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Consider a measure space $(X, \Sigma, \mu)$ and another measure $\nu$ on the same space. I'm interested in the conditions under which $\nu$ can be represented by a density function $f$ on $X$, so for measurable $S$, $\nu(S) = \int_S f \, \text{d}\mu$. A necessary condition is that for any $S$, if $\mu(S) = 0$ then $\nu(S) = 0$. Is this a sufficient condition? If not, what other conditions need to be included?

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Do you know about the Radon-Nicodym theorem? –  Mariano Suárez-Alvarez Nov 6 '12 at 16:49
    
Yes to $\sigma$-finiteness, and the Radon-Nikodym theorem is exactly what I was looking for. Thanks. –  Jeff Russell Nov 6 '12 at 17:24

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up vote 2 down vote accepted

If the measures are $\sigma$-finite, Radon-Nikodym's theorem applies. We can find it in any textbook about measure theory, see Athreya's book for example.

If one of the measures is not $\sigma$-finites, the result may not hold. For example, consider $[0,1]$ with $\mu$ Lebesgue measure and $\nu$ counting measure (which is not $\sigma$-finite). Then $\mu\ll\nu$ but each singleton has measure $0$ for $\mu$ (hence if a density $f$ existed, $f(x)=0$ for all $x$, which is not possible.

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