Non-standard measure 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 $\frac{1}{\omega}$.
 A: 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 $\sum_{n=1}^{\infty}x=0$. 
