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Let $(X_1, d_1)$,..., $(X_n, d_n)$ be separable metric spaces, and define $X = X_1 \times ... \times X_n$ to be the produce space with the metric

$$d((x_1,...,x_n),(y_1,...,y_n)) = \sqrt{d_1^2 (x_1,y_1)+...+ d_n^2 (x_n, y_n)}$$

Prove that $\mathcal{B}(X) = \mathcal{B}(X_1) \times...\times\mathcal{B}(X_n)$

Some toolsets I have are:

Definition: The $\sigma$-algebra $\mathcal{F}_1 \times ... \times \mathcal{F}_n$ is defined as the minimal $\sigma$-algebra which contains all the sets of the form $A_1 \times ... \times A_n$, where $A_i \in \mathcal{F}_i , 1 \leq i \leq n$.

If $X$ is separable, then any open set can be represented as a countable union of open balls. Thus, for a separable space $X$, we could define the Borel $\sigma$-algebra of $X$ as $\sigma (\mathcal{A})$, with $\mathcal{A}$ being the family of all open balls.

Jonathen
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2 Answers2

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This is actually true for countably infinite product of separable spaces with the suitable product metric defined by it. Explicitly if the metrics on $X_{i}$ are $d_{i}$ , then (one such) product metric(a metric which induces the product topology) is given by $\displaystyle d(x,y)=\sum_{i=1}^{\infty}\frac{1}{2^{i}}\frac{d_{i}(x,y)}{1+d_{i}(x,y)}$

Note that for the finite case, the euclidean metric is itself is enough to induce the product topology .

If you are comfortable with topology, then it is best to consider a topological proof.

Some notation:- I prefer the "tensor" notation for the product $\sigma$-algebra as it distinguishes from the mere cartesian product of the sigma algebras(which by the way is not a $\sigma$-algebra) . So the product sigma algebra will be denoted by $\displaystyle\otimes_{i=1}^{n}\mathcal{B}(X_{i})$ . Note that it is just a notation(a standard one) and there is by no means any "tensor product" involved here.

Since all the spaces are separable, prove that the product space is also separable. By definition of the product topology, you have sets of the form $\prod_{i=1}^{n} U_{i}$ where $U_{i}$'s are open in $X_{i}$ form a basis for the product topology. Now each $X$ has a countable basis of open balls . So prove that the product of these balls form a basis for the product topology. Explicitly , consider dense subsets $\{x_{n}^{i}\}_{n=1}^{\infty}$ for each $i$ . Then $\mathcal{C}=\{\prod_{i=1}^{n}B_{i}\,\,:\, B_{i}=B_{d_{i}}(x_{k}^{i},r_{m})\,,r_{m}\in\Bbb{Q}\,,m,k\in\Bbb{N}\}$ is a countable basis for the product topology. Now prove that the metric topology induced by $d$ is equivalent to the product topology (This is a standard excercise in any real analysis text if you don't know it already).

Now this means that any open set in the metric topology $U$ can be written as a countable union of sets in $\mathcal{C}$. But all the sets in $\mathcal{C}$ are in $\otimes_{i=1}^{n}\mathcal{B}(X_{i})$ . . Hence their countable union $U$ is in $ \otimes_{i=1}^{n}\mathcal{B}(X_{i})$. This holds for any open set $U$ in the product metric topology. This means that $\mathcal{B}(X)\subset \otimes_{i=1}^{n}\mathcal{B}(X_{i})$ .

Conversely consider $\mathcal{F}_{n}=\{E\in\mathcal{B}(X_{n})\,: \Pi_{n}^{-1}(E)\in \mathcal{B}(X)\}$ where $\Pi_{n}$ denotes the projection map. Now by definition all open sets in $X_{n}$ are contained in $\mathcal{F}_{n}$. Prove that $\mathcal{F}_{n}$ is a sigma algebra and conclude that $\mathcal{F}_{n}=\mathcal{B}(X_{n})$ . This holds for all $n$.

Thus $\otimes_{i=1}^{n}\mathcal{B}(X_{i})\subset \mathcal{B}(X)$ .

So we have equality.

In general $\otimes_{i=1}^{n}\mathcal{B}(X_{i})\subset \mathcal{B}(X)$ this is always true irrespective of whether the metric spaces are separable or not. But the other inclusion might fail without separability.

