# Axiomizing volume over rationals

Is there a nonnegative extended real valued function on subsets of $$\mathbb{Q}^3$$ that is finitely additive on disjoint sets, translation invariant, and outputs $$(\text{length} \times \text{width} \times \text{height})$$ for boxes? It is a remarkable result of Lebesgue measure theory that no such function that is countably additive exists over $$\mathbb{Q}^3$$ or $$\mathbb{R}^3$$. I am wondering if the same is true for finitely additive set functions over $$\mathbb{Q}^3$$. If so, it would seem to suggest that there is some fundamental obstacle to formalizing the intuitive notion of volume.

I guess for each natural $$n$$ we can define a required function $$\mu$$ on $$\Bbb Q^n$$ as follows. Let $$G=\Bbb Q^n/\Bbb Z^n$$ be the quotient group and $$q:\Bbb Q^n\to G$$ be the quotient homomorphism. The group $$G$$ is Abelian and so it is amenable (see, for instance Corollary 2.9 from [Ban]), that is $$G$$ admits an shift-invariant finitely additive measure $$\nu$$, such that $$\nu(G)=1$$. For each subset $$A$$ of $$\Bbb Q^n\cap [0,1)^n$$ put $$\mu(A)=\nu(q(A))$$. Now let $$A$$ be any subset of $$\Bbb Q^n$$. Then $$A=\bigcup \{A_{x}:x\in \Bbb Z^n\}$$, where $$A_x=(x+[0,1)^n)\cap A$$ for each $$x\in\Bbb Z^n$$. Put $$\mu(A)=\sum_{x\in\Bbb Z^n}\mu(A_x-x)$$.

I guess if $$A$$ is a box then the correct value of $$\mu(A)$$ can be shown as follows.

Lemma. For each $$k=1,\dots,n$$, each $$y\in Q$$, $$\mu(A\cap H)=0$$, where $$H$$ is a hyperplane $$\{(x_1,\dots, x_n)\in\Bbb Q^n:x_k=y\}$$.

Proof. For each $$x\in\Bbb Z^n$$ and each natural $$N$$, the group $$G$$ contains $$N$$ disjoint copies $$g_1+q((A\cap H)_x-x), g_2+q((A\cap H)_x-x),\dots, g_N+q((A\cap H)_x-x)$$ of a set $$q((A\cap H)_x-x)$$ for some elements $$g_1,\dots,g_N\in G$$. Since $$\mu$$ is shift-invariant and finitely additive, it follows that $$\mu q((A\cap H)_x-x)\le 1/N$$.

Now if $$A$$ is a box then each $$A_x$$ is a box (possible, without its “boundary”) so the correct value of $$\mu(A)$$ should follow from finiteness of a set $$X=\{ x\in\Bbb Z^n: A_x\ne\varnothing\}$$, finite additivity of $$\mu$$, and correct values of $$\mu(A_x)$$ for each $$x\in X$$. The latter should follow from finite additivity of both $$\mu$$ and $$\nu$$, correct values of $$\mu(B)$$ of basic blocks $$B$$ of the form $$B=\Bbb Q^n\cap \prod [r_i/s_i, (r_i+1)/s_i$$, where for each $$i$$, $$0\le r_i are any integers, and a partition of the set $$A_x$$ into a union of basic blocks (up to their boundaries).

References

• I cannot find the property $\nu(G)=1$ in the definition of amenable in that paper. Also, since $\nu$ is a measure and $\nu({a})=0$ for all $a\in G$, It would follow that $\nu(G)=0$, which would not be desirable. – supinf Sep 14 '20 at 10:09
• Also, this answer could be improved by explaining why the resulting measure provides the correct value for boxes, as asked in the question. – supinf Sep 14 '20 at 13:37
• @supinf The condition $\nu(G)=1$ follows from a condition $\mu(G)=1$ in the definition of density at the very beginning of the paper. Also there measures are defined to be finitely additive, but not necessarily $\sigma$-additive. – Alex Ravsky Sep 14 '20 at 22:52
• @supinf I expanded the answer. – Alex Ravsky Sep 14 '20 at 23:32