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Let $P$ be the set of irrational numbers in $[0,1]$. The set $P$ has Lebesgue measure $1$ and by regularity, there is for every $\epsilon>0$ a compact set $C\subseteq P$ satisfying $\lambda(C)>1-\epsilon$. The proof I know for this fact gives no indication as to how the approximating compact sets might look like. This leaves a big gap in my intuition and I would therefore like to know:

Is there an explicit way to construct for a given $\epsilon>0$ a compact set $C$ of irrational numbers between $0$ and $1$ such that $\lambda(C)>1-\epsilon$?

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up vote 3 down vote accepted

Enumerate $\Bbb Q\cap[0,1]$ as $\{q_n:n\in\Bbb Z^+\}$. Let $$V_{\epsilon}=\bigcup_{n\in\Bbb Z^+}B\left(q_n,\frac{\epsilon}{2^n}\right)\;,$$ and let $K_{\epsilon}=[0,1]\setminus V$.

Note that $K_\epsilon$ is nowhere dense, so it is necessarily a Cantor set.

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Thanx. I should have remembered this construction. –  Michael Greinecker Mar 10 '12 at 13:38

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