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 ?
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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.
Stella Biderman
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Is translation invariant property true for measurable sets only ? – User Feb 14 '16 at 08:35
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In the last line you said, again we are using $A$ is measurable, can you explain please, where you are using measurability of $A$. – User Feb 14 '16 at 08:37
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1@Member No: translation invariance is a property of the Lebesgue outer measure, which is defined on every subset of $\Bbb{R}$. The measurability of $A$ is used to conclude that $1= \sum_i m(A)$ by additivity of the Lebesgue measure. – Crostul Feb 14 '16 at 08:38
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@Crostul, IS that the only place where we are using measurability of $A$? – User Feb 14 '16 at 08:41
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1@Member Yes, it is. – Crostul Feb 14 '16 at 08:42
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@Crostul, okie thank you so much. – User Feb 14 '16 at 08:43
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Edited for clarity at that point. – Stella Biderman Feb 14 '16 at 15:28