Integration against a measure that has an integral form

Question

Suppose that $(X, \mathcal{X})$ is a measurable space and $(Y,\mathcal{Y}, \mu)$ is a measure space (in my particular application, they are Polish spaces endowed with their Borel $\sigma$-algebra). Suppose we have a collection of measures $(\rho_y)_{y\in Y}$ defined on $(X,\mathcal{X})$ indexed by $Y$, such that $$ \rho : \Gamma \in \mathcal{X} \mapsto \int_Y \rho_y(\Gamma) \, d\mu $$ makes sense and defines a measure on $X$. I am looking for conditions under which we can have an ''interchange'' theorem like this : $$ \int_X f \, d\rho = \int_X f \, d\left( \int_Y \rho_y \, \mu(dy) \right) = \int_Y \int_X f \, d\rho_y \, \mu(dy). $$ I don't find references for this particular application. I tried to rely on interchange theorems based on weak convergence of measures, by approximating $\rho$ with an increasing sequence of measures that converges set-wise but I don't get the interchange. Thanks in advance for your suggestions.

Answer 1

The condition $$\int_X f\,d\rho=\int_Y\int_X f\,d\rho_y\,\mu(dy) \tag{1}$$ holds, by the definition of $\rho$, in the case when $f$ is the indicator of any $\Gamma\in\mathcal X$. By the linearity in $f$, (1) continues to hold for all nonnegative simple $f$. Next, by the monotone convergence theorem, (1) still holds for all nonnegative measurable $f$. Thus, (1) holds for all measurable $f$ such that either for $f_+:=\max(0,f)$ or for $f_-:=\max(0,-f)$ in place of $f$ the right-hand side or the left-hand side of (1) is $<\infty$.