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Let $(X,\mathscr{X},P)$ be a probability space, and $f:X\mapsto\mathbb{R}^m$ be an (element-wise) integrable function. Let $\Gamma:X\twoheadrightarrow \mathbb{R}^n$ be a measurable correspondence, and $g:\mathbb{R}^{n}\mapsto \mathbb{R}^m$ be a measurable function.

Suppose, for any $x\in X$, $f(x)\in co\{g(u):u\in\Gamma(x)\}$ where $co$ stands for the convex hull. Let $\mathscr{B}$ be the Borel Sets in $\mathbb{R}^n$. My question is, can we find a probability kernel $\mu:X\times \mathscr{B}\mapsto [0,1]$ such that for almost every $x\in X$, $$f(x)=\int g(u)\mu(x,du)$$ and $\mu(x,\Gamma(x))=1$.

My thoughts: Since $f(x)\in co\{g(u):u\in\Gamma(x)\}$, it's easy to find a mapping $\lambda:X\times \mathscr{B}\mapsto[0,1]$ such that $\lambda(x,\cdot)$ is a measure on $(\mathbb{R}^n,\mathscr{B})$, $f(x)=\int g(u)\lambda(x,du)$ and $\lambda(x,\Gamma(x))=1$. However, how to ensure $\forall B\in \mathscr{B}$, $\lambda(\cdot,B)$ is a measurable function? If it helps, $\Gamma(x)$ may be a closed convex set in $\mathbb{R}^n$ for any $x$. Thanks a lot!

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    Looks like you need a measurable selection theorem. Have you looked at [Bertsekas and Shreve](http://web.mit.edu/dimitrib/www/soc.html)? – 2017-01-10
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    @TheoreticalEconomist After reading your reference, I think I can prove the result if $g$ is a bounded continuous function. Thanks. – 2017-01-11

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