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If $X: \Omega_1 \times \Omega_2 \to \mathbb{R}$ is measurable then each of the slices $X_{\omega_1} : \Omega_2 \to \mathbb{R}$, $X_{\omega_1} : \Omega_2 \to \mathbb{R}$ defined by $X_{\omega_1}(\omega_2) = X(\omega_1,\omega_2)$ etc. are measurable.

I guess that the converse need not be true. Is it true that if $X_{\omega_1}$ is $\textit{continuous}$ for all $\omega_1$ while $X_{\omega_2}$ is $\textit{measurable}$ for all $\omega_2$ (e.g. for $\Omega_1 = \Omega_2 = \mathbb{R}^d$) then $X$ is measurable?

More generally, if $X : \Omega_1 \times \Omega_2 \cdots \Omega_n \to \mathbb{R}$ and $X_{\omega_1},X_{\omega_2},\cdots,X_{\omega_{n-1}}$ are continuous and $X_{\omega_n}$ is measurable can we say that $X$ is measurable?

Thanks, Phanindra

Alex Ravsky
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jpv
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  • It seems that the claim “If $X: \Omega_1 \times \Omega_2 \to \mathbb{R}$ is measurable then each of the slices ... etc. are measurable” is not true. For instance, let $\Omega_1=\Omega_2=\Bbb R$, $A$ be a non-measuruble subset of $\Bbb R$ and $X=\chi({0}\times A)$ be a characteristic function of the set ${0}\times A.$ Then $X$ is measurable, but $X_0(\omega_2)$ is not. So it seems that a weaker property $\mu({\omega_1:X_{\omega_1}(\omega_2)$ is not measurable$ })=0$ etc. is satisfied. – Alex Ravsky Dec 26 '14 at 15:16
  • It is not clear to me why $X$ is measurable in your example as $X^{-1}({1}) = {0}\times A$ which seems to be non-measurable in $\mathbb{R}^2$. The proof of this claim can be found in Loeve (Page 130 - 140) for example. – jpv Dec 26 '14 at 16:15
  • @AlexRavsky: This depends on which kind of measurability you require. If you use the Borel sigma algebra (which is the product sigma algebra of the Borel sigma algebras) then the claim is true (and your counterexample is not valid). If we use Lebesgue measurability, then the claim is false and your counterexample is valid. The other claim with continuity/measurability is true, but I don't have the time to write the proof right now. – PhoemueX Dec 26 '14 at 17:31
  • @PhoemueX It seems the following. It depends, which sets we consider as measurable. The sets $A$ and ${0}\times A$ are not Borel. Nevertheless, if the set $A$ is bounded then for each $\varepsilon>0$ there exists a finite family $\mathcal F$ of squares with total area less than $\varepsilon$ such that $\bigcup\mathcal F\supset {0}\times A$. So Jordan measure of the set ${0}\times A$ is zero. – Alex Ravsky Dec 26 '14 at 19:06
  • PhoemueX: It would be great if you could provide a proof. The claim can be translated into slices of sets, i.e. if slices in one direction are measurable and continuous in the other then the set is measurable. Ofcourse, measurability here is in the Borel sense. I did not have much luck in pursuing this approach. – jpv Dec 27 '14 at 13:43
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    I found this: http://math.stackexchange.com/questions/661087/a-function-which-is-continuous-in-one-variable-and-measurable-in-other-is-jointl – jpv Dec 27 '14 at 17:33
  • @jpv: This question includes the question you reference, but it also asks about a generalization. Therefore it is not a duplicate. (I realize you did not say that it is.) – Rory Daulton Jan 03 '15 at 22:18

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