[proofplan] The proof uses only the defining pointwise values of an indicator. First, the expectation of $X=\mathbb{1}_A$ is the integral of an indicator over the [probability space](/page/Probability%20Space), hence equals $\mathbb P(A)$. Next, the identity $r^2=r$ for $r\in\{0,1\}$ gives $X^2=X$ pointwise, so the second moment is the same as the first. Finally, substituting these two moments into the variance identity gives the stated formula. [/proofplan]

[step:Compute the expectation by integrating the indicator over $A$]Define $p \in [0,1]$ by $p:=\mathbb P(A)$. Since $A\in\mathcal F$, the indicator map $X=\mathbb{1}_A: (\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable and bounded. Hence $X$ is integrable with respect to $\mathbb P$. By [citetheorem:10139], applied to the [measure space](/page/Measure%20Space) $(\Omega,\mathcal F,\mathbb P)$ and the measurable set $A$, we have \begin{align*} \mathbb E[X]=\int_\Omega \mathbb{1}_A\,d\mathbb P=\mathbb P(A)=p. \end{align*}[/step]

[guided]Define $p \in [0,1]$ by $p:=\mathbb P(A)$. The event $A$ belongs to $\mathcal F$, so the indicator $X=\mathbb{1}_A$ is the measurable map \begin{align*} X: (\Omega,\mathcal F) \to (\mathbb R,\mathcal B(\mathbb R)) \end{align*} given by $X(\omega)=1$ when $\omega\in A$ and $X(\omega)=0$ when $\omega\notin A$. Since $0\le X\le 1$ pointwise, $X$ is bounded and therefore integrable with respect to the [probability measure](/page/Probability%20Measure) $\mathbb P$. Expectation of a real-valued integrable [random variable](/page/Random%20Variable) is its integral over the probability space. Thus \begin{align*} \mathbb E[X]=\int_\Omega X\,d\mathbb P=\int_\Omega \mathbb{1}_A\,d\mathbb P. \end{align*} Now [citetheorem:10139] applies because $(\Omega,\mathcal F,\mathbb P)$ is a measure space and $A\in\mathcal F$. It gives \begin{align*} \int_\Omega \mathbb{1}_A\,d\mathbb P=\mathbb P(A). \end{align*} Therefore \begin{align*} \mathbb E[X]=\mathbb P(A)=p. \end{align*}[/guided]

[step:Use the pointwise idempotence of indicators to compute the second moment] Define $X^2: \Omega\to\mathbb R$ by $X^2(\omega):=X(\omega)^2$. For every $\omega\in\Omega$, the value $X(\omega)$ belongs to $\{0,1\}$, so $X(\omega)^2=X(\omega)$. Hence $X^2=X$ pointwise, and therefore \begin{align*} \mathbb E[X^2]=\mathbb E[X]=p. \end{align*} [/step]

[step:Substitute the two moments into the variance identity] Since $X$ is bounded, $X$ is square-integrable. By the definition of variance, \begin{align*} \operatorname{Var}(X)=\mathbb E[X^2]-(\mathbb E[X])^2. \end{align*} Using $\mathbb E[X^2]=p$ and $\mathbb E[X]=p$, we obtain \begin{align*} \operatorname{Var}(X)=p-p^2=p(1-p)=\mathbb P(A)(1-\mathbb P(A)). \end{align*} Together with the first two moment computations, this proves all three asserted identities. [/step]