On page 11 of Rudin's real and complex analysis,
Let $X$ be a measurable space. If $E$ is measurable set in $X$ and if
\begin{equation} \chi_{E}(x)=\begin{cases}
1, & x\in E \\
\\
0, & x\notin E.
\end{cases} \end{equation}
then $\chi_E$ is a measurable function.
Do we prove $\chi_E^{-1}(V)$ is a measurable set in $X$ for every open set $V$ in $\{0,1\}$? But $\{0\}$ is an open set in $\{0,1\}$，isn't it? So $\chi_E^{-1}(\{0\})$ is not a measurable set in $X$?
Best Answer
$\mathcal{X}_E^{-1}(\{0\}) = E^C$, which is perfectly measurable.
Take any $A \in \mathcal{B}(\mathbb{R})$:
$\mathcal{X}_E^{-1}(A) =\begin{cases} X, & 0,1 \in A \\ E, & 1 \in A, 0 \notin A \\ E^C, & 1 \notin A, 0 \in A \\ \emptyset, & o.w. \end{cases}$
all those sets are measurable, since $E$ is measurable.