Conditional Expectation of Composite Function Preliminaries
Let $(\Omega, \mathcal{G}, \mathbb{P})$ be a complete probability space.
Let $D$ be a complete, separable, metrizable topological space with Borel $\sigma$-algebra $\mathcal{B}(D)$ (such as $D = \mathbb{R}^q$ with $\sigma$-algebra $\mathcal{B}(D) = \mathcal{B}(\mathbb{R}^d)$).
Let $\mathbb{R}$ be equipped with its canonical Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$.
Let $g: \Omega \times D \rightarrow \mathbb{R}$ be a bounded $(\mathcal{G} \otimes \mathcal{B}(D) ) / \mathcal{B}(\mathbb{R})$-measurable function.
Let $\Pi: \Omega \rightarrow D$ be a $\mathcal{G}/\mathcal{B}(D)$-measurable random variable.
Let $H : \Omega \rightarrow \mathbb{R}$ be a $\mathcal{G}/\mathcal{B}(\mathbb{R})$-measurable random variable, defined by $$ H(\omega) := g(\omega, \Pi(\omega)).$$ Note, that, since $g$ is bounded, we have $H \in \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P})$.
Let $j: D \rightarrow \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P}) $ be defined by $$ j(\pi)(\omega) := g(\omega, \pi)  $$
For all $\pi \in D$, let $j(\pi)$ be independent of $\Pi$.
Question
I am interested in the conditional expectation $$\mathbb{E}[H \mid \Pi] :\Omega \rightarrow \mathbb{R}$$ of $H$ with respect to $\Pi$. More specifically, I suspect that (a $\mathbb{P}$-unique version of) this condititional expectation is given by
$$ \mathbb{E}[H \mid \Pi] (\omega) = \mathbb{E}[j(\Pi(\omega))], \quad (\dagger)  $$
whereby $\mathbb{E}[j(\Pi(\omega))]$ can of course also be written as
$$\mathbb{E}[j(\Pi(\omega))] = \int_{\Omega} j(\Pi(\omega))(\tilde{\omega}) d\mathbb{P}(\tilde{\omega}) . $$
How can I prove, that $(\dagger)$ is the case? I have tried, tracking the definition of conditional expectation and using Fubini, but with little success so far.
Thanks for any advice! 
 A: $\def\om{\omega}$
$\def\Om{\Omega}$
$\def\bR{\mathbb{R}}$
$\def\si{\sigma}$
$\def\cB{\mathcal{B}}$
$\def\cF{\mathcal{F}}$
My original answer (below) contains an error, since $\Phi$ is not necessarily measurable. In fact, that original proof sketch does not use the fact that $g$ is a measurable stochastic process, only that it is a stochastic process. Right now, I cannot see a way to fix this without adding additional assumptions on $g$. In fact, I do not believe it is true without additional assumptions.
Let $\Om=[0,1]$ with $\cF$ the Lebesgue $\si$-algebra and $P$ Lebesgue measure. Let $D=[0,1]$. Let $G(\om,t)=1_{\{\om=t\}}$ and $\Pi(\om)=\om$. For fixed $t\in D$, we have $G(t)=0$ a.s., so the random variable $G(t)$ is independent of everything, and $h(t):=E[G(t)]=0$ for all $t$. On the other hand, $G(\Pi)=1$ a.s. So $G(\Pi)$ is independent of everything, which gives
$$
E[G(\Pi)\mid\Pi]=E[G(\Pi)]=1.
$$

Original (flawed) answer:
First, let me point out a small confusion in notation. Under normal usage,
$$
E[j(\Pi)] = \int j(\Pi(\omega))(\omega)\,dP(\omega),
$$
without any tildes, which is of course not what you want. One way of carefully notating what you intend is to say that $E[H\mid\Pi]=h(\Pi)$, where $h(\pi)=E[j(\pi)]$.
This is indeed the correct answer. Heuristically, $g$ and $\Pi$ are independent, so in the conditional expectation, you can treat $\Pi$ like a constant and just use the ordinary expectation. For a rigorous formulation of this, you can do the following.
First, we may regard $g$ as a function from $\Omega$ to $\mathbb{R}^D$, the set of functions from $D$ to $\mathbb{R}$, with $g(\omega)(\pi)=g(\pi,\omega)$. With this identification, it follows that $g$ is $\mathcal{G}/\mathcal{B}(\mathbb{R})^D$-measurable. Here $\mathcal{B}(\mathbb{R})^D=\bigotimes_{\pi\in D}\mathcal{B}(\mathbb{R})$ is the product $\sigma$-algebra.
Next, show that since $j(\pi)$ and $\Pi$ are independent for all $\pi\in D$, it follows that $g$ and $\Pi$ are independent. (The $\pi$-$\lambda$ theorem should do the trick here.)
Now define $\Phi:\mathbb{R}^D\times D\to\mathbb{R}$ by $\Phi(f,\pi)=f(\pi)$, so that $H=\Phi(g,\Pi)$, and verify that $\Phi$ is $(\mathcal{B}(\mathbb{R})^D \otimes \mathcal{B}(D))/\mathcal{B}(\mathbb{R})$-measurable.
Finally, use the following.

Theorem. Let $(\Omega,\mathcal{F},P)$ be a probability space and $(S,\mathcal{S})$ a measurable space. Let $X$ be an $S$-valued random variable, $\mathcal{G}\subset\mathcal{F}$ a $\sigma$-algebra, and suppose $X$ and $\mathcal{G}$ are independent. Let $(T,\mathcal{T})$ be a measurable space and $Y$ a $T$-valued random variable. Let $f:S\times T\to\mathbb{R}$ be $(\mathcal{S}\otimes\mathcal{T},\mathcal{B}(\mathbb{R}))$-measurable with $E|f(X,Y)|<\infty$. If $Y$ is $\mathcal{G}/\mathcal{T}$-measurable, then
    $$
  E[f(X,Y) \mid \mathcal{G}] = \int_S f(x,Y)\,\mu(dx)
    \quad\text{a.s.},
  $$
  where $\mu$ is the distribution of $X$.

This theorem is a special case of Theorem 6.66 in these notes: http://math.swansonsite.com/19s6245notes.pdf.
