I have this strange question: Let $A\subseteq [0,1]$ be Lebesgue measurable and suppose there exists $c>0$ such that, for all subintervals $J\subseteq [0,1]$, it holds $\lambda(A\cap J)\geq c\lambda(J)$. Prove $\lambda(A)= 1$.
My thoughts: Choosing $J=[0,1]$ we have $\lambda(A)\geq c$. So we need just show $c=1$. I begin by letting $c<1$ but I cannot see how to get a contradiction... Is this the right approach? Any help/hints?
Thanks in advance...
I think this requires Lebesgue's Theorem. Let $f=\chi_A$. Then $\frac 1 {b-a} \int_a^{b} f \geq c$ whenever $0<a<b<1$. Letting $b \to a$ in this and using Lebesgue's Theorem we get $f(a) \geq c$ almost everywhere. Thus $\chi_A (a) \geq c$ for almost all $a$. Since $c>0$ this means $A^{c}$ has measure $0$, i.e. $\lambda (A)=$1.