Non measurable set In first answer of this question The construction of a Vitali set, how did we came to conclusion that $A$ is not measurable ?
How that series doesn't converge $\Rightarrow A$ is not measurable ? 
 A: If a set, $A$, is measurable, then it has certain properties. It's a proof by contradiction. First lets copy the relevant portion

Why is the set not measurable? Well, note that if we take the
  translations $A_q=\{a+q\pmod 1\mid a\in A\}$ for every $q\in\mathbb
> Q\cap[0,1)$ then we have a countably family of pairwise disjoint sets
  whose union is exactly $[0,1]$. Since the Lebesgue measure is
  translation invariant all sets must have the same measure.
We have that $m([0,1])=\sum m(A_q)=\sum_{i=1}^\infty m(A)$. However
  the sum of a constant positive number can either be $0$ or $\infty$,
  neither of which is the measure of $[0,1]$. This shows that the set
  $A$ cannot be measurable .

Assume $A$ is measurable then it follows, from properties of the Lebesgue Outer Measure, that the union in the first quoted paragraph has measure equal to $1$, and that each set in the union has the same measure, since they are translation invariant (this holds for all sets). Then we use the fact that measure is additive for disjoint sets to conclude that $$m([0,1])=\sum m(A_q)=\sum_{i=1}^\infty m(A)$$ again, we are assuming that $A$ is measurable, however, this sum is impossible, since there is no number $N$ such that $\sum_0^\infty N=1$, so we conclude that $A$ was not measurable.
