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.

  • 2
    $\begingroup$ Look for: Lebesgue density theorem... en.wikipedia.org/wiki/Lebesgue%27s_density_theorem $\endgroup$
    – GEdgar
    Commented Jun 29, 2020 at 17:38
  • $\begingroup$ @GEdgar How would I apply the theorem in this particular case? I'm having trouble seeing how to. $\endgroup$
    – Newman
    Commented Jun 29, 2020 at 18:02
  • 1
    $\begingroup$ Look at ${1 \over \epsilon} \int_x^{x+\epsilon} 1_A = {1 \over \epsilon} m(A \cap [x,x+\epsilon]) $. $\endgroup$
    – copper.hat
    Commented Jun 29, 2020 at 18:03
  • $\begingroup$ 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 $0$. $\endgroup$
    – GEdgar
    Commented Jun 29, 2020 at 18:06

1 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 $m(A)=0$.


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