Vitali set of outer measure 1 How to construct a Vitali set of outer measure 1. I couldn't understand the argument given here. Isn't there any easier way? I would also like if someone explains that to me.
Thank you in advance!
 A: Since  $\mathbb{R}$ is a vector space over $\mathbb{R}$, there exists a
$\mathbb{Q}$ vector subspace $V$ of $\mathbb{R}$ such that $V\oplus \mathbb{Q} = \mathbb{R}$ ( it uses the existence of a basis of $\mathbb{R}$ extending $\{1\}$, so it uses the Zorn lemma).
Now, $V$ is a Vitali set, and moreover $\mathbb{Q}^{\times}\cdot  V = V$.  This is the only thing we will use.
Now, since $\cup_{q\in \mathbb{Q}} (V+ q) = \mathbb{R}$, the exterior measure of $V$ is $>0$. (in fact, since $V$ is invariant under multiplication by non-zero rationals, $\mu^{*}(V) = \infty$). Nevertheless, since $\mu^{*}(V)>0$, for every $\epsilon > 0$ there exists an interval of form $I=[\frac{m}{n}, \frac{m+1}{n}] $ such that $\mu^{*}(V\cap I) > (1-\epsilon) \mu(I)$. Now, since $n V= V$, we get
$$\mu^{*}(V\cap I) > (1-\epsilon) \mu(I)$$
for $I=[m, m+1]$.
Now consider the Vitali set $\tilde V$ consisting of all the fractional parts of the elements of $V$
$$\tilde V \colon = \{ \{x\} \ | \ x \in V\}$$
that is
$$\tilde V = \bigcup_{m\in \mathbb{Z}} (V\cap [m, m+1)) - m$$
Clearly $\tilde V \subset [0,1)$, and moreover, from the above, for every $\epsilon > 0$, we have $\mu^{*}(\tilde V) > 1-\epsilon$. We conclude that $\mu^*(\tilde V) = 1$.
Note that if we define $\tilde V_q \colon= \{ \{x\} \ | \ x \in V+ q\}$, then we have the partition of $[0,1)$ into countably many disjoint subsets of exterior measure $1$
$$[0,1) = \bigcup_{q \in \mathbb{Q}\cap [0,1)} \tilde V_q$$
A: create a vitali set like $V$ in $\mathbb{R}$. It's exterior measure $m_*(V)$ is none-zero. now the set $\frac{1}{m_*(V)}V$ that is a scale of elements of $V$ is a set with exterior measure 1.
