[proofplan] We write the covariance in terms of expectations and then reduce the product of the two indicators to the indicator of the intersection. Since indicators are bounded real-valued random variables, all expectations involved are finite. The expectation formula for indicators then converts $\mathbb E[\mathbb{1}_A]$, $\mathbb E[\mathbb{1}_B]$, and $\mathbb E[\mathbb{1}_{A\cap B}]$ into the corresponding probabilities. [/proofplan] [step:Expand the covariance of the two bounded indicator random variables] Define the maps \begin{align*} X:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)) \end{align*} by $X(\omega)=\mathbb{1}_A(\omega)$, and \begin{align*} Y:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)) \end{align*} by $Y(\omega)=\mathbb{1}_B(\omega)$. Since $A,B\in\mathcal F$, both $X$ and $Y$ are measurable. Since $0\le X\le 1$ and $0\le Y\le 1$, both are bounded and therefore integrable. Let $a=\mathbb E[X]$ and $b=\mathbb E[Y]$. By the definition of covariance for integrable bounded random variables, \begin{align*} \operatorname{Cov}(X,Y)=\mathbb E[(X-a)(Y-b)]. \end{align*} Using linearity of the integral with respect to $\mathbb P$ and $\mathbb P(\Omega)=1$, we get \begin{align*} \operatorname{Cov}(X,Y)=\mathbb E[XY]-a\mathbb E[Y]-b\mathbb E[X]+ab. \end{align*} Since $a=\mathbb E[X]$ and $b=\mathbb E[Y]$, this becomes \begin{align*} \operatorname{Cov}(X,Y)=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]. \end{align*} [/step] [step:Identify the product of indicators with the indicator of the intersection] For each $\omega\in\Omega$, the product $X(\omega)Y(\omega)$ equals $1$ exactly when both $\omega\in A$ and $\omega\in B$, and equals $0$ otherwise. Hence \begin{align*} XY=\mathbb{1}_{A\cap B}. \end{align*} The event $A\cap B$ belongs to $\mathcal F$ because $\mathcal F$ is closed under finite intersections. [guided] The only pointwise identity needed is the multiplication rule for indicators. Fix $\omega\in\Omega$. If $\omega\in A\cap B$, then $\mathbb{1}_A(\omega)=1$ and $\mathbb{1}_B(\omega)=1$, so \begin{align*} \mathbb{1}_A(\omega)\mathbb{1}_B(\omega)=1. \end{align*} If $\omega\notin A\cap B$, then $\omega\notin A$ or $\omega\notin B$. In the first case $\mathbb{1}_A(\omega)=0$, and in the second case $\mathbb{1}_B(\omega)=0$. In either case, \begin{align*} \mathbb{1}_A(\omega)\mathbb{1}_B(\omega)=0. \end{align*} Thus the function $\omega\mapsto \mathbb{1}_A(\omega)\mathbb{1}_B(\omega)$ has value $1$ precisely on $A\cap B$ and value $0$ outside $A\cap B$. Therefore \begin{align*} XY=\mathbb{1}_{A\cap B}. \end{align*} Since $A,B\in\mathcal F$ and $\mathcal F$ is a $\sigma$-algebra, $A\cap B\in\mathcal F$, so $\mathbb{1}_{A\cap B}$ is a valid [indicator random variable](/page/Indicator%20Random%20Variable) on $(\Omega,\mathcal F,\mathbb P)$. [/guided] [/step] [step:Convert the three indicator expectations into probabilities] By [citetheorem:10139] applied to the events $A$, $B$, and $A\cap B$, we have \begin{align*} \mathbb E[X]=\mathbb E[\mathbb{1}_A]=\mathbb P(A). \end{align*} Also, \begin{align*} \mathbb E[Y]=\mathbb E[\mathbb{1}_B]=\mathbb P(B). \end{align*} Using $XY=\mathbb{1}_{A\cap B}$, \begin{align*} \mathbb E[XY]=\mathbb E[\mathbb{1}_{A\cap B}]=\mathbb P(A\cap B). \end{align*} Substituting these three identities into the covariance formula gives \begin{align*} \operatorname{Cov}(\mathbb{1}_A,\mathbb{1}_B)=\mathbb P(A\cap B)-\mathbb P(A)\mathbb P(B). \end{align*} This is the desired identity. [/step]