Pushforwards of Radon Measures under Mappings 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). 
 A: 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.
