Integral form of a conditional expectation

Question

I have seen this line in a computation and I wonder if that makes sense : $\mathbb{E} (X | Y) = \int \mathbb{E} (X | Y = y) dP(Y = y)$ where P is a probability measure, $X,Y$ are random variables, so is $\mathbb{E} (X | Y)$. But I'm not familiar with this notation. If this is right, how can I understand it ?

Answer 1

You have that $\mathbb{E}(X|Y)$ is, by definition, $\sigma (Y)$-measurable, and it can be shown that there is a measurable function $h:\mathbb{R}\to \mathbb{R}$ such that $\mathbb{E}(X|Y)=h(Y)$.

Clearly $h(Y)dP\ll dP$, and so $h(y)dP_Y(y)\ll dP_Y(y)$, where $dP_Y$ is the measure induced by $Y$ in $\mathbb{R}$. Then we set $\mathbb{E}(X|Y=y):=h(y)$. However, your identity is false, according to the above we have that

$$ \int_{\mathbb{R}}\mathbb{E}(X|Y=y)dP_Y(y)=\int_{\Omega }\mathbb{E}(X|Y)dP=\int_{\Omega }XdP=\mathbb{E}(X) $$