# Several conjectures on measurable functions defined on a product space

Let $$X$$ be equipped with a $$\sigma$$-field $$\mathcal{A}$$, and $$Y$$ be equipped with a $$\sigma$$-field $$\mathcal{B}$$, and $$f$$ be a function from $$X\times Y$$ to $$\mathbb{R}$$.

It can be shown that if $$f(x,y)$$ is $$\mathcal{A}\otimes\mathcal{B}\backslash\mathcal{B}(\mathbb{R})$$-measurable, then the map $$x\mapsto f(x,y)$$ is $$\mathcal{A}$$-measurable for each fixed $$y\in Y$$ and $$y\mapsto f(x,y)$$ is $$\mathcal{B}$$-measurable for each fixed $$x\in X$$, using an analog of $$\pi$$-$$\lambda$$ theorem for classes of functions.

My first question is whether its converse is true. That is, is it possible to show that if both $$x\mapsto f(x,y)$$ and $$y\mapsto f(x,y)$$ are measurable then $$f(x,y)$$ is product measurable?

My second question is whether the proposition is still true for more general measurable functions. $$f(x,y)$$ need not be a real valued function. Let it be a function from $$X\times Y$$ to a topological space, say $$Z$$, equipped with a $$\sigma$$-field $$\mathcal{C}$$. Then does the $$\mathcal{A}\otimes\mathcal{B}\backslash\mathcal{C}$$-measurability of $$f(x,y)$$ imply the $$\mathcal{A}\backslash\mathcal{C}$$-measurability of $$x\mapsto f(x,y)$$ and $$\mathcal{B}\backslash\mathcal{C}$$-measurability of $$y\mapsto f(x,y)$$? And converse?

Not true. Let $$B$$ be a subset of $$\mathbb R$$ which is not a Borel set and $$f(x,y)=1$$ if $$x=y$$ and $$x \in B$$, $$0$$ otherwise. You can easily check that $$x\to f(x,y)$$ is measurable for each $$y$$ and $$y\to f(x,y)$$ is measurable for each $$x$$ but $$f$$ is not Borel measurable [ Let me know if you need some details].
• Hint for seeing that $f$ is not measurable: Consider $x \mapsto f(x,x)$, which would be measurable if $f$ was. Commented Sep 24, 2019 at 10:30
• Thank you! I see it now. And what about the second question? What if $f$ is not a real valued function? Does the product measurability still imply the measurability of $x\mapsto f(x,y)$ and $y\mapsto f(x,y)$? Commented Sep 24, 2019 at 11:03
The answer to your second question is yes. For fixed $$y \in Y$$, define $$i_y : X \to X \times Y$$ by $$i_y(x) = (x,y)$$. Now verify, as an exercise, that $$i_y$$ is $$\mathcal{A} \backslash \mathcal{A} \otimes \mathcal{B}$$-measurable. Finally, note that the map $$x \mapsto f(x,y)$$ is simply $$f \circ i_y$$, and the composition of measurable maps is measurable.
• Thank you Nate! You helped me a lot today:) I can show that $i_y$ is $\mathcal{A}\backslash\mathcal{A}\otimes\mathcal{B}$-measurable by defining $f(x,y)$ as an indicator function of the set $\{(x,y)\in D\}$ for each $D$ in product $\sigma$-field. $f$ is a real valued and clearly product measurable function so I can use the theorem for real function to prove the measurability of $i_y$. Another straight way is, of course, the $\pi$-$\lambda$ theorem. Commented Sep 24, 2019 at 13:02
• @North: I don't think you even need $\pi$-$\lambda$ here. If $A \in \mathcal{A}$, $B \in \mathcal{B}$, then $i_y^{-1}(A \times B)$ is clearly measurable (it is either $A$ or $\emptyset$). Then, note that $\{C \in \mathcal{A} \otimes \mathcal{B} : i_y^{-1}(C) \in \mathcal{A}\}$ is a $\sigma$-algebra (just uses basic properties of unions and complements, it's unimportant what $i_y$ actually is in this step). Now you are done, because $\mathcal{A} \otimes \mathcal{B}$ is by definition the $\sigma$-algebra generated by all sets of the form $A \times B$. Commented Sep 24, 2019 at 13:20
• @North: The $\pi$-$\lambda$ argument is useful if you need to say something about the measurability of $i_y$ with respect to $y$, but here that is not needed. Commented Sep 24, 2019 at 13:21