Welcome to MathOverflow

Question

14

5

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 $X \times Y$?

flag	asked Sep 24 2010 at 18:11 Bill Johnson 14.4k●1●33●60

Gerald Edgar · Accepted Answer · 2010-09-24 18:20:10Z UTC

11

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.

answered Sep 24 2010 at 18:20

Gerald Edgar
15.2k●1●28●67

1

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 2010 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 2010 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 2010 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 2010 at 18:39

Here is an accesible proof: david.efnet-math.org/?p=16 – Michael Greinecker Sep 24 2010 at 19:31

show 1 more comment

Martin · Answer 2 · 2012-11-30 21:15:52Z UTC

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.

Michael Greinecker · Answer 3 · 2011-11-21 10:37:43Z UTC

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.

Byron Schmuland · Answer 4 · 2010-09-24 18:38:34Z UTC

11

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.

answered Sep 24 2010 at 18:38

Byron Schmuland
1,905●8●18

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 2010 at 19:00

3

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 2010 at 19:05

Yes, it is very nice. I had never looked at it before reading your answer. Thanks again. – Bill Johnson Sep 24 2010 at 19:08

Product of Borel sigma algebras

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

4 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.