For a set $X$, let $\mathcal P(X)$ denote its power set and let $\mathcal P(X)\otimes\mathcal P(X)$ denote the product $\sigma$-algebra in $X^2$. When $|X|\leq\aleph_0$ then $\mathcal P(X)\otimes\mathcal P(X)=\mathcal P(X^2)$ but when $|X|>2^{\aleph_0}$ this equality is known to fail. What happens when $\aleph_0<|X|\leq 2^{\aleph_0}$?
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2$\begingroup$ See also mathoverflow.net/q/39882/454 $\endgroup$– Gerald EdgarCommented Jul 13, 2017 at 20:09
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1$\begingroup$ In particular, the result of Rao mentioned by Michael Greinecker in his answer there mathoverflow.net/a/81491/1946 provides a positive answer for $X$ of $\omega_1$. So this handles $X$ of size continuum under CH. $\endgroup$– Joel David HamkinsCommented Jul 14, 2017 at 1:28
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1$\begingroup$ Call an infinite cardinal $\kappa$ Kunen if the sigma algebra generated by $\{A \times B : A, B \subseteq \kappa\}$ contains every subset of $\kappa \times \kappa$. The set of Kunen cardinals forms an initial segment of cardinals, closed under countable suprema, with the psuedointersection number (denoted $\mathfrak{p}$) as a member. In the Cohen real model (obtained by adding $\aleph_2$ Cohen reals to $L$) the continuum is not Kunen (see Kunen's 1968 thesis). This also holds in the random real model (obtained by adding $\aleph_2$ random reals to $L$). $\endgroup$– AshutoshCommented Jul 16, 2017 at 17:28
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1$\begingroup$ @Ashutosh Cool! It would be nice to see some details as an answer. $\endgroup$– Andrés E. CaicedoCommented Jul 20, 2017 at 18:44
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1 Answer
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The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here.
It is consistent (assuming large cardinals) that there is an extension of Lebesgue measure defined on all sets of reals. Here, $X=\mathbb R$ and $\Sigma=\mathcal P(\mathbb R)$. Since $\nu$ extends Lebesgue measure, the space satisfies the assumptions of the result just stated, and $\mathcal P(\mathbb R)\otimes\mathcal P(\mathbb R)$ is not $\mathcal P(\mathbb R^2)$.
By the way, there is a recent article in the Monthly dealing precisely with this problem and discussing how $\mathsf{CH}$ implies that $\mathcal P(\mathbb R)\otimes\mathcal P(\mathbb R)$ is $\mathcal P(\mathbb R^2)$ while the existence of extensions of Lebesgue measure gives a negative answer:
MR3626256 Avilés, Antonio; Plebanek, Grzegorz. A little ado about rectangles. Amer. Math. Monthly 124 (2017), no. 4, 345–350.
answered Jul 13, 2017 at 21:22
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$\begingroup$ (I assume there are more down to Earth counterexamples, maybe even along the same lines.) $\endgroup$ Commented Jul 13, 2017 at 21:23
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2$\begingroup$ Andrés, could you clarify what you mean by saying "the answer is no in general"? At first I took you to mean that all uncountable $X$ are counterexamples, but the rest of your argument seems to show only that if certain large cardinals are consistent, then it is consistent that there is a counterexample of size continuum. $\endgroup$ Commented Jul 14, 2017 at 0:59
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$\begingroup$ Yes, Joel, that's all I meant. $\endgroup$ Commented Jul 14, 2017 at 1:08
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$\begingroup$ Thanks a lot for your replies. \\ So let me see if get it rightly (not being a set theorist myself): Under ZFC the equality $\mathcal P(X)\otimes\mathcal P(X)=\mathcal P(X^2)$ is true for $|X|\leq\aleph_1$ and is untrue for for $|X|>2^{\aleph_0}$. For $\aleph_1<|X|\le 2^{\aleph_0}$ (unless one assumes CH) the equality may fail if for example one assumes large cardinals. $\endgroup$ Commented Jul 14, 2017 at 9:06
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1$\begingroup$ @Emmanuel Yes, precisely. I'm sure much more can be said for $X $ uncountable of size at most $\mathfrak c $, but that's what we have so far. $\endgroup$ Commented Jul 14, 2017 at 13:09