# Simplify $\int_{\Omega} \int_{\Omega} 1_{A}(\omega) g(\tilde{\omega}, \Pi(\omega)) d\mathbb{P}(\tilde{\omega}) d\mathbb{P}(\omega)$

Let $$(\Omega, \mathcal{G}, \mathbb{P})$$ be a (complete) probability space and $$D$$ be a compact topological space, equipped with its canonical Borel $$\sigma$$-algebra $$\mathcal{B}(D)$$.

Furthermore, let $$\Pi : \Omega \rightarrow D$$ be a $$\mathcal{G}/\mathcal{B}(D)$$-measurable random variable and $$g : \Omega \times D \to \mathbb{R}$$ be a bounded and $$(\mathcal{G} \otimes \mathcal{B}(D))/\mathcal{B}(\mathbb{R})$$-measurable function, such that, for each fixed $$\pi \in D$$, the random variable $$g(\cdot, \pi)$$ is independent of $$\Pi$$.

Now let $$A \in \sigma(\Pi)$$. I want to show, that

$$\int_{\Omega} \int_{\Omega} 1_{A}(\omega) g(\tilde{\omega}, \Pi(\omega)) d\mathbb{P}(\tilde{\omega}) d\mathbb{P}(\omega) = \int_{\Omega} 1_{A}(\omega) g(\omega, \Pi(\omega)) d\mathbb{P}(\omega).$$

I am convinced, that under our assumptions, this should be the case, but I am completely stuck. I would be extremely grateful for any advice!