Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There's an interesting property that if $(X,\mathcal{T})$ and $(Y,\mathcal{S})$ are topological spaces, then the Borel $\sigma$-algebra of $X\times Y$ with the product topology includes the product $\sigma$-algebra of the Borel $\sigma$-algebras of the respective spaces, and if the spaces are actually second countable, then these two $\sigma$-algebras coincide.

Does this still hold if one of the spaces is only known to be a metric space? Let's say $(X,\mathcal{T})$ is second countable, but $(Y,d)$ is only known to be a metric space. Does it still hold that the Borel $\sigma$-algebra on $X\times Y$ is the same as the product $\sigma$-algebra of the Borel $\sigma$-algebras on $X$ and $Y$?

I tried taking an arbitrary open set $A$ in $X\times Y$. Since $X$ is second countable, there exists a countable basis $\{U_n\}_{n\in\mathbb{N}}$. So for any $n$ and $\epsilon>0$, set $$V_{n,\epsilon}=\{y\in Y\mid\text{ for some }\delta>0, U_n\times B_{\epsilon+\delta}(y)\subset A\}.$$ [Here $B_r(s)$ is the ball centered at $s$ of radius $r$.] I was then trying to show that $A$ can be written as an arbitrary union of sets of form $U_n\times V_{n,\epsilon}$, with $\epsilon$ possibly varying, to prove it. Is there a way to flesh out this idea?

share|cite|improve this question
See also this related MO-thread – t.b. Nov 23 '11 at 2:48
Thanks for thread t.b. – Sunny Nov 23 '11 at 3:12
Hello, Sunny! If you put an "@" before the t.b., then the user is informed about your reply. You should say: "Thanks for the thread @t.b." – André Caldas Nov 23 '11 at 4:01
Thanks @AndréCaldas, I get it now. – Sunny Nov 23 '11 at 19:18
+1, nice question! I wonder what reference the property at the beginning comes from? Just like to know its background. Thanks! – Tim Nov 30 '12 at 18:53
up vote 6 down vote accepted

Let $X$ be a second countable metric space and let $Y$ be an arbitrary topological space. Then the product $\sigma$-algebra $\mathcal{B}(X) \otimes \mathcal{B}(Y)$ on $X \times Y$ is equal to the Borel $\sigma$-algebra $\mathcal{B}(X \times Y)$ of the product $X \times Y$.

Indeed, it is easy to show that $\mathcal{B}(X) \otimes \mathcal{B}(Y) \subset \mathcal{B}(X\times Y)$ for all topological spaces $X$ and $Y$ without any assumptions, so I'll only show the reverse inclusion. It suffices to prove that every open set $U \subset X \times Y$ belongs to $\mathcal{B}(X) \otimes \mathcal{B}(Y)$.

Your idea for proving this works:

Let $\{B_n\}_{n=1}^\infty$ be a countable base for the topology of $X$.

Let $U$ be a non-empty open set in $X \times Y$ and put $C_n = \bigcup \{ C \subset Y\,:\,C \text{ is open and } B_n \times C \subset U\}$, (so $C_n$ is the largest open set such that $B_n \times C_n \subset U$).

Claim: $U = \bigcup_n B_n \times C_n$.

Proof. By definition $U \supset B_n \times C_n$, so this also holds for the union. For the reverse inclusion, let $(x,y) \in U$. Since $U \subset X \times Y$ is open there are open sets $B,C$ such that $x \in B$ and $y \in C$ and $B \times C \subset U$. Since $\{B_n\}_{n=1}^\infty$ is a basis, there is some $n$ such that $x \in B_n \subset B$, so $C \subset C_n$ and hence $(x,y) \in B_n \times C_n$, and this establishes the reverse inclusion.$\qquad\qquad \square$

Since $B_n \times C_n \in \mathcal{B}(X) \otimes \mathcal{B}(Y)$ it follows that $U$ is in $\mathcal{B}(X) \otimes \mathcal{B}(Y)$, as desired.

If we drop the requirement that $X$ be second countable then what you ask about becomes false in general, see this MO-thread:

If $X$ is a discrete space and $|X| \gt \mathfrak{c}$ then the Borel $\sigma$-algebra on $X \times X$ is strictly finer than the product $\sigma$-algebra. As Gerald Edgar points out in the linked thread, the diagonal is a Borel set (since it is closed) but it is not in the product $\sigma$-algebra.

share|cite|improve this answer
Thank you t.b.. – Sunny Nov 23 '11 at 19:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.