Covariance and indicator function relations

Question

Let $\mathcal A$ and $\mathcal B$ be two (sub) sigma-algebras of some probability space. If $A\in\mathcal A$ and $B\in\mathcal B$, why the following equality holds: $$Cov(1_A-1_{A^c},1_B)=E((P(B\lvert \mathcal A)-P(B))(1_A-1_{A^c}))?$$ where the function $1_W$ stands for the indicator function given a measurable set $W$.

Answer 1

\begin{align} \operatorname{Cov}(1_A-1_{A^c},1_B)&=\mathsf{E}[(1_B-\mathsf{E}1_B)(1_A-1_{A^c})] \\ &=\mathsf{E}\!\left[\mathsf{E}[(1_B-\mathsf{E}1_B)(1_A-1_{A^c})\mid \mathcal{A}]\right] \\ &=\mathsf{E}\!\left[\mathsf{E}[(1_B-\mathsf{E}1_B)\mid \mathcal{A}](1_A-1_{A^c})\right] \\ &=\mathsf{E}\!\left[(\mathsf{E}[1_B\mid \mathcal{A}]-\mathsf{E}1_B)(1_A-1_{A^c})\right]. \end{align}