# 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? – Conrado Costa Jul 21 '15 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. – shall.i.am Jul 21 '15 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/…) – shall.i.am Jul 21 '15 at 3:54

## 1 Answer

No. The following example is due to R. O. Davies, Separate approximate continuity implies measurability, Mathematical Proceedings of the Cambridge Philosophical Society, Vol. 73, Issue 03, May 1973, pp 461.

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.

• In "4. A general problem"? Now this sounds a bit fancy for me. But thank you this is very helpful! – shall.i.am Jul 29 '15 at 1:49