[proofplan] The expectation $\mathbb{E}[\mathbb{1}_A]$ is by definition the Lebesgue integral $\int_\Omega \mathbb{1}_A \, d\mathbb{P}$, which requires $\mathbb{1}_A$ to be a measurable random variable. We first verify $\mathcal{F}/\mathcal{B}(\mathbb{R})$-measurability by an exhaustive case analysis on the preimage of an arbitrary Borel set. We then observe that $\mathbb{1}_A$ is a non-negative simple function admitting the canonical representation $\mathbb{1}_A = 1 \cdot \mathbb{1}_A + 0 \cdot \mathbb{1}_{A^c}$ over the measurable partition $\{A, A^c\}$ of $\Omega$. The Lebesgue integral of a non-negative simple function is by definition the weighted sum of measure values, which evaluates directly to $\mathbb{P}(A)$. [/proofplan]

[step:Verify that $\mathbb{1}_A$ is $\mathcal{F}/\mathcal{B}(\mathbb{R})$-measurable]Since $\mathbb{1}_A$ takes values only in $\{0, 1\}$, for any $B \in \mathcal{B}(\mathbb{R})$ precisely one of four cases holds: \begin{align*} \mathbb{1}_A^{-1}(B) = \begin{cases} \Omega & \text{if } 0 \in B \text{ and } 1 \in B, \\ A & \text{if } 0 \notin B \text{ and } 1 \in B, \\ A^c & \text{if } 0 \in B \text{ and } 1 \notin B, \\ \varnothing & \text{if } 0 \notin B \text{ and } 1 \notin B. \end{cases} \end{align*} Since $A \in \mathcal{F}$ by hypothesis, the $\sigma$-algebra axioms give $A^c \in \mathcal{F}$, and $\Omega, \varnothing \in \mathcal{F}$ always. Therefore $\mathbb{1}_A^{-1}(B) \in \mathcal{F}$ in every case, and $\mathbb{1}_A$ is $\mathcal{F}/\mathcal{B}(\mathbb{R})$-measurable.[/step]

[guided]We need to check that $\mathbb{1}_A$ qualifies as a random variable on $(\Omega, \mathcal{F}, \mathbb{P})$, which means it must be $\mathcal{F}/\mathcal{B}(\mathbb{R})$-measurable: the preimage of every Borel set must lie in $\mathcal{F}$. Since $\mathbb{1}_A$ takes only the values $0$ and $1$, for any $B \in \mathcal{B}(\mathbb{R})$ we have \begin{align*} \mathbb{1}_A^{-1}(B) = \{\omega \in \Omega : \mathbb{1}_A(\omega) \in B\}. \end{align*} We check the four mutually exclusive and exhaustive cases: - **Both $0, 1 \in B$:** Every $\omega \in \Omega$ satisfies $\mathbb{1}_A(\omega) \in \{0,1\} \subseteq B$, so $\mathbb{1}_A^{-1}(B) = \Omega \in \mathcal{F}$. - **$1 \in B$, $0 \notin B$:** Only $\omega \in A$ (where $\mathbb{1}_A(\omega) = 1$) satisfies $\mathbb{1}_A(\omega) \in B$, so $\mathbb{1}_A^{-1}(B) = A \in \mathcal{F}$ (given by hypothesis). - **$0 \in B$, $1 \notin B$:** Only $\omega \notin A$ (where $\mathbb{1}_A(\omega) = 0$) satisfies $\mathbb{1}_A(\omega) \in B$, so $\mathbb{1}_A^{-1}(B) = A^c$. Since $A \in \mathcal{F}$ and $\mathcal{F}$ is closed under complementation, $A^c \in \mathcal{F}$. - **$0, 1 \notin B$:** No $\omega \in \Omega$ can satisfy $\mathbb{1}_A(\omega) \in B$, so $\mathbb{1}_A^{-1}(B) = \varnothing \in \mathcal{F}$. In every case $\mathbb{1}_A^{-1}(B) \in \mathcal{F}$, establishing measurability.[/guided]

