# Continuous and measurable in each variable $\implies$ product measurable?

Consider a metric space $A$ with a metric $d$, and consider the measurable space $(A,\mathcal{B}(A))$ with the Borel $\sigma$-algebra generated by $d$-open sets. Let $(\Omega,\mathcal{F})$ be a measurable space. Consider the product $\sigma$-algebra $\mathcal{B}(A)\otimes\mathcal{F}:=\sigma(\{B\times F\mid B\in\mathcal{B}(A),\,F\in\mathcal{F}\})$

Suppose a function $f:A\times\Omega\to \mathbb{R}$ satisfies

1. $f(\cdot,\omega):A\ni a\mapsto f(a,\omega)$ is a continuous function for each $\omega\in\Omega$.
2. $f(a,\cdot):\Omega\ni \omega\mapsto f(a,\omega)$ is a $\mathcal{B}({A})$ measurable.

Question: Can we say that $f$ is $\mathcal{B}(A)\otimes\mathcal{F}$-measurable?

My guess is we can. I am aware that there are similar questions where $A:=\mathbb{R}$, e.g.,

and I think what I should do is to something in line with constructing a sequence $f_n$, step function in $a\in A$, and whose pre-image is a rectangle, then use the continuity in $a$. But I am not really sure how to write down explicitly.

• Is your space $A$ separable? Commented Jul 21, 2015 at 3:38
• @ConradoCosta Oh, I see, because if A is separable I can use books.google.at/… ? Thank you, I noticed that after asking this. Commented Jul 21, 2015 at 3:54
• Then I wonder 1. if assuming $A$ is not separable there is a counter example and 2. if assuming $A:=\mathbb{R}^n$ or $A$: sphere there is more concrete (in a way that is a natural generalisation of $A:=\mathbb{R}$) way of showing this. I guess the rule is to post a new question? (meta.math.stackexchange.com/questions/3561/…) Commented Jul 21, 2015 at 3:54

Let $\kappa$ be a real valued measurable cardinal and $m: \mathcal{P}(\kappa) \to [0, 1]$ be a $\kappa$-additive diffused probability measure. Put discrete metric on $\kappa$. Then the characteristic function of $\{(\alpha, \beta) : \alpha < \beta < \kappa\}$ is continuous in each coordinate but not $m \otimes m$-measurable. This easily follows from Fubini's theorem.