[proofplan] The indicator function takes only the two values $0$ and $1$, so its law is determined by the probabilities of the fibers $\{X=1\}$ and $\{X=0\}$. We first verify that $X$ is measurable as a map from $(\Omega,\mathcal F)$ to $(\mathbb R,\mathcal B(\mathbb R))$. Then we compute the two point probabilities and compare them with the defining masses of the Bernoulli distribution with parameter $\mathbb P(A)$. [/proofplan] [step:Verify that the indicator function is a real random variable] Let $A^c:=\Omega \setminus A$. Since $A \in \mathcal F$ and $\mathcal F$ is a $\sigma$-algebra, $A^c \in \mathcal F$. The function $X:\Omega \to \mathbb R$ satisfies \begin{align*} X^{-1}(\{1\})=A. \end{align*} It also satisfies \begin{align*} X^{-1}(\{0\})=A^c. \end{align*} Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$. For an arbitrary Borel set $B \in \mathcal B(\mathbb R)$, since $X(\Omega)\subset \{0,1\}$, the preimage $X^{-1}(B)$ is one of $\varnothing$, $A$, $A^c$, or $\Omega$. Each of these sets belongs to $\mathcal F$. Hence $X$ is $\mathcal F/\mathcal B(\mathbb R)$-measurable, so $X$ is a real [random variable](/page/Random%20Variable). [guided] We must first check that $X$ is actually a random variable. Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$. In this setting, that means $X:\Omega \to \mathbb R$ must be measurable from $(\Omega,\mathcal F)$ to $(\mathbb R,\mathcal B(\mathbb R))$. Let $A^c:=\Omega \setminus A$. Because $A \in \mathcal F$ and $\mathcal F$ is a $\sigma$-algebra, the complement $A^c$ also lies in $\mathcal F$. By the definition of the indicator function, $X(\omega)=1$ exactly when $\omega \in A$, so \begin{align*} X^{-1}(\{1\})=A. \end{align*} Similarly, $X(\omega)=0$ exactly when $\omega \in A^c$, so \begin{align*} X^{-1}(\{0\})=A^c. \end{align*} Now let $B \in \mathcal B(\mathbb R)$ be any Borel set. Since the only possible values of $X$ are $0$ and $1$, the set $X^{-1}(B)$ depends only on whether $B$ contains $0$, contains $1$, contains both, or contains neither. Therefore $X^{-1}(B)$ is one of $\varnothing$, $A$, $A^c$, or $\Omega$. All four sets belong to $\mathcal F$. Thus $X^{-1}(B)\in\mathcal F$ for every $B\in\mathcal B(\mathbb R)$, which proves that $X$ is measurable. [/guided] [/step] [step:Compute the masses at $1$ and $0$] Define $p:=\mathbb P(A)$. Since $\mathbb P$ is a probability measure, $p \in [0,1]$. From $X^{-1}(\{1\})=A$, we get \begin{align*} \mathbb P(X=1)=\mathbb P(A)=p. \end{align*} Since $A$ and $A^c$ are disjoint and $A \cup A^c=\Omega$, finite additivity of $\mathbb P$ gives \begin{align*} \mathbb P(A^c)=\mathbb P(\Omega)-\mathbb P(A)=1-p. \end{align*} Using $X^{-1}(\{0\})=A^c$, we obtain \begin{align*} \mathbb P(X=0)=\mathbb P(A^c)=1-p. \end{align*} [/step] [step:Identify the law as Bernoulli with parameter $\mathbb P(A)$] The Bernoulli distribution $\operatorname{Ber}(p)$ is the probability distribution on $\mathbb R$ assigning mass $p$ to $1$ and mass $1-p$ to $0$. The preceding step shows that the distribution of $X$ has exactly these two masses with $p=\mathbb P(A)$. Therefore \begin{align*} X \sim \operatorname{Ber}(\mathbb P(A)). \end{align*} This proves the theorem. [/step]