I'm reading through a text that contains a more abstract definition of expectation than what I'm used to. The definition I'm used to is: $E(X|Y)(y)=\int xf_{X|Y}(x,y)dx$, where $f_{X|Y}$ is the conditional probability density function $f_{X|Y}(x,y)=\frac{f(x,y)}{f_Y(y)}$. The definition in the text says $E(X|\mathcal{F})=Z$, where $\mathcal{F}$ is a sub-$\sigma$-algebra of the original $\sigma$-algebra and $Z$ is a $\mathcal{F}$-measurable random variable such that $\int_A Zd\mu=\int_A Xd\mu$. The new definition should reduce to the old as follows: If $Y$ is a random variable, take $\mathcal{F}$ to be the coarest $\sigma$-algebra in which $Y$ is measurable; this is denoted $\sigma(Y)$.
To show that this works, one has to show that $h(Y)$ is $\sigma(Y)$-measurable, where $h(y)=\int xf(x,y)dx/\int f(x,y)dx$. I guess this amounts to showing that for any sub-$\sigma$-algebra, plugging a measurable function into a two-variable function and taking a partial integral yields a measurable function.
So my question is: How do you do that?
I think the closest existing question to this On the Preservation of Product Measurability under Partial Conditional Expectation.. But I guess mine should be simpler.
