# Is the push forward of a sigma-finite measure equivalent to a sigma-finite measure?

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 $\varphi_*\mu$?

• You mean "mutually absolutely continuous with $\varphi_* \mu$", I suppose. Jan 10 '15 at 20:00
• Yes, because $\mu$ is equivalent to a finite measure. Jan 10 '15 at 20:03

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 $(\varphi_* \nu)(B) = \nu(\varphi^{-1}(B)) = 0$.