Mr.Gandalf Sauron
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  • What is the open set in the product space X? I am confused. Can I say the open set in the X has the form $V_1 \times V_2 ... \times V_n$, where each open set $V_i \subset X_i$? – Jonathen Sep 07 '22 at 03:43
  • Those sets $V_{1}\times V_{2}...\times V_{n}$ are indeed open, but not all open sets are of that form. The collection $\mathcal{O}={U_{1}\times U_{2}...\times U_{n}: U_{i}\in\tau_{i}}$ forms a basis for the topology on the product. That is you take arbitrary union of sets of that form to get all open sets in the product topology. For example in $\Bbb{R}^{2}$, the set $(0,1)\times (0,1)\cup (9,10)\times (e,\pi))$ is an open set, but not of the form $U_{1}\times U_{2}$. Instead it is the union of two sets(product of sets) of that form. You should read up on the product topology. – Mr.Gandalf Sauron Sep 07 '22 at 08:12
  • For infinite product $\prod_{i=1}^{\infty}X_{i}$, you need to make a subtle change in the definition. In that case the basic open sets will be of the form $\prod_{i=1}^{\infty} U_{i}$ where $U_{i}=X_{i}$ for all but finitely many $i$ . – Mr.Gandalf Sauron Sep 07 '22 at 08:16
  • For finite product space, the basis for the topology is $U_1 \times ... \times U_n$, how to interpret these $U_i$? It seems these $U_i$ are a union of open sets in $X_i$. Am I correct? – Jonathen Sep 07 '22 at 14:03
  • It seems that you are not very familiar or have lost touch with basic topology. Ofcourse $U_{i}$'s are open. A basic google search can lead you to a lot of resources. In addition you can open up any basic topology book like Munkres or Armstrong to find everything that you need about product topology. As far as this problem is concerned, any open set in the product topology can be written as the countable union of sets of the form $U_{1}\times...\times U_{n}$ . – Mr.Gandalf Sauron Sep 07 '22 at 15:09
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    @Jonathen: I think you're making it harder than it is. $B(X_{1}\times X_{2})$ is the smallest sigma algebra that contains $\tau_{X_1\times X_2}.$ An arbitrary open set in $B(X_{1}\times X_{2})$ is of the form $\bigcup_{i,j=1}^{\infty}U_i\times V_j=\bigcup_{i=1}^{\infty}U_i\times \bigcup_{j=1}^{\infty}V_j.$ By definition, $B(X_1)\times B(X_2)$ contains these sets. So it must be at least as large as $B(X_{1}\times X_{2}).$ That is, $B(X_{1}\times X_{2})\subseteq B(X_1)\times B(X_2).$ – Matematleta Sep 07 '22 at 16:04
  • @Jonathen: and you do not need to know that the product space is metrizable. All you need are the projections, and the basic definitions. I have a pdf on my computer that gives a very good, clean exposition of these ideas. But I do not know how to post it here. – Matematleta Sep 07 '22 at 16:28
  • @Matematleta You can post the pdf to your google drive. Then you can share the link here. That will be great helpful! – Jonathen Sep 07 '22 at 16:38
  • @Jonathen I hope this helps. It's really just a question of following the definitions. – Matematleta Sep 07 '22 at 17:00
  • @Matematleta It is true that it holds for countable product of separable topological spaces and not necessarily metric spaces. I have added the product metric stuff as we are in the metric space situation and it might help to have on the back of the mind that no matter how complicated the sets in the product topology are , there is always a metric upon which you can rely on to characterize them in terms of just open balls under it . Btw I think you might have added jonathen's name and not my user id in your tags haha . – Mr.Gandalf Sauron Sep 07 '22 at 17:54
  • @Mr.GandalfSauron: no offense intended. Happy to upvote your answer. I just think the issue the OP has is very easily cleared up in two or three lines. Excuse the ignorance, but what does "I think you might have added jonathen's name and not my user id in your tags haha" mean? I will gladly amend or delete my comment(s) if I have breached some netiquette. – Matematleta Sep 07 '22 at 18:41
  • No no ....I meant that I think one of your comment was intended for me but by mistake you added the "@" with Jonathen's name and not mine. – Mr.Gandalf Sauron Sep 07 '22 at 19:15
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You can find the proof by an easy google. But if you want to do it on your own with some hints, then here are some for the $n=2$ case. The extension to finite products is no harder.

