Conditional Expectation of Two Random Variables - Integrals Consider two independent random variables $X$ and $Y$ along with a measurable function $f: \mathbb{R}^2 \to \mathbb{R}$.  To be concrete, consider $f(X, Y) = X + Y$.  Then I know, for fixed $\omega \in \Omega$, that
$$
E(X + Y \mid X)(\omega) = X(\omega) + E(Y),
$$
but I'm trying to understand this more throughly by writing out the integrals.  Here's what I've tried:
\begin{align}
E(X+Y \mid X)(\omega) & = \int_{\Omega} X(\omega) + Y(\omega^\prime) \, P(d\omega^\prime \mid X(\omega)) \\
& = \int_{\mathbb{R}} X(\omega) + y \, \Lambda_{Y \mid X}(dy \mid X(\omega)) \\
& = \int_{\mathbb{R}} X(\omega) + y \, \Lambda_Y(dy) \\
& = X(\omega) + E(Y).
\end{align}
The first line is the definition of conditional probability where $P$ is the regular conditional probability given $X$.  Here, I am thinking of $X(\omega)$ as a fixed value in $\mathbb{R}$, so I can use the change of variables theorem in the second line to pushforward $P^X$ to $\Lambda_{Y \mid X}$, the conditional distribution of $Y$ given $X$.  The third is due to independence and the final line is again because of thinking of $X(\omega)$ as some fixed real number.
I'm really unsure about


*

*The correct way to think of and write $X(\omega)$ in the integrals and

*Using the change of the variables theorem when it seems there are two functions being composed ($X$ and $Y$) and I pushforward only one measure (although this may be okay because $X(\omega)$ is just some fixed constant?)


To better understand my questions, consider my attempt at the general case:
\begin{align}
E(f(X, Y) \mid X)(\omega) & = \int_{\Omega} \left( f \circ (X, Y) \right)(\omega^\prime) \rvert_{X(\omega^\prime) = X(\omega)} \, P(d\omega^\prime \mid X(\omega)) \\
& = \int_{\mathbb{R}} f(X(\omega), y) \, \Lambda_{Y \mid X}(dy \mid X(\omega)) \\
& = \int_{\mathbb{R}} f(X(\omega), y) \, \Lambda_Y(dy).
\end{align}
This seems quite awkward to me.
 A: In term of conditional distribution given by Markov kernels i would have written conditional expectation $\omega$-wise in the following way.
Since $Y$ is a random variables in $\mathbb{R}$ we know that there exists a Markov kernel $(P_x)_{x\in \mathbb{R}}$ which is a conditional distribution of $Y$ given $X$. That is 
$$
P(X \in A, Y \in B )= (X,Y)(P)(A\times B) = \int_A P_x(B) dX(P)(x).
$$
Analougusly we have that there exists a Markov kernel $(\tilde{P}_x)$, which is a conditional distribution $f(X,Y)$ given $X$. By theorem 2.1.1. in Conditioning and Markov properties, we know that the relationsship between these conditional distributions is given by 
$$
\tilde{P}_x= (f\circ \iota_x)(P_x),
$$
where $\iota_x(y)=(x,y)$. If $E|f(X,Y)|<\infty$ then theorem 2.2.1 yields that 
$$
E(f(X,Y)|X) = \phi(X) \quad \quad P-a.s.
$$
where $\phi(x)=\int z \tilde{P}_x(z)$ for $X(P)$-almost all $x\in \mathbb{R}$. Now we simply realise that
$$
\phi(x)=\int z \, d (f\circ \iota_x)(P_x)(z) = \int f \circ \iota_x(y)\, dP_x(y) =\int f(x,y)\, dP_x(y)  
$$
for $X(P)$ almost all $x$. Thus for $P$-almost all $\omega\in \Omega$ we have that
\begin{align*}
E(f(X,Y)|X)(\omega) &= \phi(X(\omega)) \\
&=\int z \, d\tilde{P}_{X(\omega)}(z) \\
&= \int f(X(\omega),y) dP_{X(\omega)}(y) \\
&= \int f(X(\omega),y) dY(P)(y) \\
&= E(f(X(\omega),Y),
\end{align*}
where we used that $P_x=Y(P)$ for all $x\in\mathbb{R}$ since $X$ and $Y$ are independent (theorem 1.4.3).
