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I was wondering. Suppose that $(X,\Sigma_X)$ and $(Y,\Sigma_Y)$ are measure spaces and $f:X\to \Bbb R$ and $g:Y\to \Bbb R$ are measurable. Does this imply that $h:X\times Y\to \Bbb R $ defined $h(x,y)=\min(f(x),g(y)) $ is measurable? Of course $X\times Y$ is equipped with product $\sigma$-algebra $\Sigma_X\otimes\Sigma_Y$ and $\Bbb R$ with the Borel $\sigma$-algebra.

I think to show this we need the measurability of the set $\{(x,y):f(x)\leq g(y)\}$, am I correct?

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I don't quite see where you would need that thing you said. For all $t\in\Bbb R$, $$\{(x,y)\in X\times Y\,:\, h(x,y)\ge t\}=\{x\in X\,:\, f(x)\ge t\} \times\{y\in Y\,:\, g(y)\ge t\}$$

and therefore $h$ is measurable.

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  • $\begingroup$ Because I was thinking that you can write $\{ f \wedge g\in B\}=\left(\{(x,y):f(x)\in B\}\cap \{f\leq g \}\right) \cup \left(\{(x,y):g(x)\in B\}\cap \{f > g \} \right)$ where $B$ is a Borel subset of $\Bbb R$. $\endgroup$
    – UserA
    Mar 11, 2021 at 23:35

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