I am working on a problem looks like this:
If a simple function $s$ is measurable, show that its characteristic function $\mathcal X_{X_i}$ is measurable.
Here are the ways I have been working along, and correct me if I am wrong in any of the steps here.
(1) I know a simple function $s$ is this: $$s := s \circ \mathcal X_{X} : (X, \mathcal A) \rightarrow (\{0, 1\}) \rightarrow (\mathbb R, \mathcal B),$$ since the $a_i$ of simple function is element of $\mathbb R$, and the "natural" algebra of $\mathbb R$ is Borel $\sigma$-algebra.
(2) Since the simple function $s$ is measurable, this implies that $$\forall B \in \mathcal B, \quad s^{-1}(B) \in \mathcal A.$$
(3) The $\sigma$-algebra of $\{0, 1\}$, let's say $\sigma(\{0, 1\}),$ is $\{\{\emptyset\},\{0\},\{1\},\{0, 1\}\}$. Correct me again if I am wrong here.
(4) To prove the characteristic function $\mathcal X_X$ is measurable, then I need to show that $$\forall C \in \sigma(\{0, 1\}), \quad \mathcal X_X^{-1}(C) \in \mathcal A.$$
(5) And indeed it is the case, since $\mathcal X_X^{-1}(C) = s^{-1}(B)$ and thus: $$\mathcal X_X^{-1}(C) \in \mathcal A.$$
Thank you very much for your time and effort.