How can I prove a function on product space lebesgue measurable?

Question

Let $X$ and $Y$ be two measurable spaces. I want to construct a function $f : X\times Y \to \mathbb{R}$ such that $f(x,\cdot) : Y \to \mathbb{R}$ is measurable for all $x \in X$ and $f(·,y) : X \to \mathbb{R}$ is measurable for all $y \in Y$, yet $f$ is not measurable in the product space. How can I construct such a function?

Answer 1

I will provide an answer if we assume the continuum hypothesis.

Consider $X = Y = \alpha,$ where $\alpha$ is the first ordinal. For every $u \in \alpha,$ declare the sets $\{u\}$ to be measurable, and then consider the sigma algebra generated by the family $(\{u\})_{u \in \alpha}$ (the minimal sigma algebra that makes every $\{u\}$ measurable). Define the set $A = \{(x,y) \in X \times Y \mid x \leq y\}$ and with it the function $f = \mathbf{1}_A,$ which is bounded. The two sets $\{f(x, \cdot) = 0\} = \{z \in \alpha \mid z \leq x, z \neq x\}$ and $\{f(\cdot, y) = 1\} = \{z \in \alpha \mid z \leq y, z \neq y\}$ are countable and hence, measurable; this claim is a bit tricky but recall that $\alpha$ was chosen to be the first ordinal, and since $x$ and $y$ are members of $\alpha$, they are different from $\alpha$, so they are countable themselves, and since $\leq$ means "membership" in $\alpha$, the claim follows. This makes $f(x, \cdot)$ and $f(\cdot, y)$ measurable.

So far, I have not used the continuum hyphotesis. Using it, one guarantee the existence of a bijection $\varphi:\alpha \to [0,1].$ Put in $\alpha$ the sigma algebra $\{\varphi^{-1}(A) \mid A \text{ is a Borel set in } [0,1]\}$ (since $\varphi$ is a bijection, this sigma algebra makes every set $\{u\}$, for $u \in \alpha$, measurable, hence there is consistency with the previus paragraph). Consider the measure $\mu$ in $\alpha$ as the image measure of Lebesgue by $\varphi$. Notice now that $\displaystyle \int_\alpha d\mu(x) \int_\alpha d\mu(y) f(x, y) = 1 \neq 0 = \int_\alpha d\mu(y) \int_\alpha d\mu(x) f(x,y),$ thus, $f$ cannot be measurable (otherwise, it would contradict Fubini's theorem).