Hypothesis in the Radon-Nikodym theorem.

Question

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?

Answer 1

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.

Answer 2

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 $d\nu = fd\mu$.