If $f:[0,1]\to\mathbb{R}$ where $S\subseteq [0,1]$, and we define the following
-
$I=[0,1]$
-
$\left(I_k\right)_{k=1}^{m}$ are $m$ open sub-intervals of $I$
-
$\ell(I)=1$ is the length of $I$
-
$\ell(I_k)=c\in\mathbb{R}^{+}$ is the length of $I_k$ for $k=1,…,m$
$$\Omega(S\cap I_k)=\begin{cases}
0 & S\cap I_k \ \text{is countable}\\
1 & S\cap I_k \ \text{is uncountable} \\
\end{cases}$$
$\mu^{*}(c,S)$ is the outer set function defined as
\begin{align*}
& \mu^{*}(c,S)= \inf\limits_{m\in\mathbb{N}}\left\{ \sum\limits_{k=1}^{m}c\ \Omega\left(S\cap I_{k}\right): S\subseteq\bigcup\limits_{k=1}^{m}I_{k}\right\}
\end{align*}
The inner set function $\mu_{*}(c,S)$ is
$\mu_{*}(c,S)=\mu^{*}(c,[0,1])-\mu^{*}(c, [0,1]\setminus S)$
And $\lim\limits_{c\to 0}\mu(c,S)$ is defined when
$$\lim_{c\to 0}\mu(c,S)=\lim_{c\to 0}\mu^{*}(c,S)=\lim_{c\to 0}\mu_{*}(c,S)$$
Then
Is $\lim\limits_{c\to 0}\mu(c,S)$ equivelant to the Lebesgue Measure of $S$?
Best Answer
The answer is negative. In fact there is a set $S\subset[0,1]$ of Lebesgue measure $0$ such that $S\cap(a,b)$ is uncountable for any $0\leqslant a<b\leqslant 1$ (thus, for such an $S$, we have $\mu^*(c,S)=1$ and $\mu_*(c,S)=0$ for any $0<c<1$). The idea to build $S$ is to take the Cantor set and "fill in the holes" by its homothetic copies.
We define it explicitly. Let $S$ be the set of all numbers $x\in[0,1]$ representable by a ternary fraction with only a finite number of $1$s, i.e. such that there are $a_n\in\{0,1,2\}$, $n\geqslant 1$, with $x=\sum_{n=1}^\infty 3^{-n} a_n$, and the set $\{n : a_n=1\}$ is finite.
$S$ is clearly uncountable. Moreover, for any $n>0$ and $0\leqslant k<3^n$, the map $x\mapsto f_{n,k}(x):=3^{-n}(k+x)$ is a bijection between $S$ and $S\cap I_{n,k}$, where $I_{n,k}:=[3^{-n}k,3^{-n}(k+1)]$ (informally, we prepend the $n$-digit ternary representation of $k$ to $x$ for $x\neq 1$); thus, $S\cap I_{n,k}$ is uncountable too. But any open interval $(a,b)$, $0\leqslant a<b\leqslant 1$, contains an $I_{n,k}$ for some $n$ and $k$; thus, $S\cap(a,b)$ is uncountable as well.
Further, as promised again, $S=\bigcup_{n=0}^\infty\bigcup_{k=0}^{3^n-1}f_{n,k}(C)$, where $C$ is the Cantor set (for, take $1\neq x\in S$, let $n$ be the greatest with $a_n=1$ ($n=0$ if there's none), and let $k=\lfloor 3^n x\rfloor$; the reverse inclusion is clear). That is, $S$ is a countable union of sets of Lebesgue measure $0$. Then, $S$ itself has Lebesgue measure $0$.