Converse to Lebesgue Differentiation Theorem

Question

Suppose we are given two finite positive Borel measures $\mu$, $\nu$ on $\mathbb{R}^n$ such that the function $$ f(x) := \limsup_{r\to 0} \frac{\mu(B(x,r))}{\nu(B(x,r))} $$ is in $L^1(\nu)$. Is it true that $\mu \ll \nu$? Note that in a sense, this would be a converse of the Lebesgue differentiation theorem, since if we started off assuming that $\mu \ll \nu$, then $\nu$-a.e., $f$ would be equal to the Radon-Nikodym derivative of $\mu$ with respect to $\nu$, which is in $L^1(\nu)$.

If it helps, I think I'm OK with also assuming that $\nu \ll \mu$, or even $\nu \leq \mu$.

For some context, this is coming up in my research constructing equilibrium states in dynamics. I tried a google search and I couldn't find very much, so any leads would be appreciated, even if not a full response. Thank you!

Answer 1

Counterexample: if $\mu$ is the Dirac's delta in $0$ and $\nu$ is a Gaussian measure (say, with parameters $\mu = 0$ and $\sigma^2 = 1$), then $f$ is the zero function, apart from the point $0$ where it equals to $+\infty$. It follows that $f$ is in $L^1(\nu)$. However $\mu$ is not absolutely continuous with respect to $\nu$.

I don't know if there are counterexamples with the further requirement that $f$ is well-defined and finite everywhere.

Answer 2

This answer is intended as a complement to Bob's.

In "Geometry of sets and measures in euclidean spaces" by P. Mattila, the following theorem is proved (theorem 2.12 in my edition):

Let $\mu,\nu$ be Radon measures on $\mathbb{R}^n$. Define $\underline{D}(\mu,\nu,x)=\liminf \frac{\mu(B(x,r))}{\nu(B(x,r))}$ and $\overline{D}$ in the same way with $\limsup$. Then

1. $D=\overline{D}=\underline{D}$ exists finite $\nu-$almost everywhere
2. $Dd\nu\le d\mu$ (as measures) and the equality holds iff $ \mu\ll \nu$
3. $\mu\ll \nu$ iff $\underline D$ is finite $\mu-$a.e.