# Sufficient condition for a probability measure to be a pushforward measure

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 $$f$$?

• 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? May 17, 2021 at 18:03
• 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. May 17, 2021 at 18:40
• 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. May 17, 2021 at 20:02
• 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. May 21, 2021 at 0:50

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.
• 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 $E$. May 21, 2021 at 1:06