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Just a bit of a strange question. Modern formulations of probability theory rest upon measure theory. This poses an issue for non-measurable sets. Typically, one simply excludes these sets from the analysis and considers only measurable subsets of, for example, the real numbers.

It can be shown that the following four assumptions cannot all be true:

$P_0$: If a set has a measure, it is a value $0 \leq x \leq \infty$ in the extended reals.

$P_1$: If a set $P$ has measure $x$, then the set $P' = E(P)$ also has measure $x$, where $E$ represents an arbitrary element of the full Euclidean symmetry group of rotations and translations.

$P_2$: Measure is a sigma-additive function. If $P$ and $P'$ are disjoint sets with measures $x$ and $x'$, respectively, then the measure of $P \cup P'$ is $x + x'$.

$P_3$: Every subset of $\mathbb{R}^n$ has a measure.

In standard analysis, it is usually $P_3$ which is rejected. One then does measure theory, and therefore probability theory, with the sigma algebra of measurable subsets. One then defines the Lebesgue measure as the unique function satisfying all postulates $P_0-P_2$ for $\mathbb{R}^n$.

My question is this. Is it possible to produce a consistent measure theory with $P_1$, $P_2$, and $P_3$ but rejecting $P_0$? In particular, if we let the measure of a set be given, in general, by a non-negative surreal number, would this allow the other axioms to hold?

As an example, one could imagine a Vitali set as having a measure of $\frac{1}{\omega}$.

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E8xE8
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    Your text says "sigma-additive" but then you only give the definition of regular additive. Actual sigma additivity may lead to unwanted behavior because infinite sums of surreals are not as nice as those of reals. – Mark S. Apr 26 '17 at 20:49
  • Just a remark: Although it is only vaguely related to your question, you should know about dream mathematics, if you don't already. – Stefan Perko Apr 26 '17 at 20:57
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    You surprised me at the end: I thought you were going to reject $P_1$, rather than $P_0$ (since $P_1$ is not a part of probability theory anyway). – Michael Apr 26 '17 at 21:01
  • Not completely related, but in the same neighbourhood: if we reject full Choice for Dependent Choice, there is a model of the reals in which every set is measurable. This does do some weird things, however... – Chappers Apr 26 '17 at 21:06
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    Minor additional comment: The proof I know that properties $P_0$-$P_3$ cannot co-exist considers the special case of the unit interval $[0,1]$ and assumes the measure of $[0,1]$ is a (finite) positive number. It likely can be extended to more general cases, but I wonder if some additional structure is needed? Such as (i) The measure must not be the trivial all-zero or all-infinity measure. (ii) We require use of the axiom of choice. Perhaps we also need the measure to be either finite or sigma-finite? Or just require the measure of bounded sets to be finite. – Michael Apr 26 '17 at 21:30
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    I heard two years ago that there was such a (finitely additive) measure for $n = 2$ but I have never checked. Are you sure about what you say? Regarding sigma additivity, it is hard to define it elsewhere than in the reals or in some ultrapower of the reals because making sense of arbitrary infinite sums or suprema in arbitrary ordered groups is hard - if not impossible. – nombre Apr 26 '17 at 21:31
  • The Solovay model has the property that every set of reals is measureable. I don't know if $P_1$ fails – Ross Millikan Jun 28 '20 at 00:13
  • Using the surreals won't do much - the main phenomenon the Banach Tarski paradox involves is that you can rotate these pieces of a sphere to get a proper superset of what you started with. The principle is not hard to see: note that in $\Bbb R^2$, you can wrap a copy of N around the circle by rotating at some irrational angle. Thus, rotating the circle can sort of draw more points out of infinity, much like the Hilbert hotel. Banach Tarski is the same, but you are wrapping a "Hilbert resort" around a sphere instead. What measure do you give to something like that? – Mike Battaglia Apr 17 '25 at 03:00
  • To be clear: every point on the circle can be identified as belonging to some "bidirectional wrapped hotel" which is a copy of Z. So then as long as we can pick an "origin" for each hotel, we can partition the circle into three parts: the origins, the points that are some CCW num of steps from their origin, and the points that are some CW of it. We then take the "CCW" partition, rotate it clockwise by our angle, and have the union of the CCW plus the origins. We have pulled a copy of the origins out of infinity as in the Hilbert hotel. So do you really want rotation invariance for ALL sets? – Mike Battaglia Apr 17 '25 at 03:12

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If we keep property $P_1$, I don't think using surreals helps, since I think it is reasonable to assume that if $x$ is a nonnegative surreal number then $\sum_{n=1}^{\infty} x$ is either nonpositive or infinite. Assuming this, we can run through the standard proof for the unit interval but allow the measure of constructed sets to be surreal:

