# Proving a set $E$ satisfying that $m^*(E\cap(a,b))<b-a$ for all $(a,b)$ has zero Lebesgue measure.

Let $m^*$ denote the outer measure corresponding to the Lebesgue measure on $\mathbb{R}$, i.e.,

$$m^*(A)=\inf\{\sum_{n=1}^\infty l(I_n):A\subset\bigcup_{n=1}^\infty I_n\},$$

where $A\subset\mathbb{R}$, $I_n\subset\mathbb{R}$ is a bounded open interval for $n=1,2,\dots$ and $l((a,b))$ is the length of the interval $(a,b)$.

Let $0<\rho<1$. Proof that if $E\subset\mathbb{R}$ and for all intervals $(a,b)$ we have that $m^*(E\cap(a,b))\leq\rho(b-a)$, then $E$ has zero Lebesgue measure.

Commonly, I would add some comments and thoughts about the question, but I'm pretty stuck on this one.

• You might want to see the theorem proven in the first answer here. – Michael Lee Aug 13 '17 at 0:30

Assume first that $E$ is bounded (so $m^*(E)<\infty$) and that $m^*(E)>0$. Let $\varepsilon>0$. Then there exists $I_n$ such that $E\subset\bigcup_{n=1}^\infty I_n$ and $$\sum_{n=1}^\infty l(I_n)\le m^*(E)+\varepsilon\le \sum_{n=1}^\infty m^*(E\cap I_n)+\varepsilon\le \rho \sum_{n=1}^\infty l(I_n)+\varepsilon,$$which implies that $$(1-\rho)\sum_{n=1}^\infty l(I_n)\le \varepsilon.$$ Note that we used the fact that the series is convergent. It follows that $$(1-\rho)m^*(E)\le \varepsilon$$ and so letting $\varepsilon\to 0$ you get $m^*(E)=0$.
If $E$ is not bounded, you apply the previous argument to $E\cap (k-1,k]$ for every $k$.
Lebesgue's Density Theorem states that $m^*$ -almost every point in $E$ is a point of density, where
$x\in E$ is a point of density $\Leftrightarrow\liminf_{y\downarrow x}\frac{m^*(E\cap(x,y))}{y-x}=1$ and $\liminf_{y\uparrow x}\frac{m^*(E\cap(y,x))}{x-y}=1.$
Thus, it is enough to show that $E$ has no points of density, but this is trivial because by hypothesis
$\limsup_{y\downarrow x}\frac{m^*(E\cap(x,y))}{y-x}\le \rho<1$ and $\limsup_{y\uparrow x}\frac{m^*(E\cap(y,x))}{x-y}\le \rho<1.$