Let $X$ and $Y$ be Hausdorff topological spaces and let $\mu$ be a positive Radon measure on $X$
A map $\pi$ from $X$ to $Y$ is said to be $\mu$-measurable in the sense of Lusin if for every compact subset $K$ of $X$ and every $\varepsilon>0$, there exists a compact subset $L$ of $K$ such that $\mu(K\setminus L)\leq\varepsilon$ and such that the restriction of $f$ to $L$ is continuous.
It is well-known that if $\pi$ is a map from $X$ to $Y$ that is $\mu$-measurable in the sense of Lusin and if for every $y\in Y$ there is a neighbourhood $V$ of $y$ in $Y$ such that $\pi^{-1}(V)$ has finite essential outer $\mu$-measure, then the pushforward of $\mu$ by $\pi$ is a Radon measure in the following sense: there exists one and only one positive Radon measure $\nu$ on $Y$ such that for every map $f$ from $Y$ to $[0,+\infty]$, the essential upper $\nu$-integral of $f$ is equal to the essential upper $\mu$-integral of $f\circ\pi$.
So now I would like to know if Lusin mesurability is necessary for the pushforward measure to be a Radon measure. More precisely, my question is:
If $\pi$ is a map from $X$ to $Y$ such that there exists a positive Radon measure on $Y$ such that for any map $f$ from $Y$ to $[0,+\infty]$, the essential upper $\nu$-integral of $f$ is equal to the essential upper $\mu$-integral of $f\circ\pi$, then is $\pi$ necessarily $\mu$-measurable in the sense of Lusin?
This seems to be true when $Y$ is metrizable and separable (and maybe even when $Y$ is just second countable) but that's a little too restrictive to be satisfying…
