Pushforward of a $\sigma$-finite measure Let $\phi: X \rightarrow Y$ be a measurable function between measurable spaces. Let $\mu$ be a $\sigma$- finite measure on $X$. In general I have seen that the push forward measure $\phi_*\mu$ need not be $\sigma$- finite. What are assumptions we need to make $\phi_*\mu$ also $\sigma$- finite?
I saw a remark that if $\phi$ is injective and $\mu$ is non atomic then $\phi_*\mu$ is $\sigma$- finite. I couldn't prove it. Do we need any more assumption?
 A: Suppose

$\phi: X \rightarrow Y$ is a measurable function between measurable spaces, and $\mu$ is a $\sigma$- finite measure on $X$.

First: even if we assume that that $\phi$ is injective  and $\mu$ is non-atomic, it may happen that $\phi_*\mu$ is not $\sigma$- finite.
Example: Let $X=(\Bbb R, \mathcal{B})$ where $\mathcal{B}$ is the Borel $\sigma$-algebra. Let $Y=(\Bbb R, \mathcal{A})$ where $\mathcal{A}=\{\emptyset, \Bbb R \}$. Let $\lambda$ be the Lesbegue measure defined on $\mathcal{B}$. Let $\phi: X \rightarrow Y$ be defined as $\phi(x)=x$.
Clearly:

*

*$\lambda$ on $X$ is a $\sigma$- finite measure

*$\phi$ is injective

*$\lambda$ on $X$ has no atoms.

However, $\phi_*\lambda$ is a measure such that $\phi_*\lambda(\emptyset)=0$ and $\phi_*\lambda(\Bbb R)=+\infty$. The measure  $\phi_*\lambda$ on $Y$ is not $\sigma$- finite, not even semi-finite.
Second: Suppose $\phi$ is injective  and, for all $A$ measurable set in $X$, $\phi(A)$ is measurable. Then $\phi_*\mu$ is $\sigma$- finite on $Y$.
Proof: Since $\mu$ is a $\sigma$- finite measure on $X$, there are $X_n$ measurable sets in $X$ such that $X = \bigcup_n X_n$ and $\mu(X_n)<+\infty$. So, $\phi(X) = \bigcup_n \phi(X_n)$. So
$$ Y = (Y \setminus \phi(X)) \cup \bigcup_n \phi(X_n) $$
and
$\phi_*\mu(Y \setminus \phi(X)) = \mu( \phi^{-1}(Y \setminus \phi(X)))= \mu(\emptyset)=0$
and, for all $n$, $\phi_*\mu(\phi(X_n))= \mu( \phi^{-1}(\phi(X_n))=\mu(X_n)< +\infty$
So, $\phi_*\mu$ is $\sigma$- finite on $Y$.
A: It's been a while since a worked with measure theory, but try like this (it's a bit naive).
Consider subsets $X_n$ of $X$ with finite measure and such that $X = \bigcup_n X_n$. Suppose $\phi$ to be injective and suppose $\phi(X_n)$ to be measurable. Then $Y_n = \phi(X_n) \cup (Y \setminus \phi(X) ) $ are measurable sets with finite measure whose union is $Y$.
