By definition of $\mathbb E(X\mid \sigma(Y))$ calculate $\mathbb E(X\mid Y=y)$ when $(X,Y)$ is absolutely continuous

Question

By definition of $\mathbb E(X\mid \sigma(Y))$, I want to show that $$\mathbb E(X\mid Y=y)=\int xf(x\mid Y=y) dx.$$ I want to know is my steps right or no. It is important to me since in preliminary probability books we define $\mathbb P(A\mid B)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}$ so if $\mathbb P(B)=0$ we can not use it to calculate some case $f(X\mid Y=y)$, when $Y$ is continuous. I also want to find the new definition (based on conditional expectation with respect to sigma field) exactly same as old definition ($\mathbb E(X\mid Y=y) =\int x f(x\mid y) dx$)

By definition of $\mathbb E(X\mid\sigma(Y))$, $\forall A\in \sigma(Y)$, $$\mathbb E\big(\mathbb E(X\mid \sigma(Y))1_A \big)=\mathbb E(X\, 1_A )$$ I think $1_A$ is a function of $Y$ (??) since $A\in \sigma(Y)$. So I think I can write $$\mathbb E\bigg(\mathbb E(X\mid \sigma(Y))1_B(Y) \bigg)=\mathbb E(X\,1_B(Y) )$$ Since $\mathbb E(X\mid \sigma(Y))$ is a function of $Y$ $$LHS=\mathbb E\bigg(\mathbb E(X\mid \sigma(Y))1_B(Y) \bigg)=\mathbb E(g(Y))=\int g(y)f_Y(y) dy\\ =\int \mathbb E(X\mid Y=y) 1_B(y) f_Y(y) dy=\int_B \mathbb E(X\mid Y=y) f_Y(y) dy;\\ RHS=\mathbb E(X\,1_B(Y) )=\mathbb E(h(X,Y))=\int \int h(x,y) f(x,y) dx \, dy \\ =\int \int x\, 1_B(y) f(x,y) dx \, dy =\int_B \int x f(x,y) dx \, dy,$$ so as $RHS=LHS$ I think (since it is for all $B$ ??) $$\mathbb E(X\mid Y=y) f_Y(y)=\int x f(x,y) dx \\ \\\Leftrightarrow\\ \mathbb E(X\mid Y=y) =\frac{1}{f_Y(y)}\int x f(x,y) dx=\int x \frac{f(x,y)}{f_Y(y)} dx =\int x f(x\mid y) dx.$$

Thanks in advance for any help you are able to provide or any clarification.

Answer 1

This is essentially correct; there are just some technical details: exchanging order of integration, meaning of equalities (a.e. with respect to distribution of $Y$) etc.

I think $1_A$ is a function of $Y$ (??) since $A\in \sigma(Y)$

Indeed, the $\sigma$-algebra generated by a random variable is $$\sigma(Y) = \big\{\{Y\in B\}, B\in \mathcal B(\mathbb R)\big\},$$ in other words, any $A\in \sigma(Y)$ has a form $A = \{Y\in B\}$, where $B\in \mathcal B(\mathbb R)$, so $1_A = 1_B(Y)$.

as $RHS=LHS$ I think (since it is for all $B$ ??)

Indeed, if $\int_B f_1(x) dx = \int_B f_2(x) dx$ for all $B\in \mathcal B(\mathbb R)$, then $f_1 = f_2$ a.e.