(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? 



A question regarding a common critique of Freiling's Axiom of Symmetry
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.


