If $A\times B$ is an element of a product $\sigma$-algebra $\mathcal A\times\mathcal B$ then do we have $A\in\mathcal A$ and $B\in\mathcal B$?

Question

If $\mathcal A$ is a $\sigma$-algebra on set $X$ and $\mathcal B$ is a $\sigma$-algebra on set $Y$ then it is well known that $\mathcal A\times\mathcal B$ is the notation for a $\sigma$-algebra on $X\times Y$ that is generated by sets $A\times B$ with $A\in\mathcal A$ and $B\in\mathcal B$.

So evidently we have: $$A\in\mathcal A\text{ and }B\in\mathcal B\implies A\times B\in\mathcal A\times\mathcal B\tag1$$

Now my question:

Is the converse of $(1)$ also true?

I have always believed it is without bothering, but when I tried to find a proof for this "obvious" fact I regretfully failed.

Answer 1

Both implications are true and they are proved in the beginning of Fubini's Theorem. For example you look at the discussion of 'sections' of measurable sets in Rudin's RCA.

Answer 2

Fortunately I found a proof for it myself after all.

---

Let $p_{1}:X\times Y\to X$ and $p_{2}:X\times Y\to Y$ be prescribed by $\left(x,y\right)\mapsto x$ and $\left(x,y\right)\mapsto y$ respectively.

Characteristic for product space $\left(X\times Y,\mathcal{A}\times\mathcal{B}\right)$ is that for any measurable space $\left(Z,\mathcal{C}\right)$ and every function $f:Z\to X\times Y$ we have: $$f:Z\to X\times Y\text{ is measurable if and only if }p_{1}\circ f\text{ and }p_{2}\circ f\text{ are both measurable}\tag1$$

The case is trivial if $B=\varnothing$ so let it be that $B\neq\varnothing$.

Now for some $y\in B$ let $u_{y}:X\to X\times Y$ be prescribed by $x\mapsto\left(x,y\right)$.

Then based on $\left(1\right)$ we find easily that $u_{y}$ is measurable.

Then consequently $A=u_{y}^{-1}\left(A\times B\right)\in\mathcal{A}$.

Similarly it can be proved that $B\in\mathcal{B}$.