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A question I recently saw while studying measures on $\mathbb R$:

Let $A$ be a set such that $A\cap K\neq \emptyset$ for any compact subset $K$ of real line of positive measure. Show that $A$ has infinite measure.

Intuitively, it feels as though this restriction on $A$ means that it must 'spread out' evenly throughout $\mathbb R$, but it is hard for me to formalize this in a way that helps tackle the problem. I know that $A$ must be dense in $\mathbb R$, though that doesn't really tell me much. I know that if $m(A)<\infty$, then $m(A-[-n,n])\rightarrow 0$ as $n\rightarrow \infty$ and it feels like this cannot be true under the conditions stated, but I can't figure out exactly why.

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If $m(A) <\infty$ then $m(A^{c}) >0$. By regularity of Lebesgue measure there is a compact set $K \subset A^{c}$ with $m(K)>0$ and this contradicts the hypotheis. Ref. for regularity: Inner regularity of Lebesgue measure for $\mathbb R$

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  • $\begingroup$ Is there a way to answer this question without directly using this fact? Or do you know a slick proof of it? The definition of Lebesgue measurability I have been working with so far is that with an outer measure $m^*(A)$ defined as $\inf\{\sum_{i=1}^{\infty} b_i-a_i| A\subset \cup_{i=1}^{\infty} (a_i,b_i)\}$ and then $E$ is defined as measurable iff $m^*(A)=m^*(A\cap E)+m^*(A-E)$ for every subset $A$ of $\mathbb R$. $\endgroup$ Dec 22, 2023 at 10:23
  • $\begingroup$ @CardioidAss22 I don't see how you can do this from the definition. $\endgroup$ Dec 22, 2023 at 23:11

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