Suppose that both $X$ and $Y$ have joint pdf $f(x,y)$ and $Y$ has mariginal density $f(y)=\int_{\mathbb R}f(x,y)dx$. The conditional density of $X$ given $Y$ is given by $f(x|y)=\dfrac{f(x,y)}{f(y)}$ and $E(X|Y=y)=\int_{\mathbb R}xf(x|y)dx$. Let $\sigma(Y)$ be the sigma field generated by $Y$ and show that $E(X|\sigma(Y))=\theta(Y)$ where $\theta(y)=E(X|Y=y)$
My question here is what is $\theta(Y)$? Is it $E(X|Y=Y)$ and if so what does this mean?
I know I need to show that $$ \int_AE(X|\sigma(Y)dP=\int_A\theta(Y)dP \;\;\; \forall\; A\in\sigma(Y) $$ but im unsure how to start since im unclear on what $\theta(Y)$ means.
Here $\theta:\mathbb R\to\mathbb R$ is prescribed by:$$y\mapsto\mathbb E[X\mid X=y]=\int xf(x\mid y)\;dx$$
So actually: $$\theta(Y)=\int xf(x\mid Y)\;dx$$
From here it is our intention to prove that $\theta(Y)$ has the characteristics of $\mathbb E[X\mid Y]$ which are:
Be aware that $\mathbb E[X\mid Y]$ is the same thing as $\mathbb E[X\mid \sigma(Y)]$.
It is immediate that $\theta(Y)$ is measurable wrt $\sigma(Y)$.
Further we have: $$A\in\sigma(Y)\iff A=\{Y\in B\}\text{ for some Borel set }B$$
With this in mind we find:
$$\mathbb E\mathbf1_A\theta(Y)=\mathbb E\mathbf1_B(Y)\theta(Y)=\int f(y)1_B(y)\theta(y)\;dy=\int f(y)1_B(y)\left[\int xf(x\mid y)\;dx\right]\;dx\;dy=$$$$\int\int 1_B(y)xf(x,y)\;dx\;dy=\mathbb E\mathbf1_B(Y)X=\mathbb E\mathbf1_AX$$
Now we are done and this together justifies to state that $$\mathbb E[X\mid \sigma(Y)]=\mathbb E[X\mid Y]=\theta(Y)$$