Prove a composite function involving cantor set is measurable

Question

Let $\cal K$ be the Cantor set. Let $f : \mathbb R \to \mathbb R$ be defined by

$$f(x)= \begin{cases} 1, &\mbox{ if }x \in \mathcal K\\ 0, &\mbox{ otherwise }\\ \end{cases} $$

and let $g : \mathbb R \to \mathbb R$ be continuous. Prove that $f \circ g = h$ is measurable.

I know that $f$ is a measurable function. Than I proceed by

$$h^{-1}([a, \infty])= \begin{cases} \mathbb R, &\text{ if $a \le 0$} \\ B, &\text{ if $0<a\le 1$} \\ \varnothing, &\text{ if $a>1$.} \end{cases} $$ where $B = \{x \mid g(x) ∈ K\}$.

I feel it very hard to show $B$ is measurable. I know $m(g(B)) = 0$ is. As it is subset of Cantor set which is measure zero. But than I cannot proceed. What should I do?

Answer 1

For a Lebesgue set $L$:$$h^{-1}(L)=\begin{cases}g^{-1}(\mathcal K) & \text{if $1\in L$ and $0\notin L$} \\ g^{-1}(\Bbb R)& \text{if $1\in L$ and $0\in L$} \\g^{-1}(\mathcal K^c)& \text{if $1\notin L$ and $0\in L$} \\ g^{-1}(\varnothing)& \text{if $1\notin L$ and $0\notin L$} \\\end{cases}$$ Since $\mathcal K$ is closed, $\mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(\mathcal K)$ is closed and $g^{-1}(\mathcal K^c)$ is open. The other two cases are trivial.

Answer 2

If $B = \{x : g(x) \in K\}$, then $B = g^{-1}(K)$

$g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set