Let $\mathcal{A}$ be a $\sigma$-algebra on a set $X$, and $\mathcal{B}$ be a $\sigma$-algebra on a set $Y$. A map $h\colon\mathcal{B}\to\mathcal{A}$ is called a $\sigma$-homomorphism if $h(\emptyset)=\emptyset$, $h(Y\setminus B)=X\setminus h(B)$ for any $B\in\mathcal{B}$, and $h\big(\bigcup B_n\big)=\bigcup h(B_n)$ for any sequence $\{B_n\}_{n\in\omega}$ in $\mathcal{B}$.

Let us consider a category where objects are pairs of the form $(X,\mathcal{A})$, $\mathcal{A}$ being a $\sigma$-algebra on $X$, and morphisms $(X,\mathcal{A})\to(Y,\mathcal{B})$ are $\sigma$-homomorphisms $h\colon\mathcal{B}\to\mathcal{A}$. Note that the arrows are reversed.

Question: Are there products in this category?

Note: Original version of this question confusingly uses the term "measurable space" for the objects of the category in concern. The question is now rewrtitten to make clear the difference.

There is a standard notion of "the category of measurable spaces" where objects are the same pairs as above and morphisms $(X,\mathcal{A})\to(Y,\mathcal{B})$ are measurable maps $f\colon X\to Y$. Measurability means that $f^{-1}[B]\in\mathcal{A}$ for any set $B\in\mathcal{B}$; it is easy to check that then $f^{-1}\colon\mathcal{B}\to\mathcal{A}$ is $\sigma$-homomorphism. However, not all $\sigma$-homomorphisms are of this form; see here. So our category has the same objects but more morphisms than the standard category of measurable spaces.

A natural candidate for a product of $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ in our category of $\sigma$-algebras is $(X\times Y,\mathcal{A}\times\mathcal{B})$, where $\mathcal{A}\times\mathcal{B}\,$ is the $\sigma$-algebra generated by rectangles of the form $A\times B\,$ for $A\in\mathcal{A}$ and $B\in\mathcal{B}$, together with $\sigma$-homomorphisms $f_1\colon\mathcal{A}\to\mathcal{A}\times\mathcal{B}$ and $f_2\colon\mathcal{B}\to\mathcal{A}\times\mathcal{B}$ defined by $f_1(A)=A\times Y$ and $f_2(B)=X\times B$. Taking another $\sigma$-algebra $\mathcal{C}$ and $\sigma$-homomorphisms $g_1\colon\mathcal{A}\to\mathcal{C}$, $g_2\colon\mathcal{B}\to\mathcal{C}$, we have to prove that there exists a unique $\sigma$-homomorphism $h\colon\mathcal{A}\times\mathcal{B}\to\mathcal{C}$ such that $g_1=h\circ f_1$ and $g_2=h\circ f_2$. We could define $h(A\times B)=g_1(A)\cap g_2(B)$ and then try to extend the definition of $h$ to the whole $\sigma$-algebra $\mathcal{A}\times\mathcal{B}$. Is this approach working? How one can prove the existence and the uniqueness of $h$?

  • $\begingroup$ The key thing that seems to be missing here is that a morphism of measurable spaces has to involve a map of the is underlying sets. Do you really not want that, here? If not, then you're not actually talking about a category of measurable spaces, but about a category of $\sigma$-algebras. $\endgroup$ Oct 27, 2018 at 16:56
  • $\begingroup$ @KevinCarlson You are right, I do not assume any map of underlying sets. I did not know that mentioning "the category measurable spaces" already implies that measurable maps of underlying sets are used as morphisms. If you really think that talking about measurable spaces in this context is confusing, I will reformulate the question. $\endgroup$ Oct 27, 2018 at 18:46
  • 1
    $\begingroup$ I absolutely think it's confusing. Notice that you even notate all your morphisms as going backwards, from the perspective of the spaces! An edit would be nice although it's not absolutely necessary now that we agree on the situation. You are asking whether the opposite of the category of $\sigma$-algebras admits binary products, nothing about measurable spaces at all. $\endgroup$ Oct 27, 2018 at 21:12

2 Answers 2


The products you're looking for are just coproducts in the category of $\sigma$-algebras. These can be shown to exist by various high-level arguments: for instance, $\sigma$-algebras are models of a generalized algebraic theory, and such categories always admit all limits and colimits.

