3
$\begingroup$

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?

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ I brought in topological concerns as, in my context, they are present, so I figured they might help. If I read your answer right, the main source of failure is when (in some coordinate system), $\phi$ acts like a projection and measure $> 0$ of the conditional measures are not integrable. I think I can work with that. In my case though, some of the spaces are not normal! How does that alter the answer? $\endgroup$ Aug 15, 2015 at 14:27
  • $\begingroup$ I have added a comment to my answer. $\endgroup$
    – R W
    Aug 15, 2015 at 23:39
  • $\begingroup$ Thanks. The context is a probabilistic programming language, so that the measures being defined are sometimes a superposition of (absolutely continuous wrt) Lebesgue and some atoms. So I do need the general case. Thanks. $\endgroup$ Aug 16, 2015 at 14:41
  • $\begingroup$ So if I understand this correctly then if $\phi$ is injective and $\mu$ is non-atomic then $\phi_{\star}(\mu)$ must be $\sigma$-finite? $\endgroup$
    – ABIM
    Mar 3, 2020 at 13:39
  • $\begingroup$ @ProbablyAHuman - If $\phi$ is injective, then it is one-to-one between $X$ and its image, and therefore $\mu$ and its image are $\sigma$-finite (or not) simultaneously. $\endgroup$
    – R W
    Mar 6, 2020 at 21:29
3
$\begingroup$

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 $\lambda '$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.