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Let $(E,d),(F,d')$ be separable metric spaces endowed with their Borel algebra, $f:E\rightarrow F$ a continuous surjective function, and $Q$ a probability measure on $F$ with separable support.

Question: Does there exist a probability measure $P$ on $E$ such that $Q$ is the pushforward measure of $P$ by $f$?

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    $\begingroup$ For an example we could try: $F = [0,1]$, $Q$ Lebesgue measure, $E$ a wild subset of $[0,1]\times[0,1]$, $f$ projection onto the $x$-axis. Can we make $E$ wild enough (but still project onto all of $[0,1]$) to defeat this? Maybe a Bernstein set? Or something constructed with CH? $\endgroup$
    – Gerald Edgar
    Commented May 17, 2021 at 18:03
  • $\begingroup$ For Radon measures the necessary and sufficient condition is that $Q$ be concentrated on $f(E)$ in the sense defined by Bourbaki (the complement is locally negligible). See Intégration, Chap. V, Exercise 11 p. 125. This result I think goes back to Heinz Bauer and is further discussed in Bogachev, Measure Theory, vol. 2, p. 458. So for $f$ surjective as you assume the answer is yes. $\endgroup$
    – Francois Ziegler
    Commented May 17, 2021 at 18:40
  • $\begingroup$ This was a much discussed question in the context of tight probability measures on completely regular spaces several decades ago. With your notation but for general spaces, a necessary condition is that for any $\epsilon >0$, there is a compact set $K$ in $E$ such that the measure of the complement of $f(K)$ is $ <\epsilon$. This is also sufficient in well-behaved spaces (e.g., polish). The proof uses Prokhorov's theorem. Thanks to covid I can't look up the reference but you could try Schwartz' book on Radon measures. $\endgroup$
    – bathalf15320
    Commented May 17, 2021 at 20:02
  • $\begingroup$ A nice question, but the title does not fit. The question as asked is whether there is a counterexample, not whether there is a sufficient condition. $\endgroup$
    – user95282
    Commented May 21, 2021 at 0:50

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$\newcommand\C{\mathscr C}\newcommand\de{\delta}$A sufficient condition is that $E$ be compact. Indeed, since $Q$ has a separable support, without loss of generality $F$ is separable. So, for each natural $n$ there is an (at most) countable set $\C_n$ of nonempty pairwise disjoint Borel subsets of $F$ of diameter $\le1/n$ such that $\bigcup\C_n=F$. For each $C\in\C_n$, take any $y_{n,C}\in C$ and let $$Q_n:=\sum_{C\in\C_n}Q(C)\de_{y_{n,C}},$$ where $\de_y$ is the Dirac probability measure supported on the set $\{y\}$. Then $Q_n\to Q$ (weakly).

Next, for each $n$ and each $C\in\C_n$, take any $x_{n,C}$ such that $f(x_{n,C})=y_{n,C}$, and let $$P_n:=\sum_{C\in\C_n}Q(C)\de_{x_{n,C}}.$$ Then $Q_n=P_nf^{-1}$, the pushforward measure of $P_n$ by $f$.

By the compactness of $E$, passing to a subsequence if needed, without loss of generality $P_n\to P$ for some probability measure $P$ on $E$. Since $f$ is continuous, it follows that $Q_n=P_nf^{-1}\to Pf^{-1}$. But $Q_n\to Q$. So, $Pf^{-1}=Q$. That is, $Q$ is the pushforward measure of $P$ by $f$, as desired.

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  • $\begingroup$ Another proof of the compact case by the Hahn--Banach theorem: Define the positive linear functional on the space of functions of the form $g\circ f$, $g$ continuous on $F$, in the obvious way from $Q$, and extend to a positive functional on the whole space of all continuous function on $E$. $\endgroup$
    – user95282
    Commented May 21, 2021 at 1:06

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