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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.

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1 Answer 1

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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 $X\setminus M$.

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  • $\begingroup$ Excellent answer, thank you very much! $\endgroup$
    – triple_sec
    Commented Jul 8, 2014 at 8:26

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