Is the product $\sigma$-algebra the tensor product over $\mathbb{Z}$? This question is inspired by Folland’s notation for the product of $\sigma$-algebras using the $\otimes$ symbol. More precisely, given a family $(\mathcal{M}_\alpha)_{\alpha\in I}$ of $\sigma$-algebras on some sets $(X_\alpha)_{\alpha\in I}$, the product $\sigma$-algebra $\bigotimes_{\alpha\in I} \mathcal{M}_\alpha$ on $\prod_{\alpha\in I} X_\alpha$ is the smallest $\sigma$-algebra generated by the measurable cylinders: $\{\pi_\alpha^{-1}(E_\alpha) : E_\alpha \in \mathcal{M}_\alpha\}$, where $\pi_\alpha : \prod_{\beta\in I} X_\beta \to X_\alpha$ is the canonical projection.
One can show that each $\mathcal{M}_\alpha$ is a commutative Boolean ring with identity, under symmetric difference (addition) and intersection (multiplication). Therefore each $\mathcal{M}_\alpha$ is a $\mathbb{Z}$-algebra in the abstract-algebraic sense. This means we can take the $\mathbb{Z}$-algebra tensor product of the $\mathcal{M}_\alpha$. My question is thus: is the product $\sigma$-algebra isomorphic (as a $\mathbb{Z}$-algebra, or at least as a module) to this tensor product?
To show this, one would have to generate a $\mathbb{Z}$-linear map $g : \prod_{\alpha\in I} \mathcal{M}_\alpha \to \bigotimes_{\alpha\in I} \mathcal{M}_\alpha$ so that for any $\mathbb{Z}$-module N and any $\mathbb{Z}$-linear  map $f : \prod_{\alpha\in I} \mathcal{M}_\alpha \to N$, there exists a unique module homomorphism $f’ : \bigotimes_{\alpha\in I} \mathcal{M}_\alpha \to N$ such that $f = f’ \circ g$.
I think $g$ ought to be the map $g((E_\alpha)_{\alpha\in I}) = \prod_{\alpha\in I} E_\alpha$, at least when $I$ is countable. But I am not very good at tensor products, so I’m not sure how to define the homomorphism $f’$. Does this approach work? If so, how do we define $f’$?
 A: The answer to the question in the title is "no", but you're still morally correct.
First, note a $\sigma$-algebra $\mathcal{A}$ is a boolean algebra with the bonus property that it's closed under countable meets, and joins respect these. That is
$$a \vee \left ( \bigwedge b_n \right )= \bigwedge \left ( a \vee b_n \right )$$
Notice this does not automatically follow from finite distributivity. This definition is equivalent to working with infinite joins (as is traditional) but when viewing a boolean algebra as a ring, we usually take products to be meets, so this definition is slightly more convenient. In particular, $\mathcal{A}$ becomes a ring when we define $a+b = \left ( a \wedge b^c \right ) \vee \left ( b \wedge a^c \right )$ and $ab = a \wedge b$. Then our extra operation tells us that we have countable products, $\prod a_n$, in our ring. Call such a ring a $\sigma$-ring for the purposes of this answer (a $\sigma$-ring should also satisfy even more bonus axioms telling us how infinite products behave. See Kriz and Pultr's Categorical Geometry and Integration Without Points for more). While we're at it, everywhere I say "ring" in this answer I'll mean "boolean ring".
Conversely, given a $\sigma$-ring, we get a $\sigma$-algebra in the usual way. We define $a \land b = ab$, $a^c = 1+a$, and $a \lor b = (a^c \land b^c)^c$. This is a $\sigma$-algebra since our infinite products tell us how to compute infinite meets $\bigwedge a_n = \prod a_n$.
These constructions are mutually inverse, so we see the category of $\sigma$-algebras is equivalent to the category of $\sigma$-rings. In particular, their coproducts agree.

Now there's an obvious functor $U$ from $\sigma$-rings to rings which forgets the infinite product. If we take the ordinary tensor product of two $\sigma$-rings $\mathcal{A}$ and $\mathcal{B}$, it amounts to taking the coproduct $U \mathcal{A} \otimes U \mathcal{B}$ in the category of rings.
Since general rings don't have infinite products, it's easy to see that this will not play nicely with infinite products in $\mathcal{A}$ and $\mathcal{B}$. Indeed if you recall the definition, $\prod (a_n \otimes b) \neq (\prod a_n) \otimes b$ in the tensor product (since that's not a relation we added).
However, in some sense the incompatibility with infinite products is the only issue!
Allowing for some abstract nonsense, we're guaranteed a functor $F$ from rings to $\sigma$-rings which is left adjoint to $U$ (since these are algebraic theories with arity at most $\omega$). This is the free $\sigma$-ring from a ring.
It's not hard to see that the coproduct $\mathcal{A} \otimes^\sigma \mathcal{B}$ (in $\sigma$-ring) should be $F (U \mathcal{A} \otimes U \mathcal{B} )$ subject to the ~bonus relations~
$$
\prod (a_n \otimes b) = \left ( \prod a_n \right ) \otimes b
$$
$$
\prod (a \otimes b_n) = a \otimes \left ( \prod b_n \right )
$$
which is the "completion" of the tensor product that people were mentioning in the comments.
Again, since $\sigma$-ring and $\sigma$-algebra are equivalent, this construction also gives the coproduct of $\mathcal{A}$ and $\mathcal{B}$ viewed as $\sigma$-algebras.

