Proving Vitali set is Non-measurable set The well known example of non-measurable set is the following. $\mathbb{Q}$ is normal subgroup of $\mathbb{R}$. Further, every real number $x$ can be written as $x_1+\frac{p}{q}$, where $x_1\in[0,1)$. Thus, we choose representatives $N=\{x_{\alpha}\colon \alpha\in J\}$, $J$ some indexing set, such that $\mathbb{R}=\cup_{\alpha\in J} x_{\alpha}+\mathbb{Q}$. Then it is proved that $N$ is a non-measurable subset of $[0,1)$. 
We can observe that, we don't know the exact position of elements of $N$; for they may actually lie in $[0,\frac{1}{2})$, or $[0,\frac{1}{3})$ or so many infinite possibilities of intervals where $N$ will lie. From this, can I conclude that $N$ should not be measurable?

Slight Explaination: Given any real $x$, we can write it as $x_1+\frac{p}{q}$, where $x_1$ is in $[0,\frac{1}{2})$ (simply consider the quotient of $\mathbb{R}$ by $\mathbb{Z}[\frac{1}{2}]=\{\frac{n}{2}\colon n\in\mathbb{Z}\})$, hence coset representatives of $\mathbb{Q}$ in $\mathbb{R}$ can be chosen from $[0,\frac{1}{2})$. Similarly, $\mathbb{R}$ can be written as $\cup_{\alpha} (x_{\alpha}+\mathbb{Q})$, where $x_{\alpha}\in [0,\frac{1}{n})$ where $n\in\mathbb{N}$, and is fixed. Thus, we don't know where $N=\{x_{\alpha}\}_{\alpha}$ can sit in an interval, so can we conclude that $N$ is non-measurable
 A: I wasn't able to follow your argument. Not knowing the exact position of elements of $N$ is insufficient to conclude nonmeasurability; indeed, if I choose a completely random countable subset of $\mathbb R$, then I know it is measurable (it has measure zero).
The usual argument to verify that the Vitali set is nonmeasurable is to observe that if we enumerate the rationals as $(q_i)_{i=1}^{\infty}$ and define $N_i = N + q_i$ (translate the set $N$ by the rational offset $q_i$, modulo the interval $[0,1]$), then the collection $\{N_i\}_{i=1}^{\infty}$ is a countable partition of $[0,1]$, in other words, the $N_i$'s are pairwise disjoint and $\cup_{i=1}^{\infty}N_i = [0,1]$.
Suppose $N$ is measurable. Since each $N_i$ is a translation of $N$ (modulo $[0,1]$), and Lebesgue measure is translation-invariant, all of the $N_i$'s have the same measure, namely $\mu(N)$. By countable additivity, we then have
$$1 = \mu([0,1]) = \mu\left(\bigcup_{i=1}^{\infty} N_i\right) = \sum_{i=1}^{\infty} \mu(N_i) = \sum_{i=1}^{\infty} \mu(N)$$
But this is impossible: the sum on the right hand side is either $0$ or $\infty$; it can't be $1$.
