Measurability of function in product space.

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?

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.

• 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 $\Bbb R$. Mar 11, 2021 at 23:35