Converge almost surely with sums of Bernoulli random variable

Question

Let $A_{1},A_{2},...$ be a sequence of independent events, and let $b_{n}=\sum_{i=1}^{n}P(A_{i})$. If $b_{n}\to \infty$, then
$\frac{1}{b_{n}}\sum_{i=1}^{n}I_{A_{i}}\to 1$ a.s.
How can I prove it? Thank you in advance!

Answer 1

Here are some hint.

1. Let $n_k:=\inf\{n\geqslant 1, b_n\geqslant k^2\}$. We can show via the classical Borel-Cantelli lemma that $\frac{1}{b_{n_k}}\sum_{i=1}^{n_k}I_{A_{i}}\to 1$ a.s.

2. To get the convergence for the whole sequence, we have to show the convergence to $0$ of $$ \max_{n_k\leqslant n\leqslant n_{k+1}-1}\left\lvert \frac{1}{b_{n }}\sum_{i=1}^{n }I_{A_{i}}-\frac{1}{b_{n_k}}\sum_{i=1}^{n_k}I_{A_{i}}\right\rvert. $$ What is annoying is that the normalization for $\sum_{i=1}^{n }I_{A_{i}}$ and $\sum_{i=1}^{n_k }I_{A_{i}}$ is not the same. but we can write $$ \max_{n_k\leqslant n\leqslant n_{k+1}-1}\left\lvert \frac{1}{b_{n }}\sum_{i=1}^{n }I_{A_{i}}-\frac{1}{b_{n_k}}\sum_{i=1}^{n_k}I_{A_{i}}\right\rvert\leqslant \max_{n_k\leqslant n\leqslant n_{k+1}-1}\left\lvert \frac{1}{b_{n }}\sum_{i=1}^{n }I_{A_{i}}-\frac{1}{b_{n_k}}\sum_{i=1}^{n}I_{A_{i}}\right\rvert +\max_{n_k\leqslant n\leqslant n_{k+1}-1}\left\lvert \frac{1}{b_{n_k}}\sum_{i=n_k+1}^{n}I_{A_{i}}\right\rvert $$ and apply Kolmorogov's maximal inequality to both terms.