This paragraph is at the end of Section 6.3 in the book Testing Statistical Hypotheses by Lehmann and Romano:
In most applications, $M(x)$ is a measurable function taking on values in a Euclidean space and it is convenient to take $\mathcal B$ as the class of Borel sets. If $\phi(x) = \psi[M(x)]$ is then an arbitrary measurable function depending only on $M(x)$, it is not clear that $\psi(m)$ is necessarily $\mathcal B$-measurable. This measurability can be concluded if $\mathcal X$ is also Euclidean with $\mathcal A$ the class of Borel sets, and if the range of $M$ is a Borel set. We shall prove it here only under the additional assumption (which in applications is usually obvious, and which will not be verified explicitly in each case) that there exists a vector-valued Borel-measurable function $Y(x)$ such that $[M(x), Y (x)]$ maps $\mathcal X$ onto a Borel subset of the product space $\mathcal M\times \mathcal Y$, that this mapping is $1 : 1$, and that the inverse mapping is also Borel-measurable. Given any measurable function $\phi(x)$ of $x$, there exists then a measurable function $\phi'$ of $(m, y)$ such that $\phi(x) ≡ \phi' [M(x), Y (x)]$. If $\phi$ depends only on $M(x)$, then $\phi'$ depends only on $m$, so that $\phi'$ $(m, y) = \psi(m)$ say, and $\psi$ is a measurable function of $m$.
I don't understand how to formalize the argument:
How to choose $\phi'$?
Why $\phi'$ depends only on $m$ if $\phi$ depends only on $M(x)$ ?
Where do we need $Y(x)$ to be 1:1 with measurable inverse and measurable range?
Can someone clarify the argument?
Edit: Why is $\psi(m)$ a measurable function of $m$?
Let $h$ be the function given by $h(x)=(M(x),Y(x))$. By assumption, the range of $h$ is a Borel set and has a measurable inverse $h^{-1}$ defined on the range which is where we need that $h$ is 1-to-1. For all $x$, we must have $h^{-1}(h(x))=x$. So we can define $\phi'$ as $\phi\circ h^{-1}$. Then $\phi(x)=\phi(h^{-1}(h(x)))=\phi'(h(x))=\phi'(M(x),Y(x))$. Also, $\phi'$ is measurable as a composition of measurable functions.
That $\phi$ depends only on $M(x)$ means that $M(x)=M(x')$ implies $\phi(x)=\phi(x').$ Since $\phi'(M(x),Y(x))=\phi(x)=\phi(x')=\phi'(M(x'),Y(x'))$, this implies that $\phi'(M(x),Y(x))=\phi(x)=\phi(x')=\phi'(M(x'),Y(x'))$ whenever $M(x)=M(x')$. And that means that $\phi'$ depends only on $m=M(x)$.