I want to prove that if $I$ is any finite interval, and $A\subseteq I$ satisfies $m(A)=m(I)$, then $A$ is dense in $I$.
Edit: Technically I'm given that $m(A^c) = 0$, from which it immediately follows that $A$ is measurable and that $m(A)=b-a$
The proof seemed suspiciously easy, so I just wanted to make sure that this is a legitimately easy proof, and not that it's easy because I lack sufficient imagination for pathological examples.
Proof: Let $I = [a,b]$, and let $A \subseteq I$ satisfy $m(A) = m(I) = b-a$.
We wish to show that if $(u,v)\subseteq I$, then $\exists z \in A$ such that $z\in (u,v)$.
For contradiction, assume no such $z$ exists in $(u,v)$. Then it follows that $(u,v)\subseteq A^c$ so that $m(A^c) > 0$, and thus $m(A) < b-a$.
Thanks in advance for any thoughts.