The general recipe for constructing coproducts of models of an algebraic theory is as follows. For objects $X$ and $Y$ that are free, say $X=F(S)$ and $Y=F(T)$ for sets $S$ and $T$, the coproduct $X\coprod Y$ must be the free algebra $F(S\coprod T)$, since $F$ is a left adjoint and preserves coproducts. If $X$ and $Y$ are general algebras, then we give them a presentation. Categorically, this corresponds to writing $X$ as a (reflexive) coequalizer of free algebras $X''\rightrightarrows X'$, where $X'$ is free on a generating set for $X$ and $X''$ is free on a set of relations. Now we can construct $X\coprod Y$ as the coequalizer of the coproduct of presentations for $X$ and $Y$, that is of the canonical maps $X''\coprod Y''\rightrightarrows X'\coprod Y'$. In short, we construct coproducts by taking the disjoint union of presentations. This is likely familar from the example of the free product of groups.

Your example falls into this description: we give $\mathcal A$ and $\mathcal B$ the maximal presentations with all elements generating and all possible relations listed. Then the coproduct $\mathcal A\coprod \mathcal B$ of $\sigma$-algebras is generated by the disjoint union $\mathcal A\sqcup \mathcal B$ of the underlying sets. This corresponds in your picture to generating $\mathcal A\coprod \mathcal B$ by the elements $A\times Y$ and $X\times B$ instead of $(A,B)$, which works fine. However, I'm not completely sure whether the $\sigma$-algebra you suggest is actually the coproduct of $\sigma$-algebras. I think the analogous question fails for sufficiently unusual topological spaces-the topology on the product of the spaces may not be the coproduct of the topologies. However, such a coproduct certainly does exist. It's just hard work to get an explicit handle on it, which is one key reason to prefer measurable spaces to abstract $\sigma$-algebras.

EDIT: I missed a subtlety resulting from your presentation, which is that it's not immediate that the coproduct of $\sigma$-algebras of sets should coincide with the coproduct of abstract $\sigma$-algebras. Unlike the case for Boolean algebras, not every abstract $\sigma$-algebra can be represented as a $\sigma$-algebra of sets, and so the question is serious. Apparently, every abstract $\sigma$-algebra can instead be represented as the quotient of a canonical $\sigma$-algebra of sets by a $\sigma$-ideal. It seems plausible to me that this quotient is the counit of an adjunction whose left adjoint is the inclusion of $\sigma$-algebras of sets into abstract $\sigma$-algebras. If that is the case, then the construction above gives the correct construction of the coproduct. Unfortunately, the reference I have for this claim is very dated, and I can't easily extract the desired adjunction from it. Here it is: http://www.ams.org/journals/bull/1947-53-08/S0002-9904-1947-08866-2/S0002-9904-1947-08866-2.pdf

In any case, it's unclear to me that there's any reason to think solely about $\sigma$-algebras of sets, if you're only actually using the maps of abstract $\sigma$-algebras. That's another way to avoid this worry, but maybe you have a concrete reason (no pun intended) no to do that.

  • 2
    $\begingroup$ My motivation for studying $\sigma$-algebras of sets instead of measurable spaces on one side and abstract $\sigma$-complete Boolean algebras on the other side was to learn and to try to understand the differences and similarities. I think I have learned a lot. If you are interested, I have posted a proof that there are products in the category of $\sigma$-algebras. It seems that Loomis-Sikorski Theorem is not involved. $\endgroup$ Nov 4, 2018 at 18:13

In short, yes, the category in question has products, and yes, these can be defined as suggested.

We have to use a theorem on extending maps on $\sigma$-algebras to $\sigma$-homomorphisms. Let us note that a family $\mathcal{B}\subseteq\mathcal{A}$ is called a "$\sigma$-complete subalgebra" of a $\sigma$-complete Boolean algebra $\mathcal{A}$ if $\mathcal{B}$ is nonempty and closed under complements and countable meets and joins computed in $\mathcal{A}$. We say that a family $\mathcal{S}\subseteq\mathcal{A}$ "$\sigma$-generates" a $\sigma$-complete subalgebra $\mathcal{A}$ if $\mathcal{A}$ coincides with the smallest $\sigma$-complete subalgebra $\mathcal{B}$ of $\mathcal{A}$ such that $\mathcal{S}\subseteq\mathcal{B}$. Finally, for $\varepsilon\in\{-1,1\}$ and an element $A$ of a Boolean algebra $\mathcal{A}$, we denote by "$\varepsilon A$" the element $A$ if $\varepsilon=1$, and the complement of $A$ if $\varepsilon=-1$.

