Let $X$ be a locally compact Hausdorff space with the Borel $\sigma$-algebra $\mathscr B_X$. Suppose that $\mu$ is a positive measure, $\nu$ is a finite positive measure, and $\nu\ll\mu$.

It is known that the Radon–Nikodym theorem may well fail if $\mu$ is not $\sigma$-finite. However, assume that $\mu$ and $\nu$ are both Radon measures, though $\mu$ is not necessarily $\sigma$-finite (but $\nu$ is finite). I am trying to prove a version of the Radon–Nikodym theorem, namely that there exists some $f:X\to[0,\infty)$ such that

  • $f\in L^1(\mu)$ and
  • $\nu(E)=\int_Ef\,\mathrm d\mu\quad\forall E\in\mathscr B_X$.

I have already shown the existence of such an $L^1(\mu)$ function that $\leq$ holds in the second line (using the proof of the standard Radon–Nikodym theorem), but the other direction seems to be elusive.

Any hint would be greatly appreciated.


As $\nu$ is finite and Radon, there is for each $n$ a compact subset $K_n$ such that $\nu(X\setminus K_n) <1/n$.

Note that for $M := \bigcup K_n$, we have $\nu(X\setminus M) =0$ and that $M$ is $\sigma$ finite for $\mu$, as it is $\sigma$ compact.

Apply the ordinary Radon Nikodym theorem on $M$ and extend the Radon Nikodym derivative by zero on $X\setminus M$.

  • $\begingroup$ Excellent answer, thank you very much! $\endgroup$ – triple_sec Jul 8 '14 at 8:26

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.