# Absolutely continuous, strictly positive measures

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.

## 1 Answer

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....

• 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 $\mathrm{supp}\,\mu$. – John Dougherty Nov 21 '13 at 23:54
• 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.... – Greg Martin Nov 22 '13 at 8:48