I saw this question while looking through some real analysis PhD qualifier exams.
Let $E$ be a Lebesgue measurable subset of $[0,1]$. Let $c$ be a positive constant such that for all $0\leq a,b\leq1$, we have $$m(E \cap [a,b]) \geq c(b-a).$$ Prove that $m(E)=1$.
My approach is to show that the complement $F := [0,1] \backslash E$ has measure zero. By countable additivity of the Lebesgue measure, we have
$$m(E \cap [a,b])+m(F \cap [a,b]) = b-a \implies (1-c)(b-a) \geq m(F\cap [a,b]).$$
Let $\{x_k\}$ enumerate the rational numbers in $\mathbb{Q}$, and for a fixed $\epsilon > 0$ and each $k$ we consider the interval $$I_k = [x_k – \frac{\epsilon}{(1-c)2^{k+1}}, x_k+\frac{\epsilon}{(1-c)2^{k+1}}].$$
Then
$$m(F) = m(F \cap [0,1]) = \sum ^\infty _{k=1} m(F \cap I_k)\leq (1-c)\sum _{k=1}^\infty \frac{\epsilon}{(1-c)2^k} = \epsilon.$$
Here I made the assumption that the intervals $I_k$ would cover $[0,1]$, but I'm not sure if this is actually true.
EDIT. Okay the assumption is definitely not true, since the intervals are getting smaller and thus we cannot cover all irrational numbers in $[0,1]$. How do we work around it?
Best Answer
Take any $\delta >0.$ Let $D$ be a closed set with $D\subset E$ and $m(D)>m(E)-\delta.$ Let $V$ be a family of open intervals with $\bigcup V\supset D$ and $m([0,1]\cap (\,\bigcup V\,))<m(D)+\delta.$
Now $D$ is compact so there exists a finite $W\subset V$ with $\bigcup W\supset D.$ So $m([0,1]\cap (\,\bigcup W\,))\le m([0,1]\cap (\,\bigcup V\,))<m(D)+\delta\le m(E)+\delta.$
Let $T=[0,1]\setminus \bigcup W.$ Then $m(([0,1]\setminus D)\cap E)\ge m(T\cap E).$ But the complement in $[0,1]$ of a finite union of open intervals is the union of a finite set of pair-wise disjoint intervals (including degenerate 1-point intervals), so $m(T\cap E)\ge c\cdot m(T).$ Therefore $$m(E)=m(D)+m(([0,1]\setminus D)\cap E)\ge$$ $$\ge m(D)+m(T\cap E)\ge$$ $$\ge m(E)-\delta+c\cdot m(T)=$$ $$=m(E)-\delta+c(1-m([0,1]\cap (\,\bigcup W\,))\,)\ge$$ $$\ge m(E)-\delta+c(1-m(E)-\delta))$$ which simplifies to $$\delta\ge c(1-m(E)-\delta).$$ This is not possible for every $\delta>0$ unless $m(E)=1.$