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(In what follows, Freiling's Axiom of Symmetry is simply the following:

($A_{\aleph_0}$) :( $\forall$$f$: $\mathbf R$ $\rightarrow$$\mathbf R_{\aleph_0}$)($\exists$$x_1$,$x_2$)($x_2$$\notin$$f$($x_1$) $\land$ $x_1$$\notin$$f$($x_2$)), where $\mathbf R$ are the reals, $\mathbf R_{\aleph_0}$ is the set of all countable sets of reals, and $f$ : $\mathbf R$$\rightarrow$$\mathbf R_{\aleph_0}$ is interpreted as "$f$ assigns to each real a countable set of reals".

Note that one can replace $\mathbf R$ and $\mathbf R_{\aleph_0}$ with $[$0,1$]$ and $[$0,1$]_{\aleph_0}$.

Note also that:

$A_{f}$ = {($x_1$,$x_2$): $x_2$$\notin$$f$($x_1$)}, $A^{f}$ = {($x_1$,$x_2$): $x_1$$\notin$$f$($x_2$)}, $\forall$$f$: $[$ 0,1 $]$$\rightarrow$ $[$0,1$]_{\aleph_0}$ ($\lambda^{2}$($A_{f}$)=1 $\land$ $\lambda^{2}$($A^{f}$)=1), where $\lambda$ is the Lebesgue measure on the Borel algebra $\mathfrak B$ in $[$ 0,1 $]$, $\lambda^{2}$ is the requisite product measure; that this implies that

$\lambda^{2}$($A_{f}$ $\cap$ $A^{f}$)=1 when $\lnot$$CH$, $CH$ implies that $\lambda^{2}$($A_{f}$ $\cap$ $A^{f}$)=0, which implies that ($[$ 0,1 $]$$\times$ $[$ 0,1 $]$)$\setminus$$A_{f}$=$A^{f}$.)

A common critique of Freiling's Axiom of symmetry is "...that violations of the Axiom of Symmetry are fundamentally connected with non-measurable sets, and counterexample functions $f$ to $AS$ cannot be nice measurable functions..." (this from Prof. Hamkins' answer to Mateo Mio's mathoverflow question " Axiom of Symmetry, aka Freiling's argument against $CH$"). This is also stated by Kai Hauser in his paper "What New Axioms Could Not Be", as follows:

"...the mathematical flaw in the transition from the thought experiment to $A_{\aleph_0}$ lies in its haphazard generalization of a plausable intuition about measurable subsets of $[$ 0,1 $]$ to arbitrary subsets of $[$ 0,1 $]$."

Why does he say this? Well, in the aforementioned paper, he states that:

"The formal statement of the conclusion of Freiling's thought experiment is

$\forall$$f$ : [ 0,1 $]$ $\rightarrow$ $[$ 0,1 $]_{\aleph_0}$ ($\lambda^2$($A_{f}$)=1 $\land$ $\lambda^2$($A^{f}$)=1) ...From this it follows that $\lambda^2$($A_{f}$ $\cap$$A^{f}$)=1, hence this intersection is nonempty" . Then, in the next paragraph, he states that "this chain of inferences requires the measurability of $A_{f}$ and $A^{f}$ for which there seems to be no a priori reason...under $CH$, let $\lt$ be a well-order of the continuum in order type $\omega_1$ and define $f$(x)={y: y$\le$x}...Then, by Fubini's theorem, neither $A_{f}$ nor $A^{f}$ are measurable" [this is just the Sierpinski example--my comment].

However Freiling, in his paper, has the following lemma:

Lemma ($ZFC$+$A_{null}$). There is no set on the unit square which is null on almost every vertical line and full on almost every horizontal line, where $A_{null}$ is

$\forall$$f$:$\mathbf R$$\rightarrow$$\mathbf R_{null}$($\exists$$x_1$,$x_2$($x_1$$\notin$$f$($x_2$) $\land$ $x_2$$\notin$$f$($x_1$)), where $\mathbf R_{null}$ is the set of all sets of reals with Lebesgue measure zero.

Question: Is $ZFC$+$A_{null}$+"$\mathfrak c$ is real-valued measurable" consistent if $ZFC$+ "There exists a measurable cardinal" is consistent?

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Claim: The principle $A_{null}$ follows from continuum is RVM.

Proof: Let $m$ be a total extension of Lebesgue measure. Suppose every vertical section of $A \subseteq \mathbb{R}^2$ is Lebesgue null. First check that $A$ is $m \otimes m$-null. Put $B = \{(y, x) : (x, y) \in A\}$. Then $B$ is also $m \otimes m$-null. So if $(x, y)$ is outside $A \cup B$, then $x \notin A_y$ and $y \notin A_x$.

The above argument only requires: Every countably generated sigma algebra extending the Borel algebra admits an extension of Lebesgue measure.

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  • $\begingroup$ Why is $A$ $m \otimes m$-measurable? $\endgroup$ – Monroe Eskew Sep 28 '15 at 21:21
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    $\begingroup$ For each $e > 0$, let $U = U(e)$ be a subset of plane such that each vertical section $U_x$ is an open cover of $A_x$ of length less than $e$. Next note that each $U$ is a countable union of sets of form $A_{I, U} \times I$ where $I$ is a rational interval and $A_{I, U} = \{x : I \subseteq U_x\}$. So it suffices for $m$ to measure $A_{I, U}$ for rational $e > 0$. $\endgroup$ – Ashutosh Sep 28 '15 at 22:24
  • $\begingroup$ @MonroeEskew: Have you a counterexample to this? $\endgroup$ – Thomas Benjamin Sep 28 '15 at 22:54
  • $\begingroup$ No, Ashutosh is right. $\endgroup$ – Monroe Eskew Sep 28 '15 at 23:09

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