$1).\ \mathcal{B}(X)$ is the smallest sigma algebra that contains $\tau_X, $ the open sets of $X.\ \mathcal{B}(X_1)\times \mathcal{B}(X_2)$ contains the open sets of $X.$

$2).\ $ The projections $\pi_{X_1}:X\to X_1$ and $\pi_{X_2}:X\to X_2$ are continuous hence measurable. If $U\in \tau_{X_1}$ then $\pi_{X_1}^{-1}(U)=U\times X_2\in \tau_X.$ If $V\in \tau_{X_2}$ then $\pi_{X_2}^{-1}(V)=X_1\times V\in \tau_X.$ Now, $\pi_{X_1}^{-1}(U)\cap \pi_{X_2}^{-1}(V)=U\times V.$ Conclude that if $U\in \tau_{X_1}$ and $V\in \tau_{X_2}$ then $U\times V\in \tau_X$ and note that by definition $\mathcal{B}(X_1)\times \mathcal{B}(X_2)$ is the sigma algebra generated by $\tau_{X_1}\times \tau_{X_2}.$

Matematleta
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  • Can you explain why $\mathcal{B}(X_1)\times \mathcal{B}(X_2)$ contains the open sets of $X.$ – Jonathen Sep 02 '22 at 13:17
  • It's because $\tau_{X}$ is generated by ${U\times V:U\in \tau_{X_1},V\in \tau_{X_2} }$ – Matematleta Sep 02 '22 at 15:20
  • I think this is just one form of $\tau_X$. So I think there should be some more details here. – Jonathen Sep 03 '22 at 20:03
  • @Jonathen: an arbitrary basis element of $X$ is $U\times V$ as in my comment above. These generate the topology of $X$. – Matematleta Sep 04 '22 at 15:19
  • The problem here that I am concerned that in the proof, we don't use the condition of separable metric space. I think the key is that in separable metric space, any open set can be written in a countable union of open balls. But the proof missed the key. – Jonathen Sep 04 '22 at 15:31
  • @Jonathen , I don't know why you chose to completely ignore my answer. Each of the questions you asked in the comments is actually justified in what I wrote.(It is also meant as a solution and hence twice as long whereas Matematleta's answer is more of a guided hint that would lead you to the correct answer). I also dealt with the infinite case. It is essentially the exact same proof as mine. It all depends on the topological fact that the product topology is induced by the product metric and countable product of separable spaces is separable. – Mr.Gandalf Sauron Sep 06 '22 at 14:58
  • @Matematleta Even in the finite case, you do need separability here – Mr.Gandalf Sauron Sep 06 '22 at 15:15
  • @Jonathen You need separability to conclude that any open set in the product can be written as a countable union of product of open balls ( which form the basis for the product topology). – Mr.Gandalf Sauron Sep 06 '22 at 15:17
  • @Mr.GandalfSauron Yes, you are right. I did not express correctly what I meant. Once $X_1,X_2$ are metric and separable, the proof goes through exactly as in your (and my) answer. – Matematleta Sep 06 '22 at 18:41
  • @Jonathen: every open set $O$ in $X$ is a countable union of sets of the form $U\times V$ as in my answer. But then, this means that $O\in \mathcal B(X_1)\times \mathcal B(X_2).$ Do you see now where separability is used? – Matematleta Sep 06 '22 at 19:20
  • @Jonathen Elaboration: Due to separability , there exists a COUNTABLE collection ${U_{k}\times V_{k}}{k=1}^{\infty}$ of sets such that $U{k}$ is open in $X_{1}$ and $V_{k}$ is open in $X_{2}$ such that ANY open set $O\subset X_{1}\times X_{2}$ is of the form $O=\bigcup_{j=1}^{\infty}U_{n_{j}}\times V_{n_{j}}$ for some subsequence $n_{j}$. But by definition of the product sigma algebra , the sets $U_{n_{j}}\times V_{n_{j}}$ are in the product sigma algebra as $U_{n_{j}}$ is a Borel set in $B(X_{1})$ and $V_{n_{j}}$ is a Borel set in $B(X_{2})$ – Mr.Gandalf Sauron Sep 07 '22 at 08:25
  • This means that ANY open set in the product $X_{1}\times X_{2}$ is in the product sigma algebra as it is a countable union of sets which are in the product sigma algebra. Thus the Borel sigma algebra of the product which is generated(the smallest sigma algebra containing) by the arbitrary open sets must be contained in the product sigma algebra as it already contains all open sets. Thus you get $B(X_{1}\times X_{2})\subset B(X_{1})\times B(X_{2})$ .The other way $B(X_{1})\times B(X_{2})\subset B(X_{1}\times X_{2})$ is always true(regardless of separability) as I have shown in my answer – Mr.Gandalf Sauron Sep 07 '22 at 08:35