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Is the complement of Cantor set $C$ still measure zero? Meanwhile, I know its accumulation point is $C$ itself (right?). So its closure would be $C$, correct? Why??? Notice: I am asking for the complement of Cantor set $C$ over closed interval $[0,1]$.

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Let $C$ be the Cantor set and $X=[0,1] \setminus C$ its complement in the unit interval. Then $m(X)=m([0,1])-m(C)=1$, so the complement has measure $1$. Furthermore the Cantor set is nowhere dense, so its complement must be dense and thus $\overline{X}=[0,1]$.

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