Does $\mathcal P ( \mathbb R ) \otimes \mathcal P ( \mathbb R ) = \mathcal P ( \mathbb R \times \mathbb R )$?

Question

Obviously the answer's yes if I replace $\mathbb R$ by a countable set.

If $\mathbb R$ is replaced by a set $X$ having a greater cardinality, then the diagonal $\{ (x,x) , \ x\in X\}$ is not in the product $\sigma$-field (see for instance http://david.efnet-math.org/?p=16) so the answer is no.

Answer 1

This problem appears in Kunen's thesis (1968). I mentioned it a few months back, here

Answer 2

Firstly let me mention that as far as integration theory and probability theory are concerned, I don't think my question is an interesting one. I was asked this by a student, I realized I didn't know the answer, nor did any of my colleagues, so asked it here out of curiosity.

Let me summarize the information I could get so far. First observe that whether $$ \mathcal P ( E ) \otimes \mathcal P ( E ) = \mathcal P ( E \times E ) $$ or not depends only on the cardinality of $E$.

The answer is obviously yes when $E$ is countable for $E \times E$ is also countable so every subset of $E\times E$ is a countable union of singletons, hence in the product $\sigma$-algebra.

The answer's no if $\vert E \vert > 2^{\aleph_0}$, as is explained there.

My question was: what happens in between?

It was proved by Rao that the answer's yes if $\vert E\vert=\aleph_1$. Let me sketch his (very nice) argument, some knowledge about ordinals is required to understand it. Consider the property, for a set $E$, of having a sequence of countable partitions that separate points: there exists a family $(A_{i,j})_{i,j\in \mathbb N}$ of subsets of $E$ such that

• For every $i\in \mathbb N$, the sequence $(A_{i,j})_{j\in \mathbb N}$ is a partition of $E$.
• For every $x\neq y$ in $E$, there exists $(i,j)$ such that $x\in A_{i,j}$ and $y\not\in A_{i,j}$.

Clearly $\mathbb R$ has this property (given an integer $i$, there is a partition of $\mathbb R$ into countably many intervals of length $1/i$). The smallest uncountable ordinal $\omega_1$ can be mapped to $\mathbb R$ in a one to one way, so it also has this property (incidentally I wonder if there's a way to prove this directly, say avoiding the axiom of choice).

Now if a set $E$ has this property, graphs belong to the product $\sigma$-field. By graph, I mean the graph of a function $f \colon F \to E$ where $F$ is a subset of $E$. Indeed, if $f$ is such a function then its graph equals $$ \bigcap_{i\in \mathbb N} \bigcup_{j\in \mathbb N} f^{-1} ( A_{i,j} ) \times A_{i,j} . $$ A subset of $E\times E$ whose every vertical section is countable is a countable union of graphs, so it also belongs to the product $\sigma$-field, and so does a subset whose every horizontal section is countable.

Lastly, consider $\{ (\alpha, \beta ) , \ \beta \leq \alpha \in\omega_1\}$, the below diagonal subset of $\omega_1 \times\omega_1$. Its vertical section above any given $\alpha\in \omega_1$ is $\{ \beta \in \omega_1 ,\ \beta \leq \alpha\} = \alpha +1$ which is countable. In the same way, every horizontal section of the above diagonal subset of $\omega_1\times \omega_1$ is countable. Therefore, every subset of $\omega_1 \times \omega_1$ can be written as the union of a set whose vertical sections are countable and a set whose horizontal section are countable, finishing the proof of $$ \mathcal P ( \omega_1 ) \otimes \mathcal P ( \omega_1 ) = \mathcal P ( \omega_1 \times \omega_1 ) . $$

So far, the answer to the original question is thus yes if $\vert E \vert \leq \aleph_1$ and no if $\vert E \vert > 2^{\aleph_0}$. So under the continuum hypothesis: $\aleph_1 = 2^{\aleph_0}$ the question is completely solved.

Now, what if we do not assume CH? Apparently, it is shown there that there is a model of ZFC in which spaces of cardinality $2^{\aleph_0}$ fail to have the property, but I'm afraid the paper is too much of logic and model theory for me.

To conclude, I think this is the first time I was posed an undecidable question by a student.