# The universal property of product holds only for the product sigma-algebra

Suppose $$\{X_\alpha \mid \alpha \in J\}$$ is a family of measurable spaces. Let X denote the Cartesian product $$\prod_{\alpha \in J} X_\alpha$$ and consider the natural projections $$\{\pi_\alpha: X \to X_\alpha \mid \alpha \in J\}$$. Given a sigma-algebra $$\mathcal{A}$$ on $$X$$, define the property $$P(\mathcal{A})$$ to be that for any measurable space $$W$$ and any function $$g: W \to X$$, $$g$$ is measurable relative to $$\mathcal{A}$$ iff for each $$\alpha \in J$$, the function $$f_\alpha \circ g$$ from $$W$$ to $$X_\alpha$$ is measurable.

I suspect that the property $$P(\mathcal{A})$$ is the universal property of the product sigma-algebra and that it holds only for the product sigma-algebra on $$X$$, i.e. the sigma-algebra generated by the natural projections $$\{\pi_\alpha: X \to X_\alpha \mid \alpha \in J\}$$. To prove that, it's enough to prove that the sigma-algebra generated by the natural projections satisfies $$P$$ and that any other sigma-algebra $$\mathcal{A}$$ that satisfies $$P(\mathcal{A})$$ must be at the same time finer and coarser than the one generated by the natural projections.

I am having troubles with the finer part. Can you, dear reader, prove this?

I couldn't understand Mark Saving's answer but I managed to figure it out on my own. I'll use slightly different, probably better, notation than in the question.

First, a lemma we will use. Note that the variables used in this lemma are on their own and don't correspond to any variables in the question.

Lemma. If $$f : X \to Y$$ is a function from a measurable space $$X$$ to a measurable space $$Y$$, and the sigma-algebra of $$Y$$ is generated by a family $$\mathcal{B}$$ of subsets of $$Y$$ and if for each $$B \in \mathcal{B}$$ the preimage $$f^{-1}[B]$$ is measurable, then $$f$$ is measurable. This lemma can be proved using inductive construction of the sigma-algebra generated by a set, which is described in page 40 of "Real Analysis. Modern Techniques and Their Applications. Second Edition" by Folland 1999 and (with some typos) at https://proofwiki.org/wiki/Inductive_Construction_of_Sigma-Algebra_Generated_by_Collection_of_Subsets. I am too lazy to prove it here.

Now, the main result.

Suppose $$\{X_\alpha \mid \alpha \in J\}$$ is an indexed family of measurable spaces. For any measurable space $$K$$ with the cartesian product $$\prod_{\alpha \in J} X_\alpha$$ as the carrier, define the property $$P(K)$$, which is the universal property of the product of measurable spaces, to be "For each measurable space $$W$$ and function $$g: W \to K$$, $$g$$ is measurable iff for each $$\alpha \in J$$ the composition of the $$\alpha$$-th natural projection $$\pi_\alpha: K \to X_\alpha$$ and $$g: W \to K$$ (i.e. $$\pi_\alpha \circ g : W \to X_\alpha$$) is measurable".

Let R denote the measurable space with the carrier $$\prod_{\alpha \in J} X_\alpha$$ and the sigma-algebra generated by $$\{\pi_\alpha \mid \alpha \in J\}$$.

First we show that $$P(R)$$ holds. Inside the quantifier, $$P(R)$$ consists of a bidirectional statement. The direction from left to right can be clearly seen by unpacking the definitions. The direction from right to left follows from the lemma given above.

Now, suppose $$M$$ is another measurable space with the same carrier $$\prod_{\alpha \in J} X_\alpha$$ such that $$P(M)$$ holds. We want to show that $$M$$ has the same sigma-algebra as $$R$$. We do that by showing that the sigma-algebra of $$M$$ is both coarser and finer than that of $$R$$.

To show that $$M$$ is coarser than $$R$$, it's enough to show that $$\operatorname{id} : R \to M$$ is measurable. Since $$P(M)$$ holds, choosing $$R$$ as $$W$$ and $$\operatorname{id}: R \to M$$ as $$g$$ in the text of the property $$P(M)$$ gives us that to show that $$\operatorname{id} : R \to M$$ is measurable it's enough to show that, for each $$\alpha \in J$$, the natural projection $$\pi_\alpha : R \to X_\alpha$$ is measurable. But that is true by the definition of $$R$$. So indeed, $$M$$ is coarser than $$R$$.

Now we will show that $$M$$ is finer than $$R$$ by contraposition. Taking $$M$$ as $$W$$ and $$\operatorname{id}: M \to M$$ as $$g$$ in the text of $$P(M)$$ will witness that $$M$$ is finer than $$R$$, as we will now show. Since $$\operatorname{id}: M \to M$$ is measurable, by $$P(M)$$, for each $$\alpha \in J$$, the natural projection $$\pi_\alpha: M \to X_\alpha$$ is measurable. This implies that $$M$$ contains every member of the generating set of the sigma-algebra of $$R$$, so $$M$$ is indeed finer than $$R$$.

This is indeed the universal property of the product.

Consider the definition of the universal property of the product: $$\DeclareMathOperator{Hom}{Hom}$$

Definition: Consider a category $$C$$. Consider a set $$J$$, and suppose we have a family of objects $$\{X_\alpha\}_{\alpha \in J}$$ in $$C$$. Consider some object $$P$$ and some family of morphisms $$\{p_\alpha \in \Hom_C(P, X_\alpha)\}_{\alpha \in J}$$. Then $$(P, p)$$ is said to satisfy the universal property of the product with respect to $$X$$ if and only if for all objects $$A$$ and all families of morphisms $$\{a_\alpha \in \Hom_C(A, X_\alpha)\}_{\alpha \in J}$$, there is a unique map $$a \in \Hom_C(A, P)$$ such that for all $$\alpha \in J$$, $$a_\alpha = p_\alpha \circ a$$.

