Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space, $\mathcal G$ a sub-sigma-field, and $\mathbb P^{\mathcal G}$ denote the conditional probability.
Is it correct that $$\int_A \mathbb P^{\mathcal G}(B)d\mathbb P = \int_B \mathbb P^{\mathcal G}(A)d\mathbb P$$ for all $A,B \in \mathcal F$?
This seems unintuitive to me, but the following calculation seems to confirm it: $$\int_A \mathbb P^{\mathcal G}(B)d \mathbb P = \int \mathbb E\big[1_A\mathbb P^{\mathcal G}(B) \mid \mathcal G \big]d\mathbb P = \int \mathbb P^{\mathcal G}(B)\mathbb P^{\mathcal G}(A) d\mathbb P,$$ and similarly for $\int_B \mathbb P^{\mathcal G}(A)d\mathbb P$.
Did I make a silly mistake, or, if not, could someone give some intuition for the equality?
This is true and more general form is $E[X (EY|\mathcal G)]=E[Y E(X|\mathcal G)]$ both sides being equal to $E[(E(Y|\mathcal G) (E(X|\mathcal G)]$. As you did you only have to condition on $\mathcal G$ to get this.