I'm trying to show that given $A \subseteq \mathbb{R}$ with $A$ Lebesgue measurable and given that $m(A\cap [a,b]) < \frac{b-a}{2}$ for every $a<b$, that $A$ must have measure zero. I've been trying to use continuity of measure in some way, but I've been unsuccessful so far.
Showing Lebesgue Measurable Set is Measure Zero
lebesgue-measuremeasure-theoryreal-analysis
Related Solutions
We shall be using the following result:
Lemma. if $E$ is Lebesgue measurable and $m(E)>0$, then for every $\alpha\in (0, 1)$, there is an interval $I$ such that $m(E \cap I) > \alpha \, m(I)$.
Proof. It is a consequence of the Lebesgue Density Theorem, according to which, if $m(E)>0$, then for almost every point $x \in E$ $$ \lim_{\varepsilon\to 0} \frac{m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)}{2\varepsilon} = 1. $$ Hence for such an $x$ we can choose an $\varepsilon$, such that $$ \frac{m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)}{2\varepsilon} > \alpha $$ or equivalently $$ m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)> \alpha\,m\big((x-\varepsilon,x+\varepsilon)\big)\big). \tag*{$\Box$} $$
In our case, as $m(E)>0$, let an interval $I=[a,b]$, such that $m(E\cap I)>\frac{4}{5}m(I)$, and set $t=m(I)/5>0$. Clearly, writing $E_t=E+t$, $$ m\big(E_t\cap [a+t,b]\big)=m\big(E\cap [a,b-t]\big) \ge m(E\cap I)-m\big([b-t,b]\big)\ge \tfrac{3}{5}m(I), $$ and similarly $$ m\big(E\cap [a+t,b-t]\big),\,m\big(E_t\cap [a+t,b-t]\big)\ge\frac{2}{5}m(I), $$ while $m\big([a+t,b-t]\big)=\frac{3}{5}m(I)$. Therefore $$ m(E\cap E_t)\ge m\big(E\cap E_t\cap [a+t,b-t]\big)\ge\frac{1}{5}m(I). $$
You showed that $l^*(E) = \inf l^*(\mathcal{O})$ (the infimum runs over all open sets $\mathcal{O}$ containing $E$).
Suppose $E$ is of finite exterior measure -then for any $n\in \mathbb{N}$ we have an open set $\mathcal{O_n}$ such that $$l^*(\mathcal{O_n}) \leq l^*(E)+n^{-1}<\infty$$ And since $l^*(\mathcal{O_n}) = l^*(E)+l^*(\mathcal{O_n} \setminus E)$ we have $l^*(\mathcal{O_n} \setminus E) \leq n^{-1}$ (The equality is infact a restatement of the measurability of $E$, see Equivalent Definition of Measurable set)
Taking $\mathcal{O} = \bigcap_{n=1}^\infty \mathcal{O_n}$ we have come up with a $G_\delta$ set (countable intersection of open sets) which is obviously a borel set, with the property: $$l^*(\mathcal{O}\setminus E) \leq l^* (\mathcal{O_n}\setminus E) \leq n^{-1}$$ for all $n\in\mathbb{N}$ which is the first corollary.
If $E$ is of infinte exterior measure then we denote $E_n = E \cap B(0,n)$ ($E$'s intersection with the ball of radius $n$ centered at the origin). Each $E_n$ is bounded and so has finite exterior measure (Since it obviously is encompassed within $B(0,n)$) And we may extract $G_n$ sets such that $$l^*(G_n \setminus E_n)=0$$ Putting $G=\bigcup_{n=1}^\infty G_n$ we have $$l^*(G\setminus E) = l^*\left(\bigcup_{n=1}^\infty G_n \setminus \bigcup_{n=1}^\infty E_n \right)\leq \sum_{n=1}^\infty l^*(G_n \setminus E_n ) =0$$
This shows the first corollary.
For the second: For every measurable $E^C$ we obviously have a borel set $G\supset E^C$ such that $l^*(G\setminus E^C) =0$ (Take the set from the first claim) that shows that $G\setminus E^C$ is a lebesgue measurable set as it is a null set. Then obviously $E^C= G \setminus (G\setminus E^C)= G \cap (G\cap E)^C$. Taking completements on both sides we have:$$ E = G^C \cup (G\setminus E^C)$$
$G^C$ is again a borel set (As a completement of borel set).
As a final comment on this excercise: One notices we used 2 main properties of the Lebesgue measure:
- It is outer regular meaning that for every measurable subset $E$ , for every $\varepsilon>0$ we have an open set $\mathcal{O}\supset E$ with the property $l^*(\mathcal{O} \setminus E) \leq \varepsilon$ which we had to prove.
- It is defined on borel sets, bounded sets have finite exterior-measure.
It turns out that the second implies the first and the first property is what we actually needed. In this proof I made a slight detour (as i wasn't sure which Lebesgue measurability criterion you are using).
Best Answer
By definition of outer Lebesgue measure (or by regularity, depending on how you define Lebesgue measure), given $\varepsilon>0$ there exist disjoint intervals $I_1,\ldots,I_r$, $I_\ell=(a_\ell,b_\ell)$ such that $A\subset \bigcup_\ell I_\ell$ and $$ m(\bigcup_\ell I_\ell)<m(A)+\varepsilon. $$ So \begin{align} m(A)&\leq m(\bigcup_\ell I_\ell)<m(A)+\varepsilon= m(A\cap\bigcup_\ell I_\ell)+\varepsilon =\sum_\ell m(A\cap(a_\ell,b_\ell))+\varepsilon\\[0.3cm] &<\frac12\,\sum_\ell(b_\ell-a_\ell)+\varepsilon =\frac12\,m(\bigcup_\ell I_\ell)+\varepsilon\\[0.3cm] &\leq\frac12\,m(A)+\frac{3\varepsilon}2. \end{align} So $$ m(A)\leq 3\varepsilon $$ for all $\varepsilon>0$, showing that $m(A)=0$.