A Math5011 Exercise

In a discussion with some students in this course I find that I have another solution different from the official one.

Problem. Let $\mathcal N$ denote the Vitali set in $[0,1]$, show that $m^*([0,1]\setminus \mathcal N)=1$.

Remark. Here $\mathcal N$ is a set of those representatives of classes in $[0,1]/\!\!\sim$, where $\sim$ is an equivalence relation on $[0,1]$ given by $x\sim y\iff x-y\in \Q$, and we have $[x]=(x+\Q)\cap [0,1]$. Of course we know that $\mathcal N$ being a nonmeasurable set must satisfy $m^*(\mathcal N)>0$.

Solution. This directly follows form

Claim. $m_*(A)=m(L)-m^*(L\setminus A)$ for any closed set $\bm{L\supseteq A}$ that is bounded. 

Proof. To see this, note that
\begin{align*}
m(L)-m^*(L\setminus A) &= \inf\{m(L)-\lambda(U):U\supseteq L\setminus A,U \text{ open}\}\\
&=\inf\{m(L\setminus U) : L\setminus U\subseteq A,U\text{ open}\}\\
&=\inf\{\lambda(K):K\subseteq A,K\text{ closed, contained in }L\},
\end{align*} finally note that since $A\subseteq L$ and $L$ is bounded, we have
\[
\{K\subseteq \R:K\subseteq A,K\text{ closed, contained in }L\} = \{K:K\text{ compact and contained in }A\}
\] it follows that $m(L)-m^*(L\setminus A)=m_*(A)$.$\qed$

The claim says that to do inner approximation of $A$, we can do outer approximation of $L\setminus A$ first (provided that $L\supseteq A$) and then subtract this extract part from $\lambda(L)$ to get inner approximation of $A$.

Now the problem is readily solved in the following way: Set $L=[0,1]$, $A=\mathcal N$, then \[m_*(\mathcal N) = 1- m^*([0,1]\setminus \mathcal N),\] and since measurable subset of $\mathcal N$ must have measure zero (same proof as $m(\mathcal N)$ if it were measurable), and we are done.$\qed$