I'm trying to understand a passage in the proof of Borel Cantelli Lemma part 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 $$\sum^{+\infty} P(A_n) = +\infty \qquad \quad \text{then} \qquad P(\text{lim sup}_n A_n) = 1$$
Start of Proof
Be $S_n = \sum_{k}^n \mathbb{1}_{A_k}$; $\quad$ $S = \sum_{k}^{+\infty} \mathbb{1}_{A_k}$; $\quad$ $a_n = \int S_n \text{d}P = \sum^n P(A_n)$
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 \\\\ & = \sum_{i}^n\int \left(\mathbf{1}_{A_i}- P(A_i)\right)^2d P=\sum_{i}^n P(A_i)- P(A_i)^2. \end{split} \end{equation*}
I cannot get why the very last identity is true / where it is obtained.
I know that
$$\left(\mathbf{1}_{A_i}- P(A_i)\right)^2 = \mathbb{1}^2_{A_i} - 2\mathbb{1}_{A_i} P(A_i) + P^2(A_i)$$
But cannot go on with the integration
Using the fact that $$\mathbb P(X)=\int \boldsymbol 1_X\,\mathrm d \mathbb P,$$ gives
\begin{align*} \int(\boldsymbol 1_{A_i}-\mathbb P(A_i))^2\,\mathrm d \mathbb P&=\int \boldsymbol 1_{A_i}\,\mathrm d \mathbb P+\mathbb P(A_i)^2-2\mathbb P(A_i)\int \boldsymbol 1_{A_i}\,\mathrm d \mathbb P\\ &=\mathbb P(A_i)+\mathbb P(A_i)^2-2\mathbb P(A_i)^2\\ &=\mathbb P(A_i)-\mathbb P(A_i)^2. \end{align*}