I have a question about the proof of the existence of a conditional expectation using the Radon-Nikodyn derivative.
Let $X$ be a nonnegative random variable defined on $(\Omega, \mathcal{F}_0, \mathbb{P})$ with $\mathbb{E}|X| < \infty$ and let $\mathcal{F} \subset \mathcal{F}_0$.
We can define the probability measure on $\nu$ on $\mathcal{F}_0$ s.t. $\nu(A) = \int_AXd\mathbb{P}$.
Let $\mathbb{P}'$ and $\nu'$ be the restrictions of $\mathbb{P}$ and $\nu$ to the measurable space $(\Omega, \mathcal{F})$. By Radon-Nikodyn, there exists a $\frac{d\nu'}{d\mathbb{P}'}$ that is measurable with respect to $\mathcal{F}$ and for $B \in \mathcal{F} $ we have $\nu'(B) = \int_B \frac{d\nu'}{d\mathbb{P}'} d\mathbb{P}'$.
The claim is that $\mathbb{E}(X|\mathcal{F}) = \frac{d\nu'}{d\mathbb{P}'}$.
Therefore, we need that for $B \in \mathcal{F}$, $\int_B \frac{d\nu'}{d\mathbb{P}'} d\mathbb{P}$ = $\int_B Xd\mathbb{P}$. (This is in the probability space $(\Omega, \mathcal{F}_0, \mathbb{P})$)
We know $\int_B Xd\mathbb{P}$ = $\nu(B)$.
Also, in the probability space $(\Omega, \mathcal{F}, \mathbb{P}')$, we have $\int_B \frac{d\nu'}{d\mathbb{P}'} d\mathbb{P}'$ = $\nu'(B) = \nu(B)$ (since $\nu'$ is a restriction). But it must be that $\int_B \frac{d\nu'}{d\mathbb{P}'} d\mathbb{P}$ = $\nu(B)$ and I do not see why that is necessarily the case.
Since $\mathcal{F} \subset \mathcal{F}_0$, by the construction of the Lebesgue Integral (sup of the integral of simple functions less than $\frac{d\nu'}{d\mathbb{P}'}$) we have:
$\int_B \frac{d\nu'}{d\mathbb{P}'} d\mathbb{P} \geq \int_B \frac{d\nu'}{d\mathbb{P}'} d\mathbb{P}' = \nu(B)$.
I do not understand why the reverse inequality must hold (in $\mathcal{F}_0$ we have more simple functions to work with).
I actually think that I have it. $\frac{d\nu'}{d\mathbb{P}'}$ is measurable wrt $\mathcal{F}$. If it is also simple then for $B \in \mathcal{F}$, $\int_B \frac{d\nu'}{d\mathbb{P}'} d\mathbb{P}'$ = $\int_B \frac{d\nu'}{d\mathbb{P}'} d\mathbb{P}$ by the fact that $\mathbb{P}'$ is the restriction of $\mathbb{P}$ to $\mathcal{F}$. Even if $\frac{d\nu'}{d\mathbb{P}'}$ is not simple it can be approximated from below by simple functions in $\mathcal{F}$ (since $\frac{d\nu'}{d\mathbb{P}'}$ is a.s. positive). The result follows by two applications of monotone convergence.