Conditional Expectation: How to Intergrate indicator function multiplied by the joint denisity?

Question

I am currently reading "Measure, Integral and Probability" by Capinski, Marek (see p179). It includes some motivation for the definition of the conditional expectation. For example, given two random variables $X,Y$ with joint density $f_{(X,Y)}$ (and so the marginal and conditional densities), we want to show that for any set $A \subset \Omega, A=X^{-1}(B), B$ Borel, that $$\int_A\mathbb{E}(Y|X)dP= \int_A \mathbb{E}(Y)dP.$$ This is one of the defining condition of an conditional expectation. The book shows the following calculation, \begin{align} \int_A\mathbb{E}(Y|X)dP &= \int_\Omega 1_B(X)\mathbb{E}(Y|X)dP\\ &= \int_\Omega 1_B(X(\omega))\left(\int_\mathbb{R}yf_{Y|X}(y|X(\omega))dy\right)dP(\omega)\\ &=\int_\mathbb{R}\int_\mathbb{R}1_B(x)yf_{(Y|X)}(y|x)dy f_X(x)dx\\ &=\int_\mathbb{R}\int_\mathbb{R}1_B(x)yf_{X,Y}(x,y)dxdy\\ &= \int_\Omega 1_A(X)YdP\\ &= \int_A YdP. \end{align} What I don't understand is the second to last equality immediately above, i.e. $$ \int_\mathbb{R} y \int_\mathbb{R}1_B(x)f_{X,Y}(x,y)dxdy = \int_\Omega 1_A(X)YdP .$$ I think it is a typo since $X\in \mathbb{R}$ and $A \subset \Omega$ --- however, I cant figure the correction either!

Answer 1

There's a typo, the author certainly intended to write $\int_\Omega 1_B(X)YdP $.

Indeed $$\int_\mathbb{R} \int_\mathbb{R}1_B(x)yf_{X,Y}(x,y)dxdy = \int 1_B(x)y dP_{(X,Y)}(x,y) = \int 1_B(X)YdP$$