Let $(\Omega,\mathcal{F},P)$ be a complete probability space; $X, Y$ complete separable metric spaces. The Measurable Projection Theorem says that if the set $G\in\mathcal{F}\otimes\mathcal{B}(Y)$ then its projection to $\Omega$ is measurable, i.e $$\pi_\Omega(G)=\{\omega\in \Omega| \exists x\in X: (\omega, x)\in G\}\in \mathcal{F}.$$

Now let $K\in \mathcal{F}\otimes\mathcal{B}(X)\otimes\mathcal{B}(Y)$. Then can we conclude that the projection of $K$ to $\Omega\times X$ is in $\mathcal{F}\otimes\mathcal{B}(X)$? If not, are there further conditions we can add to $X$ so that the above property hold?


The measurable projection theorem can be formulated on base of measurable space $(\Omega,\mathcal F)$ (so leaving out $P$) by saying that $\pi_{\Omega}(G)$ will be universally measurable if $G\in\mathcal F\otimes\mathcal B(Y)$.

Working with measurable space $(\Omega',\mathcal F'):=(\Omega\times X,\mathcal F\otimes\mathcal B(X))$ and Polish space $Y$ we can conclude that the projection of $K$ will also be universally measurable if $K\in \mathcal{F}\otimes\mathcal{B}(X)\otimes\mathcal{B}(Y)$.

That means that the set will be measurable wrt to the completion of $\mathcal F'$ wrt any finite measure $\mu$ on $(\Omega',\mathcal F')$.

  • $\begingroup$ Actually, I am trying to show that there exists a measurable selection of some correspondence $F:\Omega\to 2^X\backslash\{\emptyset\}$. I want to use the Aumann Measurable Selection Theorem, which says that if $Graph(F)$ is $\mathcal{F}\otimes\mathcal{B}(X)$ then there exists a measurable selection of $F$. Now, if $Graph(F)$ is just universally measurable then can we still deduce that there is a measurable selection of $F$? $\endgroup$ – Huy Phạm May 17 '18 at 15:49

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