Theorem 1. Let $\mathcal{A}$ be a $\sigma$-complete Boolean algebra and let $\mathcal{S}\subseteq\mathcal{A}$ be a family that $\sigma$-generates $\mathcal{A}$. Let $(X,\mathcal{B})$ be a $\sigma$-algebra of sets. Let $f\colon\mathcal{S}\to\mathcal{B}\,$ be such that for every sequence $\{A_n\!:n\in\omega\}$ in $\mathcal{S}$ and for every sequence $\{\varepsilon_n\!:n\in\omega\}$ in $\{-1,1\}$, if $\bigwedge\varepsilon_n A_n=0_{\mathcal{A}}$ then $\bigcap\varepsilon_n f(A_n)=\emptyset$. Then there exists a unique $\sigma$-homomorphism $h\colon\mathcal{A}\to\mathcal{B}$ such that $h(A)=f(A)$ for all $A\in\mathcal{S}$.

Proof. Let us first note that if $A\in\mathcal{S}$ and $-A\in\mathcal{S}$ then $f(-A)=X\setminus f(A)$. Indeed, we have $1(A)\wedge 1(-A)=-1(A)\wedge-1(-A)=0_{\mathcal{A}}$ and hence $$f(A)\cap f(-A)=(X\setminus f(A))\cap(X\setminus f(-A))=X\setminus(f(A)\cup f(-A))=\emptyset.$$ Without a loss of generality we may assume that $-A\in\mathcal{S}$ for all $A\in\mathcal{S}$. Otherwise we can take $\mathcal{S}'=\mathcal{S}\cup\{-A\in\mathcal{A}\!:A\in\mathcal{S}\}$ and define $f'(A)=f(A)$ for $A\in\mathcal{S}$, and $f'(A)=X\setminus f(-A)$ for $A\in\mathcal{S}'\setminus\mathcal{S}$.

Let $x\in X$ be arbitrary and let $\{A_n\!:n\in\omega\}$ be a sequence in $\{A\in\mathcal{S}\!:x\in f(A)\}$. Then $x\in\bigcap f(A_n)$, hence $\bigcap f(A_n)\neq\emptyset$ and thus $\bigwedge A_n\neq 0_{\mathcal{A}}$. It follows that for every $x\in X$ there exists a $\sigma$-complete filter $p_x$ in $\mathcal{A}$ such that $\{A\in\mathcal{S}\!:x\in f(A)\}\subseteq p_x$. If we denote $$\mathcal{A}_x=\{A\in\mathcal{A}\!:A\in p_x\text{ or }-A\in p_x\}$$ then $\mathcal{A}_x$ is a $\sigma$-complete subalgebra of $\mathcal{A}$ and $\mathcal{S}\subseteq\mathcal{A}_x$, hence $\mathcal{A}_x=\mathcal{A}$. So $p_x$ is an ultrafilter.

Let $Z$ denote the set of all $\sigma$-complete ultrafilters on $\mathcal{A}$. We have defined a mapping $\varphi\colon X\to Z$ by $\varphi(x)=p_x$. A mapping $\psi\colon\mathcal{A}\to Z$ defined by $\psi(A)=\{p\in Z\!:A\in p\}$ is a $\sigma$-homomorphism of $\mathcal{A}$ into the power-set algebra $\mathcal{P}(Z)$. Then $h=\varphi^{-1}\circ\psi$ is a $\sigma$-homomorphism from $\mathcal{A}$ to $\mathcal{B}$. For $A\in\mathcal{S}$, $$h(A)=\{x\in X\!:p_x\in\psi(A)\}=\{x\in X\!:A\in p_x\}=\{x\in X\!:x\in f(A)\}=f(A).$$

The uniqueness of $h$ follows from the fact that if $h,h'\colon\mathcal{A}\to\mathcal{B}$ are $\sigma$-homomorphisms then $\mathcal{A}'=\{A\in\mathcal{A}\!:h(A)=h'(A)\}$ is a $\sigma$-complete subalgebra of $\mathcal{A}$ such that $\mathcal{S}\subseteq\mathcal{A}'$, hence $\mathcal{A}'=\mathcal{A}$. q.e.d.

