Universally measurable sets of $\mathbb{R}^2$

$$\text{Is }{{\cal B}(\mathbb{R}^2})^u={{\cal B}(\mathbb{R}})^u\times {{\cal B}(\mathbb{R}})^u\,?\tag1$$

Is the $\sigma$-algebra of universally measurable sets on $\mathbb{R}^2$ equal to the product $\sigma$-algebra of two copies of the universally measurable sets on $\mathbb{R}$? It is not hard to see that (1) is true with $\supseteq$ instead of $=$, and I would be astonished if (1) were true, but I'm not sure.

Has anyone encountered this problem, or know a reference that might help?

Solution 1:

As you expected, the answer to your question is negative.

Recall that a set $U \subset X \times Y$ is universal for a class $\Gamma \subset \mathcal{P}(Y)$ if for every $G \in \Gamma$ there is $x \in X$ such that $G$ is the vertical section $U_x$ of $U$, that is: $G = U_x = \{ y \in Y : (x,y) \in U\}$. Recall also that analytic sets are universally measurable.

One standard way to prove that there are analytic sets in $X$ that aren't Borel sets is to exhibit an analytic set $U \subset X \times X$ which is universal for the analytic sets in $X$. The set of diagonal points $D = \{x \in X: (x,x) \in U\}$ of $U$ turns out to be an analytic set which is not Borel. The proof is a diagonal argument reminiscent of the Russell paradox and of Cantor's $\kappa \lt 2^\kappa$-theorem. See for example Srivastava, A course on Borel sets, Theorem 4.1.5, p.130.

A clever refinement of this argument yields the answer to your question:

Let $U \subset [0,1] \times [0,1]$ be an analytic set which is universal for the analytic sets in $[0,1]$. Then $U$ is not an element of the product $\sigma$-algebra $\mathcal{P}[0,1] \times \mathcal{L}[0,1]$ of the power set $\sigma$-algebra $\mathcal{P}[0,1]$ and the Lebesgue $\sigma$-algebra $\mathcal{L}[0,1]$.

In particular, being analytic, $U$ is universally measurable in $[0,1]^2$, but it does not even belong to the product $\sigma$-algebra $\mathcal{L}[0,1] \times \mathcal{L}[0,1] \supset \mathcal{B}([0,1])^u \times \mathcal{B}([0,1])^u$.

This is proved in B.V. Rao, Remarks on analytic sets, Fund. Math., 66 (1970), 237-239. A variant was independently found by R. Mansfield in the first part of The solution to one of Ulam's problems concerning analytic sets, I and II by different and more complicated methods (the argument in the second part is essentially the same as the one given by Rao). Srivastava's book, Theorem 4.3.4, page 141 also contains a proof of this. Of course, one can replace $[0,1]$ by $\mathbb{R}$ or any other uncountable Polish space.

Further elaborations and refinements can be found in Arnold W. Miller's Measurable rectangles. See in particular Theorem 1 where he proves that $\mathcal{L}[0,1]$ can be replaced by the Baire property algebra $\mathcal{BP}[0,1]$ in the Mansfield-Rao theorem.

Moreover, Miller showed in section 4 of On the length of Borel hierarchies that it is relatively consistent with ZFC that no universal analytic set belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \times \mathcal{P}[0,1]$.