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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?

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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

1 Answer 1

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.

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Thank you t.b.. –  Sunny Nov 23 '11 at 19:19

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