So we're left with an obvious question. Call a $\sigma$-algebra $\mathcal{A}$ concrete if it arises as an algebra of measurable subsets of a set $X$. That is, if $(X, \mathcal{A})$ is a measurable space. Notice that unlike boolean algebras, not every $\sigma$-algebra is of this form! Though it is the case that every $\sigma$-algebra is the quotient of a concrete $\sigma$-algebra by a $\sigma$-ideal (which you might think of as nullsets). See this blog post for more.
For two concrete $\sigma$-algebras $\mathcal{A}$ on $X$ and $\mathcal{B}$ on $Y$, is it the case that $\mathcal{A} \otimes^\sigma \mathcal{B}$ (in the sense defined above) agrees with $\mathcal{A} \otimes \mathcal{B}$ (as defined in Folland)?
Since $\mathcal{A} \otimes^\sigma \mathcal{B}$ is the coproduct, there's a natural map $\varphi : \mathcal{A} \otimes^\sigma \mathcal{B} \to \mathcal{A} \otimes \mathcal{B}$ sending $\varphi(a \otimes b) = a \times b$.
This is quickly seen to be surjective, and it suffices to check injectivity...
Unfortunately this seems to be hard, and I suspect it's false in general (though I'm also struggling to come up with an explicit counterexample).
For intuition, we know that $(X \times Y, \mathcal{A} \otimes \mathcal{B})$ is a product in the category of measurable spaces. Since a measurable map $(X, \mathcal{A}) \to (Z, \mathcal{C})$ is automatically a homomorphism $\mathcal{C} \to \mathcal{A}$, this tells us $\mathcal{A} \otimes \mathcal{B}$ is the coproduct of $\mathcal{A}$ and $\mathcal{B}$ provided we restrict attention to homomorphisms coming from measurable functions. Unfortunately, not every homomorphism is of this form, even for reasonable measure spaces (see here). Coupled with the case that, in pointless topology the corresponding theorem fails (there are topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$ where the product topology on $X \times Y$ is not the coproduct $\tau_X \otimes \tau_Y$ as frames), I suspect there's a counterexample somewhere here.
So, (probably) the answer to the modified question "is the tensor product of two (concrete) $\sigma$-algebras (in the sense of Folland) the tensor product in the category of $\sigma$-rings" is still no.

But what about "nice" spaces?
Say that $X$ and $Y$ are both second countable and hausdorff and $\mathcal{B}(X)$ and $\mathcal{B}(Y)$ are their borel $\sigma$-algebras (this can almost certainly be relaxed).
Then it's not hard to show that every homomorphism $\mathcal{B}(X) \to \{\emptyset, \{\star\} \}$ comes from a map $\{ \star \} \to X$. That is, a point in $X$.
Second countability says we can restrict attention to a countable set of gemerators $U_n$, and if $\varphi : \mathcal{B}(X) \to \{ \emptyset, \{ \star \} \}$ is a homomorphism you can check that
$$
\bigcap \big \{ U_n \mid \varphi U_n = \{ \star \} \big \}
$$
is a singleton, which corresponds to a point in $X$.
But now we're in a great place! By the universal property of the coproduct, maps $\mathcal{B}(X) \otimes^\sigma \mathcal{B}(Y) \to \{ \emptyset, \{ \star \} \}$ are in bijection with pairs of maps from $\mathcal{B}(X)$ and $\mathcal{B}(Y)$. That is, with pairs of points $(x,y)$!
Then $\mathcal{B}(X) \otimes^\sigma \mathcal{B}(Y)$ lives on $X \times Y$, and so we should recover the injectivity we needed from the previous section (though I admit I haven't checked all the details). This will mean that, in these nice settings, the answer to your question is yes!

Lastly, I'll say a philosophical word. In the setting of topological spaces, it turns out to be better constructively to work with locales instead of spaces. These are the formal dual to frames, which are lattices that look like lattices of open sets. In nice settings the two approaches agree, though they each have different pathologies.
It seems as though there should be a similar story for measurable spaces. Instead of working with the concrete measure spaces, we might instead work with the opposite category of the category of $\sigma$-algebras. We already know that nice measurable spaces form a full subcategory (though, as mentioned above, in general this embedding need not be full) so that we again agree in the nice case. This is the approach taken by Kriz and Pultr in their paper above, and one might hope that this approach handles "uncountable" measure spaces with more grace than concrete spaces (in the same way that locales handle uncountable operations more gracefully than topological spaces. We're closer to the algebra, which makes things more constructive). This is, in fact, the hope of Terry Tao and Asgar Jamneshan, who are developing this theory in the hopes of extending ergodic theory (and likely other aspects of measure theory as well) to the nonseparable setting. See Terry Tao's blog post here as well as Asgar Jamneshan's talk here. Jamneshan has used this machinery to great effect in getting uncountable analogues of classical theorems, as you can see from some of his papers.

I hope this helps ^_^
