Why the characteristic function is measurable?

Question

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 $X$?

Answer 1

$\mathcal{X}_E^{-1}(\{0\}) = E^C$, 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.

Answer 2

If you take the subspace topology of $\{0,1\}$ induced from the real line you'll note that ${0}$ is open in $\{0,1\}$ since $\{0\}=(-1/2,1/2)\cap \{0,1\}$. Likewise $\{1\}$ is open in $\{0,1\}$ for a similar reason and hence the topology on $\{0,1\}$ is the discrete topology.

Answer 3

Consider $\chi_E^{-1}((-\infty,\alpha])$ for various $\alpha$