A question on Conditional Expectation from Breiman

Question

Breiman Proposition 4.36 Let $X$ & $Y$ be random variables. Let $f(X,Y)$ be a real valued, random variable such that the expectation of the absolute value of $f(X,Y)$ is finite. If $Q(\cdot\mid Y=y)$ is a regular conditional distribution for $X$ given $Y=y$, then,

$E(f(X,Y)\mid Y=y)$=$\int_\ f(x,y) \, dQ(\cdot \mid Y=y)$ a.s. with respect to the Law of $Y$.

Quoting Brieman (Setion 4.3, page 80), "It is tempting to replace the right-hand of (4.37) (equation above) by $E(f(X,y)\mid Y=y)$. But this object cannot be defined through the standard definition of conditional expectation (4.18) (defined below)."

Definition 4.18. $E(X\mid Y=y)$ is any random variable on $R$, where $Q(B)=P(Y \in B)$, satisfying

$\displaystyle \int_B E(X\mid Y=y) dQ = \displaystyle \int_{Y \in B} X dP$ for all Borel subsets $B$.

That is essentially the relevant part of Breiman. (I have made small changes as my ability to write in scientific notation is limited.)

This should help clarify my questions below

Is $E(f(X,Y)\mid Y=y)$ = $E(f(X,y)\mid Y=y)$?

By my understanding, Brieman's Probability text (Setion 4.3, page 80) seems to suggest that$E(f(X,y)\mid Y=y)$ is not well defined, so the question is ill posed.

I was looking for an explanation of Breiman's statement, if my understanding of it is correct of course.

The intuitive answer from introductory probability would suggest that the two expressions are the same, as $Y$ can be treated as a constant when conditioning on $Y=y$.

Thanks

Answer 1

The expression $E[Z\mid Y=y]$ is not well defined when $P[Y=y]=0$. One approach to define it would be to come back to $E[Z\mid Y]$, a random variable well defined up to null events and which is $E[Z\mid Y]=u(Y)$ for some measurable function $u$, and to consider that $E[Z\mid Y=y]=u(y)$.

Unfortunately, changing the function $u$ at $y$ only yields a second measurable function $\bar u$ such that, when $P[Y=y]=0$, $u(Y)=\bar u(Y)$ almost surely, hence $E[Z\mid Y]=u(Y)$ almost surely and $E[Z\mid Y]=\bar u(Y)$ almost surely. Thus, there is no reason to consider that $E[Z\mid Y=y]=u(y)$ rather than $E[Z\mid Y=y]=\bar u(y)$.