I have an integration that looks like:

\begin{align}\label{eq1}\tag{1} \int_{f \in F} \left[\int_{x \in \mathbb{R}} \chi_{\{x \in A\}} \mathrm{d} \gamma(x|f)\right] \mathrm{d} \mu(f), \end{align} where $F$ is some interval, $\gamma(\cdot | f)$ is a distinct probability measure supported on $\mathbb{R}$ for all $f \in F$, $\mu$ is a probability measure on $F$, and $\chi$ is the indicator. I would like to change the order of integration, but this doesn't look feasible due to the $\color{red}{\text{red}}$ term \begin{align} \int_{x \in \mathbb{R}} \left[ \int_{f \in F} \chi_{\{x \in A\}} \mathrm{d} \mu(f) \right] \color{red}{\mathrm{d} \gamma(x|f)}. \end{align} Hence, I write the integration \eqref{eq1} by using an additional index: \begin{align} & \int_{f \in F} \left[\color{blue}{\int_{(f',x)\in F \times \mathbb{R}} \chi_{\{x \in A\}}\chi_{\{ f = f' \}} \mathrm{d} \gamma(x|f)}\right] \mathrm{d} \mu(f)\label{eq2}\tag{2} \\ \iff &\int_{(f',x)\in F \times \mathbb{R}} \left[\int_{f \in F} \chi_{\{x \in A\}}\chi_{\{ f = f' \}} \mathrm{d} \mu(f) \right] \mathrm{d} \gamma(x|f')\label{eq3}\tag{3} \end{align} I was wondering if equation~\eqref{eq3} is correct, which depends on the correctness of the $\color{blue}{\text{blue}}$ part in~\eqref{eq2}.

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    $\begingroup$ I think you want to know in what circumstances the disintegration theorem works, that is, when you can decompose a measure in two measures where one is conditioned to the other $\endgroup$
    – Masacroso
    Nov 20, 2021 at 0:09


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