The composition of a measurable function with a continuous function

Question

I know that the composition of a continuous function with a measurable function is measurable, however the composition of a measurable function with a continuous function is not necessarily measurable (I suppose). What example can you give? Thank you!

Detail. I mean function f is measurable if set A = $\{x: f(x) < C \}$ for any $C \in \mathbb{R}$ is measurable

Answer 1

The composition of a Lebesgue measurable function with a continuous function need not be Lebesgue measurable.

Take our measurable set to be the Cantor set $C$, and let $h:[0,1]\rightarrow [0,1]$ be the Cantor-Lebesgue function. This is continuous and $h(C)=[0,1].$ Let $g:[0,1]\rightarrow [0,2]$ be defined via $g(x)=x+h(x),$ which is both continuous and injective; hence, its inverse is continuous. Since $g(C)$ has positive measure there is a non-measurable subset $D\subset g(C)$. Call $E=g^{-1}(D).$ This must have measure zero, as it's a subset of $C$. If you define $f(x)=\chi_E(x)$, then this is a measurable function, as $E$ is measurable. However, $f\circ g^{-1}:[0,2]\rightarrow [0,1]$ is non-measurable. Indeed, it equals $\chi_D$, and you can check that $(1/2,1]$ is open in $[0,1]$, but $(f\circ g^{-1})^{-1} ((1/2,1])=D,$ which is not measurable.

Answer 2

If you mean that measurable function $f:\mathbb{R}\to\mathbb{R}$ is a function such that $f^{-1}(B)$ is a Borel set for any Borel set $B\subseteq \mathbb{R}$, then if $g$ is continuous, $g$ is measurable, and so both $f\circ g$ and $g\circ f$ are measurable.