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 $i$, $0\le r_i<s_i$ are any integers, and a partition of the set $A_x$ into a union of basic blocks (up to their boundaries).


[Ban] Taras Banakh, Extremal densities and measures on groups and $G$-spaces and their combinatorial applications.

  • $\begingroup$ 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. $\endgroup$ – supinf Sep 14 '20 at 10:09
  • $\begingroup$ Also, this answer could be improved by explaining why the resulting measure provides the correct value for boxes, as asked in the question. $\endgroup$ – supinf Sep 14 '20 at 13:37
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    $\begingroup$ @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. $\endgroup$ – Alex Ravsky Sep 14 '20 at 22:52
  • $\begingroup$ @supinf I expanded the answer. $\endgroup$ – Alex Ravsky Sep 14 '20 at 23:32

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