Consider the following two theorems:

Theorem: Consider a family of sets $$\{X_\alpha\}_{\alpha \in J}$$. Then $$(\prod\limits_{\alpha \in J} X_\alpha, \{\pi_\alpha : (\prod\limits_{\beta \in J} X_\beta) \to X_\alpha\}_{\alpha \in J})$$ satisfies the universal property of the product (in the category of sets). In other words, for all sets $$A$$ and all families of functions $$\{a_\alpha : A \to X_\alpha\}_{\alpha \in J}$$, there is a unique function $$a : A \to \prod\limits_{\alpha \in J} X_\alpha$$ such that for all $$\alpha \in J$$, $$a_\alpha = \pi_\alpha \circ a$$. $$\square$$

Theorem: Consider a family of measurable spaces $$\{X_\alpha\}_{\alpha \in J}$$. Then let $$P$$ be the product $$\sigma$$-algebra on $$\prod\limits_{\alpha \in J} X_\alpha$$; abusively write $$\prod\limits_{\alpha \in J} X_\alpha = (\prod\limits_{\alpha \in J} X_\alpha, P)$$. Then $$\prod\limits_{\alpha \in J} X_\alpha$$, together with the maps $$\{\pi_\alpha\}_{\alpha \in J}$$, satisfy the universal property of the product (in the category of measurable spaces). In other words, all the $$\pi_\alpha$$ are measurable maps, and furthermore, for all measure spaces $$A$$ and all families of measurable functions $$\{a_\alpha : A \to X_\alpha\}_{\alpha \in J}$$, there is a unique measurable function $$a : A \to \prod\limits_{\alpha \in J} X_\alpha$$ such that for all $$\alpha \in J$$, $$a_\alpha = \pi_\alpha \circ a$$. $$\square$$

Now, we can state the following theorem:

Thm. Let $$P(\mathcal{A})$$ if and only if the measure space $$(\prod\limits_{\alpha \in H} A_\alpha, \mathcal{A})$$, together with the maps $$\{\pi_\alpha\}_{\alpha \in J}$$, satisfy the universal property of the product.

Proof: Suppose the $$\sigma$$-algebra satisfies $$P(\mathcal{A})$$.

Note that since the identity map $$1_{\prod\limits_{\alpha \in J} X_\alpha} : \prod\limits_{\alpha \in J} X_\alpha \to \prod\limits_{\alpha \in J} X_\alpha$$ is measurable, each of the $$\pi_\alpha$$ is measurable, since $$\pi_\alpha = \pi_\alpha \circ 1_{\prod\limits_{\alpha \in J} X_\alpha}$$ is measurable by $$P(\mathcal{A})$$.

Suppose we have measurable functions $$\{g_\alpha : W \to X_\alpha\}_{\alpha \in J}$$. Then by the universal property of the product in the category of sets, there is a unique function $$g : W \to X$$ such that $$\forall \alpha, \pi_\alpha \circ g = g_\alpha$$. Then $$g$$ is measurable by $$P(\mathcal{A})$$. So the universal property of the product is satisfied.

Conversely, suppose the universal property of the product is satisfied. Suppose that for all $$\alpha$$, $$\pi_\alpha \circ g$$ is measurable. Consider the unique measurable function $$h : W \to X$$ such that for all $$\alpha$$, $$\pi_\alpha \circ g = \pi_\alpha \circ h$$. Therefore, $$g = h$$ by the universal property of the product of sets. Since $$h$$ is measurable and $$h = g$$, $$g$$ is measurable. Conversely, if $$g$$ is measurable, then for all $$\alpha$$, $$\pi_\alpha \circ g$$ is measurable, since the composition of measurable functions is measurable. $$\square$$

To finish the proof, note that the product $$\sigma$$-algebra $$W$$ clearly satisfies $$P(W)$$.

Now suppose we have two $$\sigma$$-algebras $$W, Z$$ on $$\prod\limits_{\alpha \in J} X_\alpha$$ which both satisfied $$P(-)$$, then consider the two measurable spaces $$M = (\prod\limits_{\alpha \in J} X_\alpha, W)$$ and $$N = (\prod\limits_{\alpha \in J} X_\alpha, Z)$$. Note that the identity maps $$M \to N$$ and $$N \to M$$ must both be measurable. Therefore, $$W = Z$$.

• It seems I don't understand the structure of your proof. Can you please write that down more explicitly? For instance, in "Suppose we have..." and "Conversely, suppose..." I am not sure what assumptions you have and which sigma-algebra you're talking about. I am not sure what you've proved at the point where you put the empty square symbol. Commented Nov 5, 2021 at 12:11
• @crabMan Let me know if the edits made things clearer or if you’re still confused. Thanks for the comment; I was definitely not being maximally clear. Commented Nov 5, 2021 at 18:59
• When you say "So the universal property of the product is satisfied.", which do you mean? Option 1: "for every measurable space $W$ and a family of measurable functions ${g_\alpha : W \to X_\alpha \mid \alpha \in J}$, there exists a unique measurable function $g$ from $W$ to the product in question such that $\forall \alpha \pi_\alpha \circ g = g_\alpha$". Option 2: The sigma algebra in question satisfies $P$. Commented Nov 9, 2021 at 13:09
• @CrabMan I mean option 1. I've updated the answer to give a bit more context. Commented Nov 9, 2021 at 20:43