> [!theorem] > > Let $(X, \cm)$ be a [[Measure Space|measurable space]]. For any $E \subseteq X$, the **characteristic/indicator function** $\chi_E$ of $E$ is equal to > $ > \chi_E(x) = \begin{cases} > 1 &x \in E \\ > 0 &x \not\in E > \end{cases} > $ > then $\chi_E$ is $\cm$-[[Measurable Function|measurable]] if and only if $E \in \cm$. > > *Proof*. Let $A \in \cb_\real$ such that $1 \in A, 0 \not\in A$, then $\chi_E^{-1}(A) = E \in \cm$. If $1 \not\in A$ but $0 \in A$, then $\chi_E^{-1}(A) = E^c \in \cm$. If $0, 1 \in A$, then $\chi_E^{-1}(A) = X \in \cm$. So $\chi_E$ is measurable only when $E \in \cm$.