# Measure of set intersecting every compact subset of R

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.

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 $\mathbb R$
• 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 $\mathbb R$. Dec 22, 2023 at 10:23