Failure of a push-forward to be sigma-finite Let $X$ and $Y$ be locally compact, second countable spaces, and let $φ:X→Y$ be a measurable function. Let $μ$ be a sigma-finite measure on $X$. In general, the push-forward $\phi_{*}\mu$ is not sigma-finite.
Question: what are the causes of this failure?
(naive?) conjecture: The only failure is when there is at least one point $y\in Y$ such that $\mu\left(\phi^{-1}\left(\{y\}\right)\right) = \infty$.
Does it make a difference if I assume $\phi$ to be piecewise continuous rather than measurable?
 A: Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^2$. Consider the continuous map $\mathrm{pr}_1:\mathbb{R}^2\rightarrow \mathbb{R}$. Set $\lambda ':=\mathrm{pr}_{1\ast}\lambda$. For every $x\in \mathbb{R}$ we have $\lambda '(x)=\lambda(\mathrm{pr}_1^{-1}(x))=\lambda(\{x\}\times \mathbb{R})=0$. But for any small open set $U\subset \mathbb{R}$ we get $\lambda '(U)=\infty$. Hence $\mathbb{R}$ is not $\sigma$-finite with respect to $\lambda '$.    
A: This is an entirely measure theory question and has nothing to do with topology, continuity etc. Your conjecture is actually true if reformulated in a more appropriate way.
The key example is the following model situation: $X=I^2$ is the unit square, $Y=I$ is the unit interval, and $\phi:(x,y)\to x$ is the vertical coordinate projection. Let $\mu$ be a $\sigma$-finite measure on $X$ absolutely continuous with respect to the Lebesgue measure $\lambda$, i.e., $d\mu(x,y)=f(x,y)d\lambda(x,y)$ for a measurable non-integrable density $f$. Then the image measure $\phi\mu$ is $\sigma$-finite iff the integrals $F(x)=\int f(x,y)\,dy$ are a.e. finite (in which case $F$ is the density of the image measure). 
In the general case, since $\mu$ is $\sigma$-finite, we may consider it as $\mu=f\cdot\lambda$ for a probability measure $\lambda$ and a density $f$. Now, if you add to your conditions on $X$ and $Y$ the requirement that they are normal, then the measure spaces $(X,\lambda)$ and $(Y,\phi\lambda)$ are so-called Lebesgue spaces, which in particular implies that there exists a family of conditional measures on the fibers of the map $\phi:X\to Y$. Then $\sigma$-finiteness of the quotient measure is equivalent to integrability of the density $f$ with respect to almost all conditional measures.
Actually the above example is almost the general case, because non-atomic Lebesgue spaces are isomorphic to the unit interval endowed with the Lebesgue measure, and any morphism between non-atomic Lebesgue spaces with non-atomic conditional measures can be realized as the quotient map from the example.
EDIT. Lebesgue measure spaces are the ones that are separable in the sense that there exists a countable family of measurable sets which separates points. In the overwhelming majority of situations one actually deals with Lebesgue spaces, so that I am wondering whether your spaces could still be Lebesgue. However, the criterion that I formulated in reality only uses conditional expectations (rather than conditional measures), so that it works in full generality. 
Namely, instead of the integrals of the density $f$ with respect to the conditional measures of the projection $(X,\lambda)\to (Y,\phi\lambda)$ just take directly the result $\mathbf E f$ of applying the conditional expectation operator $\mathbf E$ of this projection to the density $f$. The condition is that $\mathbf E f$ has to be a.e. finite.  
