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I'm trying to pinpoint the "intuitive argument" for Freiling's Axiom of Symmetry. It's meant to be a "probabilistic" argument, so thinking about what seems to me to be the probabilistic intuition, it seems natural to ask:

Is it a theorem of ZFC that there exists $c>0$ and a set $S \subset [0,1] \times [0,1]$ such that

  • for every $x \in [0,1]$, $\{y \in [0,1] : (x,y) \in S\}$ is contained in a Borel set of zero Lebesgue measure;
  • for every $y \in [0,1]$, $\{x \in [0,1] : (x,y) \in S\}$ contains a Borel set of Lebesgue measure greater than or equal to $c$?

(If the answer is yes, then the natural follow up question is whether we can take $c=1$; but this seems less important.)

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    $\begingroup$ @JosephVanName $S$ may not be measurable, even if its sections are, so Fubini does not apply. For example, under CH there is such a set, the graph of a well order, which has countable sections one way and co-countable the other. This realizes the case $c=1$ under CH. $\endgroup$ Nov 27, 2023 at 13:03
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    $\begingroup$ Related to “what forms of Fubini are consistent with ZFC”: this other question I asked a few months ago, and this paper by Harvey Friedman on “A Consistent Fubini-Tonelli Theorem”. $\endgroup$
    – Gro-Tsen
    Nov 27, 2023 at 13:42

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For $c=1$ the answer is negative (that is, the negation of the statement you give is consistent with ZFC). Indeed, Harvey Friedman's paper “A Consistent Fubini-Tonelli Theorem for Nonmeasurable Functions”, Illinois J. Math. 24 (1980), 390–395 shows that the following statement is consistent with ZFC (theorem 2, slightly rephrased):

Let $F\colon \mathbb{R} \times \mathbb{R} \to [0,\infty)$ be such that for almost all $x$ and $y$, $F_x \colon y \mapsto F(x,y)$ and $F^y \colon x \mapsto F(x,y)$ are measurable. Furthermore, assume that $x \mapsto \int F_x$ and $y \mapsto \int F^y$ are both measurable (on their respective domains of full measure) as extended real valued functions. Then $\int (x\mapsto \int F_x) = \int (y \mapsto \int F^y)$.

If a set $S$ as you describe exists with $c=1$, then its characteristic function $(x,y) \mapsto \mathbf{1}_S(x,y)$ satisfies the assumptions of the aforequoted statement (note that a subset of $[0,1]$ contained in a Borel set of zero measure, or containing a Borel set of full measure, is measurable) and violates its conclusion. So the existence of $S$ for $c=1$ cannot be proved in ZFC.

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    $\begingroup$ I guess you can extend the argument to general $c>0$: using AC, you can reduce each horizontal section to a Borel set of measure exactly $c$. $\endgroup$ Nov 27, 2023 at 14:23
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This is not a full answer, but let me just point out that the statement is relatively consistent both with CH and also with $\neg$CH.

CH implies the statement, since we can take $S$ to be the (converse of) the graph of a well order with order-type $\omega_1$. All sections in one direction are initial segments of the order, which are countable and hence Borel and measure zero, and all sections in the other direction are co-countable, hence Borel of measure one. So this realizes the case $c=1$ under CH.

Meanwhile, MA+$\neg$CH also implies the statement. Under this hypothesis, all sets of size less than continuum are measurable with measure zero. So let $S$ be (the converse of) the graph of a well order of the reals in the unit interval with order-type continuum. All sections in one direction are initial segments of the order, hence measure zero (and so contained in a Borel measure zero set). All sections in the other direction are final segments of the order, hence full measure, and so contain a full-measure Borel set. So this hypothesis realizes the case $c=1$.

Meanwhile, with only ZFC and no additional hypotheses, this graph-of-a-well-order argument doesn't seem to work, since we don't know that the initial segments of the order will be measurable. And so I am inclined to think that the property can fail in some ZFC models.

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  • $\begingroup$ Thanks. So according to the third paragraph of your answer, there is a model consistent with Freiling’s axiom of symmetry in which my statement holds. I guess that that, by itself, is not an argument against Freiling’s intuition, but just creates a “probabilistic intuition against Martin’s Axiom” (because MA plus something probabilistically intuitive implies something probabilistically counterintuitive)? $\endgroup$ Nov 27, 2023 at 14:09
  • $\begingroup$ Yes, that's right. My own view is that the prereflective probabilistic intuitions are not very important, for what we've learned from looking into probability theory more seriously is that many naive intuitions are simply wrong. The lesson I take from this is that we are advised to abandon the naive ideas and adopt a more sophisticated treatment of probability. $\endgroup$ Nov 27, 2023 at 14:36
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In any model of ZFC where there is no such set for $c=1$ (e.g., Friedman's model mentioned in Gro-Tsen's answer), there is no $S \subset [0,1]^2$ such that all vertical slices $S_x$ are null and all horizontal slices $S^y$ have positive inner measure (i.e., we're not fixing any particular $c>0$).

Suppose there is such an $S.$ Let $T= \{(x+q, y): (x,y) \in S, q \in \mathbb{Q}\},$ where addition is taken mod 1. Then every vertical slice of $T$ is null and every horizontal slice is measure 1, contradiction.

By the way, there are reasonably probabilistic analogues of Freiling's symmetry paradox provable in ZFC. E.g, there is $S \subset [0,1]^2$ such that every $S_x$ and every $[0,1] \setminus S^y$ is perfectly small (i.e., has a perfect set of disjoint translations). The graph of any well-ordering $\prec$ of $[0,1]$ of order type $\mathfrak{c}$ has this property. Perfectly small sets have intuitive claim to being "probabilistically negligible" despite ZFC not proving them to be of Lebesgue measure 0. Similarly, the initial segments of this order are absolutely negligible, i.e. for each initial segment $I,$ any translation-invariant probability measure $\mu$ on $[0,1]$ extends to a translation-invariant $\mu'$ such that $\mu'(I)=0.$

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