Boundedness of Radon-Nikodym derivative

Question

I have the following question:

Under which conditions (I mean the conditions on the measures) the Radon-Nikodym derivative is bounded ? I made some researches but I couldn't find any answer, although I find it very interesting and could have many useful applications in statistics :/ Does any one have an idea ? Thank you for your time

Answer 1

I don't know if this is useful to you, but here's one obvious formulation. Suppose $\mu$ is a measure and $\lambda$ is a measure with Radon-Nikodym derivative $f$ with respect to $\mu$. Then $f$ is bounded (up to sets of $\mu$-measure $0$) iff $\lambda/\mu$ is bounded, in the sense that there exists $C$ such that for any set $A$ such that $\mu(A)>0$, $\lambda(A)/\mu(A)\leq C$. Or more simply, there exists $C$ such that $\lambda(A)\leq C\mu(A)$ for all measurable $A$.

The proof is straightforward. First, suppose $f$ is bounded; say $f(x)\leq C$ for all $x$. Then for any measurable $A$, $\lambda(A)=\int_A f\,d\mu\leq C\mu(A)$. Now suppose $f$ is unbounded. Since we're working up to sets of measure $0$, this means that for any $C$, there is a set $A$ of positive measure on which $f\geq C$. We then have $\lambda(A)=\int_A f\,d\mu\geq C\mu(A)$. Since $C$ is arbitrary, this means $\lambda/\mu$ is unbounded.

Answer 2

One can show in the following setting that the Radon-Nikodym derivative is bounded a.e. :

If $\nu$ and $\mu$ are both finite, positive measures on the same measurable space, and $\nu$ is absolutely continuous with respect to $\mu$, and $L^1(\mu)\subseteq L^1(\nu) $, then the Radon-Nikodym derivative is in $L^{\infty}(\mu) $.