2
$\begingroup$

I am a bit confused about the exact statement of the Radon-Nikodym Theorem. Suppose that in the usual setup, $v \ll u$. Does it require both $v$ and $u$ to be sigma-finite, or only $u$ to be sigma finite for the theorem to hold?

$\endgroup$
3
$\begingroup$

You need both measures to be $\sigma$-finite. Suppose you have the collection of Lebesgue measurable sets on $[0,1]$ and let $\mu$ be the counting measure. The only set of measure zero with respect to $\mu$ is the empty set, so every measure is absolutely continuous with respect to $\mu$ and you can show that there is no nonnegative Lebesgue measurable function $f$ on $[0,1]$ for which $$ m(E)=\int_Ef\,d\mu $$ where $E$ is any Lebesgue measurable set on $[0,1]$.

$\endgroup$
  • $\begingroup$ Please clarify one thing. In your answer, the counting measure mu is not sigma finite, so I cannot see how that answers my question. Actually, I am looking for a counterexample, where only mu is sigma finite, nu is absolutely continuous w.r.t mu, but not sigma finite, and the Radon Nikodym Theorem fails. $\endgroup$ – Somabha Mukherjee Oct 7 '14 at 6:11
  • $\begingroup$ You need both measures to be $\sigma$-finite. The counting measure is not $\sigma$-finite and every other measure $\nu$ is absolutely continuous with respect to $\mu$. In particular, the Lebesgue measure is absolutely continuous with respect to $\mu$. So if the Radon-Nikodym result holds in this case, we should be able to find a Lebesgue measurable function such that $m(E)=\int_Ef\,d\mu$, but this is impossible. $\endgroup$ – Laars Helenius Oct 7 '14 at 15:31
0
$\begingroup$

Only $u$ must be $\sigma$-finite. $v$ can be any (signed) measure. The result is proved in Section 2.2 in Ash and Doleans-Dade's book Probability and Measure Theory (2nd edition, 2000).

As a simple illustration, let $\Omega$ be some set and consider the trivial $\sigma$-algebra $\mathcal{F} = \{\emptyset,\Omega\}$. Suppose $u(\Omega) = 1$, $u(\emptyset) = 0$, $v(\Omega) = \infty$ and $v(\emptyset) = 0$. Obviously $u$ is $\sigma$-finite, $v$ is not $\sigma$-finite, and $v$ is absolutely continuous w.r.t. $u$. We can express $v(A) = \int_A g d\mu$, $A \in \mathcal{F}$, where $g = \infty$ is the Radon-Nikodym derivative of $v$ w.r.t. $u$.

The Lebesgue Decomposition Theorem requires both measures to be $\sigma$-finite.

$\endgroup$

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.