# Is the product of $\sigma$-algebras a tensor product in some sense?

Let $$(X_{1},\Sigma _{1})$$ and $$(X_{2},\Sigma _{2})$$ be two measurable spaces, that is, $$\Sigma _{1}$$ and $$\Sigma _{2}$$ are sigma algebras on $$X_{1}$$ and $$X_{2}$$ respectively, and let $$\mu _{1}$$ and $$\mu _{2}$$ be measures on these spaces. Denote by $$\Sigma _{1}\otimes \Sigma _{2}$$ the sigma algebra on the Cartesian product $$X_{1}\times X_{2}$$ generated by subsets of the form $$B_{1}\times B_{2}$$, where $$B_{1}\in \Sigma _{1}$$ and $$B_{2}\in \Sigma _{2}$$. This sigma algebra is called the tensor-product σ-algebra on the product space. A product measure $$\mu _{1}\times \mu _{2}$$ is defined to be a measure on the measurable space $$(X_{1}\times X_{2},\Sigma _{1}\otimes \Sigma _{2})$$ satisfying the property:

$$(\mu _{1}\times \mu _{2})(B_{1}\times B_{2})=\mu _{1}(B_{1})\mu_{2}(B_{2})$$

for all $$B_{1}\in \Sigma _{1},\ B_{2}\in \Sigma _{2}$$.

a. I can see that the product of $$\sigma$$-algebras is a categorical product; but is it correct to say that it is also a tensor product in some analogue to the tensor product in vector spaces? In which case how? And is there a universal property that expresses it as such?

b. I don't see how to express the product of measures as a categorical product. Is it?

Though tensor products are usually defined for vector spaces, its most generally defined for abelian groups:

For $$A$$ and $$B$$ two abelian groups, their tensor product $$A\otimes B$$ is a new abelian group which is such that a group homomorphism $$A\otimes B \rightarrow C$$ is equivalently a bilinear map out of $$A$$ and $$B$$.

Then some evidence that we might be able to interpret the product of measures as a kind of 'tensor product' is that supposing $$B_1$$ and $$B_1'$$ are disjoint in $$\Sigma_1$$, we have:

$$(\mu _{1}\times \mu _{2})((B_{1} \sqcup B_1')\times B_{2})=\mu _{1}(B_{1} \sqcup B_1')\mu_{2}(B_{2})=(\mu _{1}B_{1}+ \mu _{1}B_{1}') \times \mu_{2}B_{2}$$

=$$(\mu_1 \times \mu_2)(B_1,B_2)+(\mu_1 \times \mu_2)(B_1',B_2)$$

That is it satisfies the bilinearity property for the tensor product for disjoint sets.

• are you asking about the tensor product of the sigma algebras, or of the measures? Feb 10, 2014 at 0:05
• @weiss. I've clarified the question a bit. I think - both. But the tensor part of the question is addrressed to the sigma-algebras. Feb 10, 2014 at 0:22
• People use the term "tensor product" to denote all sorts of things. Don't worry about it too much. Feb 10, 2014 at 3:54

In my opinion this should simply be called the product of the two measurable spaces. In fact, it satisfies the corresponding universal property.

To answer your general question, you should look at monoidal categories. Every category with products has a monoidal structure with $\otimes = \times$. The category of vector spaces is also monoidal with $\otimes$ being the usual tensor product.

• What about the product measure? Mar 6, 2022 at 9:11

To have a universal property one first must define a category. So, if we take the category of measurable spaces, where objects are pairs $(X,\Sigma)$ of a set and a sigma algebra on it, and morphisms $f:(X,\Sigma)\to (Y,\Sigma ')$ are functions $f:X\to Y$ which are measurable, then the tensor product you mention is nothing but the categorical product. Notice however that your question mentioned measures $\mu_1,\mu_2$ which are never used. That may indicate that what you actually had in mind was to talk about the product of the measures in some categorical way. That is much more subtle since it is entirely not clear what is a useful way to define a category of measures. Intuitively, it should have objects triples $(X,\Sigma, \mu)$, a set with a sigma algebra on it, and a measure as well. But what should the morphisms be?

• Actually there is a very natural choice of morphisms of measure spaces $(X,A,\mu) \to (Y,B,\eta)$. It is a measurable map $(X,A) \to (Y,B)$ such that $\mu$ is the pullback of $\eta$ by this map. Feb 10, 2014 at 0:21