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

(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?

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 $y \notin A_x$.
• Why is $A$ $m \otimes m$-measurable? – Monroe Eskew Sep 28 '15 at 21:21
• 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 $e > 0$. – Ashutosh Sep 28 '15 at 22:24