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 $D$ =$\overline{D} = [0,1]$.
Lebesgue outer measure and closure of a set
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real-analysis
outer-measure
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0Consider the contrapositive and use monotonicity of outer measure. – 2017-01-29
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0What if we take $D=(2,3]$? What is $m^*(D)$ and what is $\overline{D}$? – 2017-01-29
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0@ΘΣΦGenSan Sorry I should be more clear. D is a subset of [0,1]. – 2017-01-29
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0Then edit your question correctly please – 2017-01-29
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0@ΘΣΦGenSan Will do! :) – 2017-01-29
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0@Wilson i think $I_{k}$ are the countable open coverings?..Also $D$ is a subset of $[0,1]$ , then how can $\overline{D} = [0,1]$ , if $D$ would have been $(0,1)$ then we could have said that $\overline{D}=[0,1]$ ? – 2017-01-29
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0@Wilson also see..http://math.stackexchange.com/questions/1480147/outer-measure-of-set-equals-outer-measure-of-closure?rq=1 – 2017-01-29
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0@BAYMAX Say D=[0,1]nQ, then closure of D = [0,1]. – 2017-01-29
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0ok..i missed the outer measure of $D$ bieng 1. – 2017-01-29
2 Answers
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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 $1?$
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0Yeah, thanks man! This is exactly what i came up with an hour ago when i was randomly scribbling down some ideas... hahaha – 2017-01-29
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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 $$m^*(D)\leq m([0,1] \backslash J)=1-m(J)<1.$$