Let $g: \mathbb{R}^{+}_0 \longrightarrow(0,1): x \mapsto \exp (-x)$ be a function and let $\mathcal{M}=\left\{M \subseteq \mathbb{R}^{+} \mid M\right.$ denumerable or $\bar{M}$ denumerable $\}$ and $\mathcal{M}^{\prime}=\{M \subseteq(0,1) \mid M$ denumerable or $\bar{M}$ denumerable $\}$ be $\sigma$-Algebras .
I am trying to show that the function $g$ is not $M-{M}^{\prime}$ measurable, but I am missing the the counterexample. I gave it some thought and I think $g$ should be $M-\mathcal{B}((0,1))$ measurable, because in ${M}^{\prime}$ are also Vitali sets included. Any help is much appreciated
$g$ is a bijection so $E$ is denumerable if and only if $g^{-1}(E)$ is denumerable. Also $g^{-1}(E^{c})=(g^{-1}(E))^{c}$. Hence, $f$ is $\mathcal M \to \mathcal M'$ measurable.
[I am writing $A^{c}$ for the complement of $A$].
$g^{-1}(0,\frac 1 2)$ is not in $\mathcal M$ so $g$ is not $\mathcal M \to \mathbb B (0,1)$ measurable.