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Let $A \subset \mathbb{R}$ be a boundary set ($\operatorname{Int}A=\emptyset$) and Lebesgue measure $0$ ($\lambda (A) = 0$). What can we say about Lebesgue measure of $\operatorname{Cl}A$ ?

It is obviously that we have $ \lambda \left( \operatorname{Cl}A \right) \ge 0$. But what we can say about the second inequality?

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The rationals have something to say... –  David Mitra Jan 1 at 11:27
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Note that $\operatorname{Int}(\mathbb{Q}\cap[0,1])=\varnothing$, though $\lambda(\operatorname{Cl}(\mathbb{Q}\cap[0,1]))=\lambda([0,1])=1$

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So if $\operatorname{Int}A = \emptyset$ then $ \lambda (\operatorname{Cl}A) = \lambda(A)$ ? –  Thomas Jan 1 at 11:38
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No, this is not true even for my example. Look carefully! –  Norbert Jan 1 at 13:15
    
Yes, my mistake. I don't have any idea how can me help yours answer. I understand what have you wrote, but I don't know how can I use it to find bounded of Lebesgue'a measure of set $A$. Could you give me next hint? –  Thomas Jan 1 at 13:35
    
I do not understand thi sentence "how can I use it to find bounded of Lebesgue'a measure of set $A$" –  Norbert Jan 1 at 14:08
    
I mean $\operatorname{Cl}A$. Because I would like find $a$ such that $\lambda(\operatorname{Cl}A) \le a$. I don't know how can I use your answer to get it $a$. –  Thomas Jan 1 at 14:20
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