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I have a question concerning a proof of Radon-Nikodym theorem here.

Why is the hypothesis "$\nu$ is finite" necessary? The author uses it to have the measure $\sigma=\mu+\nu$ finite and then, from $|Tu|\leq \Vert u\Vert_{L^2(X,\mathcal{F},\sigma)}\sqrt{\sigma(X)}$, conclude that $T$ is bounded. By I think that it is also true that $|Tu|\leq\Vert u\Vert_{L^2(X,\mathcal{F},\sigma)} \sqrt{\mu(X)}$ (applying first Hölder's inequality and then the fact that $\mu\leq\sigma$), so $T$ is bounded in any case .

So for me, $\mu$ has to be finite, but $\nu$ not. However, this would contradict the example posted here. Any ideas?

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user39756
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  • You're right. This proof works for infinite $\nu$ as well. However, the second sentence under the link says, "The extension to the $\sigma$-finite case is a standard exercise". Apparently, the author didn't bother trying to make the proof extremely general. – zhoraster Oct 20 '16 at 12:23
  • @zhoraster So, at a first step, one proves the Radon-Nikodym theorem for $\mu$ finite measure, $\nu$ any measure, with $\nu <<\mu$. After this, one can extend the result to $\mu$ $\sigma$-finite measure, $\nu$ any measure, with $\nu << \mu$. Am I right? – user39756 Oct 20 '16 at 12:52
  • It is enough to prove the first step for both measures finite. Then one can split the space into (countably many) parts where both measures are finite. – zhoraster Oct 20 '16 at 13:46
  • @zhoraster Thank you for comments. The thing is that I don't exactly understand why we have to assume that $\nu$ is finite. Why does one work with that superfluous hypothesis? I mean, we have the result when $\mu$ is finite and $\nu$ is any measure. When $\mu$ is $\sigma$-finite and $\nu$ is any measure, we write the total space $X=\cup_{n}X_n$, where the $X_n$'s are disjoint and $\mu(X_n)<\infty$, and apply the result on $(X_n,\mathcal{F})$ (why does one need $\nu(X_n)<\infty$ as well??) – user39756 Oct 20 '16 at 15:56

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As an example, take $\mu$ to be counting measure on $[0,1]$, and take $\nu$ to be Lebesgue measure. If $\mu E =0$, then $E=\emptyset$ and, so, $\nu E = 0$. But you're not going to express $d\nu = fd\mu$.

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  • But I'm interested on the case $\mu$ finite or $\sigma$-finite and $\nu$ any measure. In particular, I'd like to know if in the proof presented here the fact that $\nu$ is finite is used. – user39756 Oct 30 '16 at 07:40
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I answer my own question, as in the end I have been able to solve it. I will use the notation from the proof.

The fact that $\nu$ is finite is used. Indeed, it is used to have the measure $\sigma=\mu +\nu$ finite, and this is needed for the statement just after formula (2): "then $\mu(A)=\int_A g\,d\sigma$ for every $A\in\mathcal{F}$". One needs $1_A\in L^2(\Omega,\mathcal{F},\sigma)$ for every $A\in\mathcal{F}$, and for that it is necessary to have $\sigma$ finite.

Hence, we do not have a contradiction with the example.

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