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If $X$ and $Y$ are separable metric spaces, then the Borel $\sigma$-algebra $B(X \times Y)$ of the product is the the $\sigma$-algebra generated by $B(X)\times B(Y)$. I am embarrassed to admit that I don't know the answers to:

Question 1. What is a counterexample when $X$ and $Y$ are non separable?

Question 2. If $X$ is an uncountable discrete metric space, does $B(X)\times B(X)$ generated the Borel $\sigma$-algebra on $X \times X$?

Question 3. If $X$ and $Y$ are metric spaces with $X$ separable, does $B(X)\times B(Y)$ generated the Borel $\sigma$-algebra on $X \times Y$?

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4 Answers 4

up vote 14 down vote accepted

Q1. Discrete spaces with cardinal > c ... then the diagonal is a Borel set, but not in the product sigma-algebra.

This also answers Q2 (no)

but not Q3.

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I think he does mean $>c$. The assertion being made is exercise 29 in the Radon measures chapter of Folland's real analysis text. –  Keenan Kidwell Sep 24 '10 at 18:25
    
Thanks, Jerry & Keenan. Simple but nice. What is the answer to Q2 when $X$ has cardinality the continuum? –  Bill Johnson Sep 24 '10 at 18:34
    
If $X$ is of size $c$, say $X=\mathbb R$, then the complement of the diagonal in $X^2$ is the union of countably many rectangles. This follows from the fact that $\mathbb R^2$ has a countable basis for the topology consisting of rectangles. Now, we are interested in the discrete topology on $X$, but the fact remains, the diagonal in $X^2$ is in the product $\sigma$-algebra. –  Stefan Geschke Sep 24 '10 at 18:35
    
Of course, this does not show that the product $\sigma$-algebra is the same as the $\sigma$-algebra on the product. It just shows that the diagonal does not distinguish the two algebras. –  Stefan Geschke Sep 24 '10 at 18:39
    
Here is an accesible proof: david.efnet-math.org/?p=16 –  Michael Greinecker Sep 24 '10 at 19:31
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The answer to question 3 is yes. At least according to Lemma 6.4.2 of the second volume of Bogachev's book "Measure Theory".

He requires both spaces to be Hausdorff and one of them to have a countable base. They need not be metric spaces.

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Oh, yes, Byron. This is simple enough that this simpleton should have seen the argument. Thanks, and sorry that I can only accept one answer. –  Bill Johnson Sep 24 '10 at 19:00
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No problem. I really didn't do anything except follow my own advice: when you have an "unusual" measure theory question, try looking in Bogachev. His book is an impressive achievement. –  Byron Schmuland Sep 24 '10 at 19:05
    
Yes, it is very nice. I had never looked at it before reading your answer. Thanks again. –  Bill Johnson Sep 24 '10 at 19:08
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To close a gap: From the answer of Gerald Edgar, we know that the answer to the second question is no if the spaces involved have cardinality larger than $\mathfrak{c}$. This leaves open what happens when they do have cardinality $\mathfrak{c}$. The answer is yes under the continuum hypothesis, and in general it holds that $2^{\omega_1}\otimes 2^{\omega_1}=2^{\omega_1\times\omega_1}$. This was shown in

B. V. Rao, On discrete Borel spaces and projective sets Bull. Amer. Math. Soc. Volume 75, Number 3 (1969), 614-617.

In Bogachev's remarkable book, it can be found as Proposition 3.10.2.

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This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.

Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.

Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.

See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.

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