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A major argument against Freiling's Axiom of Symmetry is the following (this from the wikipedia article of the same name):

"The naive probabalistic notion used by Freiling tacitly assumes that there is a well-behaved way to associate a probability to any subset of the reals. But the mathematical formalization of the notion of 'probability' uses the notion of measure, yet the axiom of choice implies the existence of nomeasurable subsets, even of the unit interval..."

However, there seems a way to so formalize Freiling's naive, pre-reflective arguments for associating a probability to any subset of the reals. I propose the following:

Definition. A cardinal $\kappa$ is real-valued measurable iff there is a $\kappa$-additive probability measure on the power set of $\kappa$ which vanishes on singletons (i.e. singletons have probability measure zero)." (This from the wikipedia article on measurable cardinals.)

Axiom. Let $\mathfrak c$ be the cardinality of the continuum. $\mathfrak c$ is real-valued measurable.

It is interesting to note that the wikipedia article on measurable cardinals states that back in 1929, Banach and Kuratowski proved that $CH$ implies that $\mathfrak c$ is not real-valued measurable.

Since the Axiom of Symmetry is equivalent to the negation of $CH$, assuming that $\mathfrak c$ is real-valued measurable as an axiom immediately implies that the axiom of symmetry holds.

Also one has from the wikipedia article on measurable cardinals, the following (at least according to my understanding at this moment):

"A real-valued measurable cardinal='$\mathfrak c$' exists iff there is a countably additive extension of the Lebesgue measure to all sets of reals iff there is an atomless probability measure on $\mathscr P$($\mathfrak c$).

Question. In what way(s) does "$\mathfrak c$ is real-valued measurable' not (if any) capture Freiling's naive, pre-reflective arguments?

Question. Which large cardinal axioms imply that $\mathfrak c$ is real-valued measurable?

Question. What happens to the axiom of choice in the presence of my proposed axiom?

Question. Why is my proposed axiom yet another example of an axiom that "could not be"?

[Addendum: Two questions in response to Monroe's answer:

Let $S$ be "There is a subset of $\mathscr R^2$ which is countable on every horizontal line and co-countable on every vertical line." It is known that $CH$ implies $S$. One can also denote $S$ by "There exists a Sierpinski example", so one should be able to express the contrapositive as follows:

" 'There is no Sierpinski example' implies $\lnot$$CH$."

Question: Does $ZFC+$$($$\bigstar$$)$$\vdash$$\lnot$$CH$?

Question: Does $ZFC+$$($$\bigstar$$)$$\vdash$"There is, on $\mathscr R^2$, a product measure that is a total measure?"]

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Real-valued measurable cardinals (RVM) are equiconsistent with 2-valued measurable cardinals. I believe this is due to Solovay and Kunen. (Solovay for the forcing direction, Kunen for the inner model $L[U]$.) They contradict CH, so their existence is not implied by standard large cardinals.

I claim that Freiling's argument applies equally against RVMs. The key part of his argument is the "symmetry" part, which seems to be the fishiest part. Suppose there is a $\kappa$-additive real-valued measure on $\kappa$. If we buy Freiling's reasoning, then it seems also for any function $f : \kappa \to [\kappa]^{<\kappa}$, it should be equally probable whether $y \in f(x)$ or $x \in f(y)$, where $x$ and $y$ are the positions where two darts land. Since $\kappa$ is RVM, the objection that arbitrary events don't have a measure goes away. (Right?) But if $f(\alpha) = \alpha$ for $\alpha < \kappa$, then we get the same paradox: After we know $x$, the probability that $x \in f(y)$ is 1, and the probability that $y \in f(x)$ is 0.

Well nonmeasurability hasn't really gone away. The underlying event space is really $\mathbb R^2$. If $\frak c$ is RVM, then the product measure generated by the total measure on $\mathbb R$ is NOT a total measure on $\mathbb R^2$. The set $\{ (\alpha, \beta ) \in \kappa^2 : \alpha < \beta \}$ is nonmeasurable.

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    $\begingroup$ Historical remark: The fact "If $I$ is a $\theta$-saturated normal ideal over $\kappa$ where $\theta < \kappa$ then $\kappa$ is measurable in $L[I]$" is due to Solovay (Real valued measurable cardinals). In his thesis, Kunen used iterated ultrapowers to show that $\kappa^+$-saturation suffices. $\endgroup$ – Ashutosh Sep 7 '15 at 2:50
  • $\begingroup$ Are there other axioms one must assume in order that if $\mathfrak c$ is real-valued measurable then $\mathbb R^2$ has a total measure that is an extension of the total measure on $\mathbb R$? $\endgroup$ – Thomas Benjamin Sep 7 '15 at 12:16
  • $\begingroup$ Also, could someone provide me with the definition of the term "measure class"? I think it might be important for a proper discussion of this answer. Thanks. $\endgroup$ – Thomas Benjamin Sep 7 '15 at 12:28
  • $\begingroup$ If the continuum is RVM then there is a total measure on $\mathbb R^2$, but it won't be the product measure. So you won't be able to apply Fubini's theorem which is what Freiling is doing implicitly. You can get a total measure on the plane extending one on a line just by taking a "disjoint union measure," but is that what you really want? (Usually we give one dimensional objects measure zero in the plane.) $\endgroup$ – Monroe Eskew Sep 7 '15 at 12:33
  • $\begingroup$ @MonroeEskew: I was thinking more along the lines of the principle ($\bigstar$): "$\mathscr R$ is not a chain of measure-zero sets" found in Joseph Shipman's paper "Cardinal Conditions For Strong Fubini Theorems" (Transaction of the American Mathematical Society, Volume 21, Number 2, October 1990, pp. 465-81, found on pp.475-477). Assuming ($\bigstar$) (according to Shipman) one cannot "construct a Sierpinski example". The question is, is ($\bigstar$) a consequence of "$\mathfrak c$ is real-valued measurable"? If so, then can one use ($\bigstar$) to construct a total measure on $\endgroup$ – Thomas Benjamin Sep 8 '15 at 2:26
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In response to the first of your augmented questions, Does ZFC + ($\bigstar) \vdash\neg$ CH?

Yes. Recall, in Shipman's paper "Cardinal Conditions for Strong Fubini Theorems" (TAMS, Volume 321, Number 2, October 1990, pp. 465-81) see p. 475, we have

($\bigstar$) If $\bf{R}$ is a union of $\gamma$ measure-0 sets, then some union of less than $\gamma$ of them is not measure 0.
(He is talking in terms of Lebesgue measure here.)

As pointed out in that paper, $\bigstar$ is consistent with ZFC. Now of course, under CH, $\bf{R}$ is a union of $\aleph_1$ singletons, so no union of lesser cardinality ($\aleph_0)$ of them can fail to to be measure zero.

Incidentally, I would not agree with the comment by Monroe Eskew that " $\frak{c}$ is RVM" implies $\neg (\bigstar)$; in the same paper of Shipman, he points out (p. 479) that RVM implies $(*_n)$ (his "cardinal conditions"); on page 475 he already had that $(*_2)$ implies $\bigstar$

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