Understanding this passage in Borel Cantelli Lemma N.2

Question

I'm trying to understand a passage in the proof of Borel Cantelli Lemma 2.

Be $(\Omega, \mathcal{F}, P)$ a probability space and $(A_n)$ a sequence in $\mathcal{F}$ such that $(A_n)$ are pair independent. If $$\sum^{+\infty} P(A_n) = +\infty \qquad \quad \text{then} \qquad P(\text{lim sup}_n A_n) = 1$$

_ Part of Proof_

We start by denotng $S_n = \sum_{k}^n \mathbb{1}_{A_k}$; $\quad$ $S = \sum_{k}^{+\infty} \mathbb{1}_{A_k}$; $\quad$ $a_n = \int S_n \text{d}P = \sum^n P(A_n)$

Then from pair independence

\begin{equation*} \begin{split} \int (S_n - a_n)^2\ \text{d}P & = \sum_{i, k}^n \int \left(\mathbb{1}_{A_i}- P(A_i)\right)\left(\mathbb{1}_{A_k}- P(A_k)\right)\ \text{d}P \\\\ & = \color{red}{\sum_{i}^n \int\left(\mathbb{1}_{A_i}- P(A_i)\right)^2\ \text{d}P} \\\\ & = \color{blue}{\sum_i^n P(A_i)(1 - P(A_i))} \leq a_n \end{split} \end{equation*}

I think I got why in the end it's $\leq a_n$ but the not-understood passage is the red coloured one that turns into blue coloured one. How does one obtain that?

Thank you!

Answer 1

The proof is much simpler: by definition $\limsup A_n=\cap_n B_n$ where $B_n=\cup_{i=n}^\infty A_i$. By the continuity of probability, since $B_n^c$ is a decreasing sequence of events, $1-P(\limsup A_n)=P(\cap B_n^c)=P(A_n^c)P(A_{n+1}^c)P(A_{n+2}^c)\dots$ by independence. But this equals $(1-P(A_n))(1-P(A_{n+1}))(1-P(A_{n+2}))\dots=0$ since $\sum_{i=n}^\infty P(A_i)<\infty$.

Answer 2

For a measurable set $A$, $$\left(\mathbf{1}_A-\mathbb P(A)\right)^2=\mathbf{1}_A^2-2\mathbf{1}_A\mathbb P(A)+\mathbb P(A)^2=\mathbf{1}_A -2\mathbf{1}_A\mathbb P(A)+\mathbb P(A)^2$$ then integrating, $$ \int \left(\mathbf{1}_A-\mathbb P(A)\right)^2d\mathbb P=\mathbb P(A)-\mathbb P(A)^2. $$

Answer 3

The integral in the red formula is just an expectation of $\mathbb{1}_{A_i}-P(A_i)$ squared, which is the variance of the Bernoulli random variable with parameter $p=P(A_i)$, which is $p(1-p)$.