Let $(X,\mathcal{X})$, $(Y,\mathcal{Y})$ and $(Z,\mathcal{Z})$ be measurable spaces. Let $(X \times Y, \mathcal{X} \otimes \mathcal{Y})$ be the product of the two spaces, where $\mathcal{X} \otimes \mathcal{Y} := \sigma(\mathcal{X} \times \mathcal{Y})$.
Let $f:X \times Y \to Z$ be a pointwise measurable function (my own term) in the sense that
- $\forall x \in X: f(x, \cdot):Y \to Z$ is measurable
- $\forall y \in Y: f(\cdot, y):X \to Z$ is measurable
Sufficient conditions for separately measurable functions being jointly measurable. claims that $f:X \times Y \to Z$ is not necessarily measurable. Why? Is there a simple counter-example?
Yes, there is. Indeed consider the Borel measure on $[0,1]$ and let $A$ be your favorite non measurable subset of the unit interval. Then define a function $f$ on $[0,1]^2$ as follows: $$f(x,y) = \begin{cases} 0 & \text{if } x \notin A \text{ or if } x \neq y , \\ 1 & \text{otherwise}. \end{cases} $$
One can easily check that $f$ is separately measurable, but if it were measurable on the product space, its composition with the function $g \colon [0,1] \to [0,1]^2$ defined via $g(x) = (x,x)$ would also be measurable (composition of Borel measurable functions is measurable). Clearly this is not the case since $$f \circ g = \chi_{A},$$ which by assumption is not measurable.