2
$\begingroup$

Let $\mu$ be a strictly positive $\sigma$-finite measure on $\mathbb{R}^{n}$ that is absolutely continuous with respect to the Lebesgue measure, $\lambda$. One way to think about the absolute continuity requirement is as a lower bound: the $\sigma$-ideal $N$ of $\mu$-null sets has to contain the $\sigma$-ideal of Lebesgue-null sets. Likewise, the requirement of strict positivity is something of an upper bound: $N$ can't contain any nonempty open sets.

My question:

Up to equivalence, how many absolutely continuous, strictly positive, $\sigma$-finite measures on $\mathbb{R}^{n}$ are there?

Intuitively, I think there can't be that many. By the Radon-Nikodym theorem, $\mu(A) = \int_{A} f\, d\lambda$ for some measurable $f$. I want to say that strict positivity requires that $f$ have full support, meaning that we could write $\lambda(A) = \int_{A} f^{-1}\, d\mu$, showing that $\mu$ and $\lambda$ are equivalent. That would mean $\lambda$ is the only such measure, up to equivalence. But I'm not happy with the step from strict positivity to full support.

$\endgroup$
2
$\begingroup$

Let $S$ be a nowhere dense subset of $\mathbb R^n$ with positive Lebesgue measure. Define $\mu(A) = \lambda(A\setminus S)$ for all Lebesgue measurable $A\subset\mathbb R^n$. Note that $\mu(A)\le\lambda(A)$ always; then certainly $\mu$ is $\sigma$-finite and absolutely continuous with respect to $\lambda$.

Furthermore, it's not hard to show that $S$ is strictly positive: for any open set $U$, we have $\mu(U) = \lambda(U\setminus S) \ge \lambda(U\setminus \bar S) > 0$ since $U\setminus \bar S$ is a nonempty open set. However, $\lambda(S) > 0$ while $\mu(S)=0$; hence $\mu$ is not equivalent to $\lambda$.

So there are at least as many inequivalent $\mu$ as there are "inequivalent" nowhere dense sets with positive measure. That's a lot....

$\endgroup$
  • $\begingroup$ In retrospect this should have occurred to me, since a topological constraint like strict positivity isn't strong enough to control behavior on nowhere dense sets. Followup: is every $\mu$ satisfying my constraints of the form $\mu(A) = \lambda(A \setminus S)$ for some (possibly empty) nowhere dense $S$? The idea would be that since $\mathrm{supp}\,\mu = \mathrm{supp}\,\lambda$, they could only differ on a set that doesn't change the closure of $\mathrm{supp}\,\mu$. $\endgroup$ – John Dougherty Nov 21 '13 at 23:54
  • 1
    $\begingroup$ I'm not sure, but you'd at least have to say equivalent to such a measure, since you could take $\mu(A) = \int_{A\setminus S} g\,d\mu$ for suitable $g$. If it were true, presumably one would start by defining $S$ to be the complement of the support of the $f$ coming from the Radon-Nikodym theorem.... $\endgroup$ – Greg Martin Nov 22 '13 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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