Product of Borel sigma algebras - MathOverflow

Product of Borel sigma algebras

2010-09-24T18:11:10Z

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 $X \times Y$?

Answer by Gerald Edgar for Product of Borel sigma algebras

2010-09-24T18:20:10Z

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.

Answer by Byron Schmuland for Product of Borel sigma algebras

2010-09-24T18:38:34Z

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.

Answer by Michael Greinecker for Product of Borel sigma algebras

2011-11-21T10:37:43Z

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.

Answer by Martin for Product of Borel sigma algebras

2012-11-30T21:10:38Z

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.