Showing Lebesgue Measurable Set is Measure Zero

Question

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.

Answer 1

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 $m(A)=0$.