Suppose $\mu([0,1])$ is surreal and $0 < \mu([0,1])< \infty$. By the standard construction, define a collection $\mathcal{C}$ of equivalence classes over $[0,1]$ so that $x,y \in [0,1]$ are in the same equivalence class if $x-y$ is rational. For each class $c \in \mathcal{C}$, use the axiom of choice to choose a representative element $x(c) \in c$. Define $R$ as the set of rationals in $[0,1]$. For each rational $r \in R$, define $$B_r = \{(x(c) + r) \mod 1 : c \in \mathcal{C}\}$$ So $[0,1]$ is a countable union of disjoint sets $$ [0,1] = \cup_{r \in R} B_r $$ where $B_r$ are rigid shifts of each other. So if we assume $\mu(B_r)$ exists as a (surreal) number then $\mu(B_r) = \mu(B_0)$ for all rationals $r \in [0,1]$ and: $$ \mu([0,1]) = \sum_{r\in R} \mu(B_r) = \sum_{r \in R} \mu(B_0) $$ The right-hand-side sum is either nonpositive or infinite, leading to the contradiction.

Note: I edited the above to assume that if $x\geq 0$ then $\sum_{n=1}^{\infty} x$ is either nonpositive or infinite. For example, it is reasonable to expect a definition for the countably infinite sum of nonnegative surreals to satisfy the following: If $x \geq 0$ and $\sum_{n=1}^{\infty} x$ does not diverge to infinity, then $$ \sum_{n=1}^{\infty} x = x + \sum_{n=2}^{\infty}x = x + \sum_{n=1}^{\infty} x $$ and so $x=0$ and $\sum_{n=1}^{\infty}x=0$.

Michael
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  • Note: My hesitancy here is not knowing how to precisely define a countable sum of nonnegative surreals (I could not find a definition online), but I would think it would be something like: $$\sum_{n=1}^{\infty} x = \lim_{N\rightarrow\infty} \sum_{n=1}^N x $$ in which case it seems the countable sum should be nonpositive whenever $x$ is nonpositive. – Michael Apr 26 '17 at 22:16
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    This is sound, but without a good definition of sum, one could allow that it is finite if $\mu(B_0)$ is infinitesimal. – nombre Apr 26 '17 at 22:16
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    (As for definition of sums of surreal numbers, there is one that coïncides with the definition of formal sums of Hahn series when seeing $No$ as a Field of Hahn series in the "natural way". It does not allow summing infinitely many identical non zero numbers however. To my knowledge, there is no sound measure theory for surreals.) – nombre Apr 26 '17 at 22:24
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    That's the thing. My intuition is if we set the measure of a single Vitali set to be something like $\frac{1}{\omega}$, a countable sum over this infinitesimal should be non-infinite.

    Though not a proof, it intuitively seems like the sum of $\frac{1}{\omega}$ over an countable indexing set of order type $\omega$ (i.e. $\mathbb{N}$) should be equal to $1$.

     – E8xE8 Apr 27 '17 at 03:33
  • I edited my above answer to yield a contradiction whenever $x \geq 0$ implies $\sum_{n=1}^{\infty} x$ is either nonpositive or infinite. If $\sum_{n=1}^{\infty}x$ does not diverge to infinity and is equal to some value $y$, it is reasonable to expect $y = (x+x+x+...) = x+(x+x+x+...) = x+y$ and so $x=0$. If $\sum_{n=1}^{\infty}x \neq \sum_{n=2}^{\infty}x$, it is like saying $\sum_{r \in R} x$ is undefined as it depends on the ordering used to list elements of $R$. It seems better to just remove property $P_1$ (or $P_3$) than to make summations of nonnegative numbers dependent on order. – Michael Apr 27 '17 at 06:18
  • I think that it probably would depend on order type in the case of surreal numbers, though. By the way x + (x + x + ...) has the same order type as (x + x + ...). They both have order type $\omega$. The order type of $\omega + 1$ is given by (x + x + ... + x) + x. A set consisting of a single element followed by a countable number of elements is order isomorphic to the natural numbers, while a set consisting of a countable number of elements followed by a single element has order type $\omega + 1$. While this set is equal in cardinality to the naturals, it is not order isomorphic to them. – E8xE8 Apr 28 '17 at 18:14
  • The reason I want to keep P3 is to construct a generalization of probability and measure theory such that all subsets of $\mathbb{R}^n$ can be included in the set of events. P2 is vital to maintaining the definitions of probability, and P1 is necessary in order to keep the notion of a "uniform" measure. The only thing which can give is the stipulation that the measure of a set is a real number. – E8xE8 Apr 28 '17 at 18:17