Let $X$ and $Y$ be locally compact, second countable spaces, and let $\varphi \colon X \to Y$ be a Borel map. Let $\mu$ be a sigma-finite measure on $X$. In general, the push-forward $\varphi\_\*\mu$ is not sigma-finite. Does there, however, exist a sigma-finite measure $\nu$ that is mutually absolutely continuous with $\varphi\_\*\mu$? 



Is the push forward of a sigma-finite measure equivalent to a sigma-finite measure?
Yes, it is even equivalent to a finite measure.


Note that $\mu$ is equivalent to a finite measure $\nu$ on $X$. For example, if we can write $X = \bigcup\_n X\_n$ where $X\_n$ is Borel with $0 < \mu(X\_n) < \infty$, set $f = \sum\_{n=1}^\infty \frac{1}{2^n \mu(X\_n)} 1\_{X\_n}$ and take $d\nu = f\,d\mu$. Then $\nu$ is a probability measure and is mutually absolutely continuous with $\mu$ (since $f>0$ everywhere). That is, $\mu(A) = 0$ iff $\nu(A) = 0$ for all Borel $A \subset X$. Now the pushforward $\varphi\_\* \nu$ is a finite measure on $Y$, and is mutually absolutely continuous with $\varphi\_\* \mu$, since for any Borel $B \subset Y$ we have $(\varphi\_\* \mu)(B) = \mu(\varphi^{-1}(B)) = 0$ iff $(\varphi\_\* \nu)(B) = \nu(\varphi^{-1}(B)) = 0$.


