Htykuut

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

$$h^{-1}([a, infty])=
begin{cases}
mathbb R, &text{ if $a le 0$} \
B, &text{ if $0<ale 1$} \
varnothing, &text{ if $a>1$.}
end{cases}
$$
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?

For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
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.

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