))\le 1$, hence by independence $$Var(S_n)=Var\left(\sum_{m=1}^n\mathbf1_{B_m}\right)=\sum_{m=1}^nVar(\mathbf1_{B_m})\le n\tag1$$ So, by Chebychev's inequality
\begin{align}P\left(\left | X_n-Y_n\right |\ge \epsilon\right)&=P(|S_n-\mathbb E[S_n]|\ge n\epsilon)\le \frac{Var(S_n)}{n^2\epsilon^2}\overset{(1)}\le\frac{1}{\epsilon^2}\cdot\frac{n}{n^2}=\frac{1}{\epsilon^2}\cdot\frac{1}{n}\longrightarrow 0 \end{align} as $n\to\infty$.