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If $q:X\to Y$ is a proper and continuous mapping of locally compact Hausdorff spaces, when does a Radon measure on $X$ pushforward to a Radon measure on $Y$.

Showing inner regularity of the pushforward measure is straightforward using the properness. I can prove that $q$ pushes forward Radon measures to Radon measures if $q$ is a quotient mapping and $X$ has a basis of saturated open sets (a subset $A\subset X$ is saturated if $f^{-1}(f(A)) = A$).

Is there a counterexample in the general case? Or is it true that proper continuous maps pushforward Radon measdures to Radon measures.

(Just to be clear, by pushforward measure, I mean the measure on $Y$ such that the measure of a set is defined to be the measure of its preimage in $X$.)

Edit: It seems that my definition of Radon is not standard; To me, a Radon measure is locally finite and both inner and outer regular (and, of course, a Borel measure).

Kyle Austin
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1 Answers1

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I'll use $f$ rather than $q$, and denote the measures on $X$ and $Y$ by $\mu$ and $f_*(\mu)$ respectively. So $f_*(\mu)(B)=\mu(f^{-1}(B))$ for measurable subsets $B$ of $Y.$

If $X$ is locally compact, $\mu$ is locally finite, and $f$ is proper, then $f_*(\mu)$ is locally finite. Proof: Given a point $y\in Y,$ pick a compact neighbourhood $K\ni y.$ Recall that a locally finite measure is finite on compact sets. So $f_*(\mu)(K)=\mu(f^{-1}(K))<\infty.$ This shows that $f_*(\mu)$ is locally finite.

By the way, properness is not needed to show that $f_*(\mu)$ inherits inner regularity from $\mu$. Given $B$ measurable in $Y,$ we have $f_*(\mu)(B)=\mu(f^{-1}(B))=\sup_{K\subseteq f^{-1}(B)}\mu(K) \leq \sup_{K\subseteq B}f_*(\mu)(K),$ where $K$ is restricted to compact sets. But clearly $f_*(\mu)(K)\leq f_*(\mu)(B)$ for any $K\subseteq B.$

Edit: to deal with outer regularity, the sensible thing to do is assume $\sigma$-compactness. But the following fact should do the job.

Fact: Let $f:X\to Y$ be a proper map between locally compact Hausdorff spaces. For any $B\subseteq Y,$ each open neighborhood of the preimage of $B$ contains a preimage of an open neighborhood of $B.$ That is, for any open set $U$ with $f^{-1}(B)\subseteq U\subseteq X,$ there is an open $V$ with $B\subseteq V$ and $f^{-1}(V)\subseteq U.$

I can't remember a reference, but it follows immediately from the standard fact that proper maps of LCH spaces are closed maps and taking $V=Y\setminus f(X\setminus U)$. Given this fact, for any Borel $B\subseteq Y,$ we have $f_*(\mu)(B)=\mu(f^{-1}(B))=\inf_{U\supseteq f^{-1}(B)}\mu(U)$ by outer regularity of $X,$ but this is at least $\inf_{f^{-1}(V)\supseteq f^{-1}(B)}\mu(f^{-1}(V))\geq\inf_{V\supseteq B}f_*(\mu)(V)\geq f_*(\mu)(B).$ Here $U,V$ are restricted to open sets.

Dap
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  • Ahhhh, this is emberassing; It seems that I am running into two different definitions of Radon measure in the literature and I did not notice! One of the sources that I have been working with requires Radon measures to be regular and locally finite while it seems that Wikipedia requires only inner regular and locally finite. Please excuse my saying so, but showing inner regularity and local finiteness are simple (though you do have a point that properness is not required to show inner regularity). I think that one cannot prove outer regularity in general; at least, I have no idea how to do so. – Kyle Austin Oct 16 '17 at 18:20
  • @KyleAustin: ok, I've added an argument for outer regularity – Dap Oct 18 '17 at 08:30
  • Aha! Indeed, the fact is correct and does immediately do the job. Proof of fact: Let $U$ be an open neighborhood of the preimage of $B$. Notice that the image of the complement $X\setminus U$ ($f(X\setminus U$) is closed in $Y$ and is disjoint from $B$. Therefore, $f^{-1}(Y\setminus f(X\setminus U))$ is an open neighborhood of $B$ which is completely contained in $U$. I checked your statement that proper maps of LCH spaces are closed maps. Everything looks like it checks out! Well done Dap! – Kyle Austin Oct 19 '17 at 09:36
  • Why is $\sup_{K\subseteq f^{-1}(B)}\mu(K)\leq\sup_{K\subseteq B}f_{*}(\mu)(K)$? – feltshire Oct 27 '24 at 18:58
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    @feltshire if $K\subseteq f^{-1}(B)$ is compact, then $K'=f(K)$ is a compact subset of $B$, and $K\subseteq f^{-1}(K')$. Now just use monotonicity of the measure and take suprema where relevant – Lorago Nov 07 '24 at 10:42