[step:Express $\mathbb{1}_A$ as a non-negative simple function and evaluate its Lebesgue integral by definition]Since $A$ and $A^c$ are disjoint members of $\mathcal{F}$ with $A \cup A^c = \Omega$, the function $\mathbb{1}_A$ admits the representation \begin{align*} \mathbb{1}_A = 1 \cdot \mathbb{1}_A + 0 \cdot \mathbb{1}_{A^c}, \end{align*} a non-negative simple function expressed as a finite linear combination of indicators of measurable sets over the partition $\{A, A^c\}$ of $\Omega$ with non-negative coefficients $c_1 = 1$ and $c_2 = 0$. By the definition of the Lebesgue integral for non-negative simple functions, \begin{align*} \mathbb{E}[\mathbb{1}_A] = \int_\Omega \mathbb{1}_A \, d\mathbb{P} = 1 \cdot \mathbb{P}(A) + 0 \cdot \mathbb{P}(A^c) = \mathbb{P}(A), \end{align*} where the last equality uses $0 \cdot \mathbb{P}(A^c) = 0$ (since $\mathbb{P}(A^c) \leq 1 < \infty$, the coefficient $0$ renders this term zero). This gives $\mathbb{E}[\mathbb{1}_A] = \mathbb{P}(A)$.[/step]

[guided]We now use the definition of expectation. For a random variable $X: \Omega \to \mathbb{R}$ on a probability space, $\mathbb{E}[X] := \int_\Omega X \, d\mathbb{P}$ is the Lebesgue integral of $X$ with respect to $\mathbb{P}$. For a **non-negative simple function** — one of the form $\phi = \sum_{k=1}^n c_k \mathbb{1}_{E_k}$ where $c_k \geq 0$, $E_k \in \mathcal{F}$ are disjoint, and $\bigcup_{k=1}^n E_k = \Omega$ — the Lebesgue integral is defined by \begin{align*} \int_\Omega \phi \, d\mathbb{P} := \sum_{k=1}^n c_k \, \mathbb{P}(E_k). \end{align*} We identify $\mathbb{1}_A$ as such a function. The sets $A$ and $A^c$ satisfy: both are in $\mathcal{F}$ (since $A \in \mathcal{F}$ by hypothesis and $\mathcal{F}$ is closed under complementation), they are disjoint ($A \cap A^c = \varnothing$), and their union is $\Omega$ ($A \cup A^c = \Omega$). Thus $\{A, A^c\}$ is a measurable partition of $\Omega$, and \begin{align*} \mathbb{1}_A(\omega) = 1 \cdot \mathbb{1}_A(\omega) + 0 \cdot \mathbb{1}_{A^c}(\omega) \quad \text{for all } \omega \in \Omega, \end{align*} since on $A$ the expression evaluates to $1 \cdot 1 + 0 \cdot 0 = 1$, and on $A^c$ it evaluates to $1 \cdot 0 + 0 \cdot 1 = 0$. This shows $\mathbb{1}_A$ is a non-negative simple function with $n = 2$, $c_1 = 1$, $c_2 = 0$, $E_1 = A$, $E_2 = A^c$. Applying the definition of the integral: \begin{align*} \mathbb{E}[\mathbb{1}_A] = \int_\Omega \mathbb{1}_A \, d\mathbb{P} = c_1 \, \mathbb{P}(E_1) + c_2 \, \mathbb{P}(E_2) = 1 \cdot \mathbb{P}(A) + 0 \cdot \mathbb{P}(A^c). \end{align*} Since $\mathbb{P}$ is a probability measure, $\mathbb{P}(A^c) \leq \mathbb{P}(\Omega) = 1 < \infty$, so the term $0 \cdot \mathbb{P}(A^c) = 0$. Therefore \begin{align*} \mathbb{E}[\mathbb{1}_A] = \mathbb{P}(A). \end{align*} This is why $\mathbb{P}(A)$ can be interpreted as an expectation: the probability of an event is exactly the expected value of its indicator. This identity is the bridge between measure-theoretic probability and expectation, and underlies results such as the law of total expectation and the Fubini-based formula $\mathbb{E}[X] = \int_0^\infty \mathbb{P}(X > t) \, d\mathcal{L}^1(t)$ for non-negative random variables.[/guided]