# Showing Lebesgue Measurable Set is Measure Zero

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 $$a, that $$A$$ must have measure zero. I've been trying to use continuity of measure in some way, but I've been unsuccessful so far.

• Look for: Lebesgue density theorem... en.wikipedia.org/wiki/Lebesgue%27s_density_theorem Commented Jun 29, 2020 at 17:38
• @GEdgar How would I apply the theorem in this particular case? I'm having trouble seeing how to. Commented Jun 29, 2020 at 18:02
• Look at ${1 \over \epsilon} \int_x^{x+\epsilon} 1_A = {1 \over \epsilon} m(A \cap [x,x+\epsilon])$. Commented Jun 29, 2020 at 18:03
• Your inequality shows that the set $A$ has density at most $1/2$ everywhere. But we also know that it has density $1$ almost everywhere on $A$. Therefore, $A$ has measure $0$. Commented Jun 29, 2020 at 18:06

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 $$m(\bigcup_\ell I_\ell) So \begin{align} m(A)&\leq m(\bigcup_\ell I_\ell) So $$m(A)\leq 3\varepsilon$$ for all $$\varepsilon>0$$, showing that $$m(A)=0$$.