Lebesgue outer measure and closure of a set

Question

I think it should be true? I was thinking to use the definition of Lebesgue outer measure saying that m*($D$) $= 1 =$ SUM of length of $I_{k}$ where $D$ is a subset of the SUM of $I_{k}$, then $D$ =$\overline{D} = [0,1]$.

Answer 1

Let's see. If $D$ is a subset of $[0,1]$ of outer measure $1,$ then $\overline D$ is a closed set of outer measure at least $1,$ right? And $\overline D$ is still a subset of $[0,1]$ (because $[0,1]$ is closed), and $\overline D$ is measurable (because closed sets are measurable), and $\overline D$ has measure exactly $1$ (because $[0,1]$ has measure $1$). Now, what can we say about a closed subset of $[0,1]$ which has measure $1?$

Answer 2

Suppose $p\in [0,1]$ \ $\bar D\; $. Then $(-r+p,r+p)\cap D=\phi$ for some $r>0$, and $J=(-r+p,r+p)\cap [0,1]$ is an interval of positive length . Now $[0,1]$ \ $J$ is an interval or the union of two intervals, and covers $D.$ So $$m^*(D)\leq m([0,1] \backslash J)=1-m(J)<1.$$