Theorem 2. The category of $\sigma$-algebras of sets and reversed $\sigma$-homomorphisms has binary products.

Proof. Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be $\sigma$-algebras of sets, and let $f_{\mathcal{A}}\colon\mathcal{A}\to\mathcal{P}(X\times Y)$ and $f_{\mathcal{B}}\colon\mathcal{B}\to\mathcal{P}(X\times Y)$ be $\sigma$-homomorphisms defined by $f_{\mathcal{A}}(A)=A\times Y$ and $f_{\mathcal{B}}(B)=X\times B$. Denote $\mathcal{S}=f_{\mathcal{A}}[\mathcal{A}]\cup f_{\mathcal{B}}[\mathcal{B}]$ and let $\mathcal{E}$ be the $\sigma$-complete subalgebra of $\mathcal{P}(X\times Y)$ $\sigma$-generated by $\mathcal{S}$. We claim that $\sigma$-algebra of sets $(X\times Y,\mathcal{E})$ is a product of $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ in the category of $\sigma$-algebras and reversed $\sigma$-homomorphisms.

Let $(Z,\mathcal{C})$ be another $\sigma$-algebra and let $g_{\mathcal{A}}\colon\mathcal{A}\to\mathcal{C}$, $g_{\mathcal{B}}\colon\to\mathcal{C}$ be $\sigma$-homomorphisms. Define $f\colon\mathcal{S}\to\mathcal{C}$ by $f(A\times Y)=g_{\mathcal{A}}(A)$ and $f(X\times B)=g_{\mathcal{B}}(B)$. To check that $f$ satisfies the assumption of the above theorem, let $\{A_n\!:n\in\omega\}$ be a sequence in $\mathcal{S}$ and let $\{\varepsilon_n\!:n\in\omega\}$ be a sequence in $\{-1,1\}$ such that $\bigcap\varepsilon_n A_n=\emptyset$. Denote $N=\{n\in\omega\!:A_n\in f_{\mathcal{A}}[\mathcal{A}]\}$. There exist $A\in\mathcal{A}$, $B\in\mathcal{B}$ such that $$\bigcap_{n\in N}\varepsilon_n A_n=A\times Y\quad\text{and}\quad\bigcap_{n\notin N}\varepsilon_n A_n=X\times B,$$ hence $$\bigcap_{n\in\omega}\varepsilon_n A_n=A\times B$$ and $$\bigcap_{n\in\omega}\varepsilon_n f(A_n)=f(A\times Y)\cap f(X\times B)=g_{\mathcal{A}}(A)\cap g_{\mathcal{B}}(B).$$ Clearly, if $A\times B=\emptyset$ then $A=\emptyset$ or $B=\emptyset$, hence $g_{\mathcal{A}}(A)\cap g_{\mathcal{B}}(B)=\emptyset$. By Theorem 1 there exists a unique $\sigma$-homomorphism $h\colon\mathcal{E}\to\mathcal{C}$ such that $h(E)=f(E)$ for all $E\in\mathcal{S}$, that is, $h(A\times Y)=g_{\mathcal{A}}(A)$ for all $A\in\mathcal{A}$ and $h(X\times B)=g_{\mathcal{B}}(B)$ for all $B\in\mathcal{B}$. It follows that $h$ is the unique $\sigma$-homomorphism satisfying $g_{\mathcal{A}}=h\circ f_{\mathcal{A}}$ and $g_{\mathcal{B}}=h\circ f_{\mathcal{B}}$. q.e.d.

Note: Theorem 1 and its proof is a modification of Theorem 34.1 from [R. Sikorski, Boolean Algebras, Springer, 1964]. It is valid for $\kappa$-complete Boolean algebras and $\kappa$-homomorphisms for arbitrary cardinal $\kappa$. The assumption that $\mathcal{B}$ is an algebra of sets can be replaced by a weaker assumption ($\kappa$-distributivity) but cannot be completely removed. Theorem 2 can be proved for arbitrary products, not only binary.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .