# Integration against a measure that has an integral form

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.

• If $\mu$ abd $\rho_y$ are probability measures you find some results of this sort as applications of the Fubini theorem. You can extend these results easily to the situation that $\mu$ is a $\sigma$-finite measure and $\rho$ a uniformly $\sigma$-finite kernel. – Dieter Kadelka Jun 18 '20 at 16:52
• for more on this theme you might want to lookup "desintegration of measures" – Abdelmalek Abdesselam Jun 18 '20 at 22:31

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 $$<\infty$$.