Why do we prefer countable subadditivity to countable additivity? 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.


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*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?

*Does this example fully describe the issue with countable additivity or are there other problems that arise due to it?


Are there
 A: 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 $\infty$.
