Why do we prefer countable subadditivity to countable additivity?

Question

I have read that countable additivity is not sufficient for assigning a measure to every subset of the reals in order to give a notion of the length of an interval. That we should instead use the weaker idea of countable subadditivity.

An example of the problem is: Take two real numbers and consider them equivalent if there difference is rational. Let $E$ be the subset of the half unit interval that contains exactly one element of each equivalence class. Assume all translates of $E$ have the same measure. Then countable additivity implies that the unit interval has measure zero or infinity.

1. I understand up to the part "Let E be the subset of the half unit interval that contains exactly one element of each equivalence class." I don't understand then how it is true that the unit interval has measure zero or infinity?
2. Does this example fully describe the issue with countable additivity or are there other problems that arise due to it?

Are there

Answer 1

The problem stated is the following: (Assuming the axiom of choice,)There is no map $\mu \colon \mathfrak P(\mathbf R)\to [0,\infty]$ satisfying the following three desirable properties for a notion of length:

(1) $\mu(A + x) = \mu(A)$ for any $A \in \mathfrak P(\mathbf R), x \in \mathbf R$ (translation invariance)
(2) $\mu\left(\biguplus_n A_n\right) = \sum_n \mu(A_n)$ for any disjoint sequence of subsets $A_n \subseteq \mathbf R$ (countable additivity)
(3) $\mu([0,1)) = 1$.

For let us suppose, that $\mu$ has (1) and (2). Define an equivalence relation on $\mathbf R$ by $$ x \sim y \iff x-y \in \mathbf Q $$ Then for every $x \in \mathbf R$, there is an equivalent number in $[0,\frac 12)$, hence we can choose a set $E \subseteq [0,\frac 12)$ that contains one element of every equivalence class (here we use choice!). Then the sets $E + q$ for $q \in \mathbf Q\cap [0,\frac 12)$ are pairwise disjoint and cover the unit intervall. Therefore $$ [0,1) = \biguplus_{q \in \mathbf Q \cap [0,\frac 12)} (E+q) $$ Countable addivity gives $$ \mu([0,1)) = \sum_{q\in \mathbf Q \cap [0,\frac 12)} \mu(E+q) = \sum_{q\in \mathbf Q \cap [0,\frac 12)} \mu(E) $$ If now $\mu(E) = 0$ the right hand side is $0$, therefore $\mu([0,1))=0$. If $\mu(E) > 0$ the right hand side is $\infty$, hence $\mu([0,1)) = \infty$. That is translation invariance and countable additivity together force the unit intervall to have "length" $0$ or $\infty$.