Let A be a bounded measurable subset of R, and let B be a proper measurable subset of A such that m(B)=m(A). Prove that B is dense in A.
Suppose that B is not dense in A. Then there is a point $\displaystyle x\in A$ and a ball U around x which meets no point of B. Since B is contained in A-U, $\displaystyle m(B)\le m(A-U)$. Now $\displaystyle m(A) = m(B) = m(A-U) + m(U) \ge m(B) + m(U)$, but m(U) must be positive, a contradiction.