Let $A\subset (0,1)$ be a Lebesgue measurable set and $\lambda>0$. Suppose that if $0\le a<b\le 1$ then $\mu(A\cap (a,b))\ge \lambda(b-a)$. Prove that $\mu(A)=1$.
It is clear that $\lambda \le 1$ and to show $\mu(A)=1$ is equivalent to show that $\mu(A^c)=0$.
by the constraction of lebesgue measure,
$\mu (A)=inf\{ \sum \mu(E_n): A \subset \cup E_n\ , E_n$ are disjoint intervals $\}$
hence for any $\epsilon >0$ there are disjoint intervals $E_1 ... E_N$ such that $\mu (A \Delta \cup^N_{n=1}E_n)< \epsilon$.
on the other hand, clearly $(\cup^N_{n=1}E_n)^c=\cup^M_{m=1}F_m$, where $\{F_m\}^M_{m=1}$ are also disjoint intervals (closed/semi-closed/open). Thus, we get:
$\epsilon>\mu (A \cap (\cup^M_{m=1}F_m)) \geq \lambda (1-\mu(\cup^N_{n=1}E_n)) >\lambda(1-\mu(A)+\epsilon)$, so
$\epsilon /\lambda -\epsilon > 1-\mu(A)$ , for any $\epsilon >0$ ,which is $\mu(A)=1$