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Let $\mu,\nu$ be two positive measures on $(X,\mathscr{A})$ and $\mu$ is finite. If $\nu \ll \mu$, then does there exist a measurable function $f: X \to [0,\infty]$ such that $$ \nu(E) = \int_E f d \mu,~\forall E \in \mathscr{A}? $$

I don’t know what happens when $\nu$ is not $\sigma$-finite.

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For the case where $\nu$ is not $\sigma$-finite the answer is yes provided $f$ is allowed to take the value $\infty$. See this answer.

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    $\begingroup$ What happens if $\nu$ is not $\sigma$-finite? $\endgroup$ – Martin Argerami Jun 27 at 3:26
  • $\begingroup$ @Martin Argerami It still holds provided the function $f$ is allowed to take the value $\infty$ (which the OP has indicated is allowed in their setting). Here is another thread where this question is addressed. $\endgroup$ – user1222 Jun 27 at 3:51
  • $\begingroup$ As it is written, your answer is quoting the Wikipedia article which requires $\nu$ to be $\sigma$-finite. And I don't really see how the other answer contributes to yours; actually, it confuses things, because it says that the Wikipedia article is wrong, but then one would have to check what the article said back in 2016. $\endgroup$ – Martin Argerami Jun 27 at 4:46
  • $\begingroup$ @Martin Argerami See the second answer in the thread I linked (the one with 1 up vote). The person who wrote it gives a detailed proof that $f$ exists in the case where $\nu$ is not $\sigma$-finite provided $f$ can take the value $\infty$. $\endgroup$ – user1222 Jun 27 at 4:53
  • $\begingroup$ Fair enough, but then link to that specific answer. $\endgroup$ – Martin Argerami Jun 27 at 5:55

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