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Let $\lambda$ be the Lebesgue measure on $\mathbb R$.

Let $A \subset \mathbb R$ be a measurable set such that $\lambda(A)>0$.

Question:

Can we always find two measurable sets $K_1, K \subset \mathbb R$ such that

  • $K_1 \subset A$
  • $K \subset \mathbb R$
  • $K_1 \subset K$
  • $K$ compact
  • $\lambda(K_1) = \lambda(K)$
  • $\lambda(K_1)>0$

?

Motivation why $K_1$ and $K$:

$K_1$ does not have to be compact... Imagine $A=[0,1] \cap \mathbb Q^c$, then there is no compact set included, but $K_1=A$ and $K=[0,1]$ would satisfy the question...

mimi
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    Is $K$ and $A$ fixed or do you want to find both $K_1$ and $K$? – John Nov 13 '14 at 10:48
  • Yes both. Sorry I ll edit that. – mimi Nov 13 '14 at 10:48
  • Your example does not work, since $A$ is numerable and thus $\lambda(K_1)=0\neq \lambda(K)=1$. Maybe take $A=[0,1]-\mathbb{Q}$. – J.A.L Nov 13 '14 at 11:02
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    I think you are missing some important restriction (for example $\lambda(K_1)=\lambda(A)$, or $\lambda(K_1)>0$). Otherwise just let $K_1=K={x}$ for some $x\in A$. – Quang Hoang Nov 13 '14 at 11:19
  • Sorry yes two errors in the question: Of cause $\lambda(K_1) >0$ and i wanted to set $A=[0,1] \cap \mathbb Q^c$... I ll change that – mimi Nov 13 '14 at 11:50
  • Why do you have two sets $K$ and $K_1$? If $\lambda(A)\gt0$ then there is a compact set $K\subseteq A$ with $\lambda(K)\gt0$. So just let $K_1=K$. – bof Nov 13 '14 at 11:56
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    Does $\mathbb Q^c$ mean the complement of $\mathbb Q$? Are you aware that the set of irrational numbers in $[0,1]$ contains a compact set of measure $0.999$? – bof Nov 13 '14 at 12:01
  • Yes the complement of $\mathbb Q$... How to define this compact set? – mimi Nov 13 '14 at 12:42
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    Construct an open set $U$ such that $\mathbb Q\subseteq U$ and $\lambda(U)\lt0.001$. (This is easily done using the countability of $\mathbb Q$ and countable additivity of Lebesgue measure.) Then $[0,1]\setminus U$ is a compact set of measure $\gt0.999$. – bof Nov 13 '14 at 12:56

1 Answers1

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Yes. You can argue it using the fact Lebesgue measure on $\mathbb{R}$ is inner regular, hence a measurable set $A$ can be approximated by compact subset from inside.

John
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  • As pointed out by Quang Hoang. I forgot to add the restriction $\lambda(K_1)>0$. I have added it now... With this additional restriction your example does NOT work. – mimi Nov 13 '14 at 11:52
  • @mimi I have edited my answer. – John Nov 13 '14 at 12:08
  • This seems to be correct... But this implies that there is a compact set $K \subset [0,1] \cap \mathbb Q^c$ of measure $1-\epsilon$.... But is this really possible? – mimi Nov 13 '14 at 12:40
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    @mimi See here – John Nov 13 '14 at 12:45