# Compact subsets of a set of positive Lebesgue measure having a distinctive property

I noticed the following exercise in a measure theory text. It strikes me as interesting and I wanted to see if any visitors to the site today could assist me in thinking through it.

Suppose $S \subseteq \mathbb{R}$ is a Lebesgue measurable set of positive measure $\mu(S) > 0$. Then for each $\varepsilon \in (0, \mu(S))$, there exists $C \subseteq S$ such that $C$ is compact and $\mu(C) = \varepsilon$.

I am aware of a result saying that the measure of a Lebesgue measurable set $S$ can be approximated by the measure of some open set containing $S$ and that of some compact set contained in $S$, and I was trying to apply that here for about an hour, but to no avail. I would really appreciate some help.

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You have to show that a particular function is continuous. $F(t)=m(B_{t}(0))$ –  checkmath Mar 2 '12 at 1:18
Find a compact set $C$ in $S$ such that $\mu(C)>\epsilon$. $C$ is in some interval $[a,b]$. Let $f(x)=\mu([a,x) \cap S$. $f$ is an increasing continuous function, so you can apply the intermediate value theorem to get what you want. –  ShawnD Mar 2 '12 at 1:20
@chessmath: What is $$B_t(0)? Is it an open or closed ball, or something else perhaps? – Malikah Mar 2 '12 at 2:00 You have to choose a compact C\subset S such m(C)>\varepsilon then the function to be considered is$$F(B(0,t)\cap C)$$. The ball is closed since you are looking for a compact set. – checkmath Mar 2 '12 at 2:03 @Shawn: I see why f is increasing, but I am having trouble with continuity. Is there a particular formulation of continuity that would be best for seeing that f is continuous? Also, is it standard in measure theory (at least with the Lebesgue measure) that you can define continuous functions in terms of the measure of a particular set, and if so, are you aware of standard texts where this is discussed (I ask since I have seen at least one other post on the site where defining such a continuous function solved a visitor's problem)? – Malikah Mar 2 '12 at 23:13 ## 1 Answer Since Lebesgue measure is inner regular, you can find a compact set K \subset S such that \mu(K)>\epsilon. Since K is compact, it is in particular bounded and we can find an interval [a,b] such that K \subset [a,b]. Now define f:[a,b]\rightarrow \mathbb{R} by f(x)=\mu([a,x] \cap K). It is clear that f is increasing. Notice that (if y>x) we have$$f(y)-f(x)=\mu((x,y] \cap K) \leq \mu((x,y])=y-x, from which it immediately follows that $f$ is continuous.

Since $f$ is continuous and $f(a)=0$ and $f(b)=\mu(K)>\epsilon$, it follows by the intermediate value theorem that there is an $x_0$ such that $f(x_0)=\epsilon$. The set $C=[a,x_0]\cap K$ will be a compact set in $S$ with measure $\epsilon$.

I'm not sure if it is a "standard technique" to do something like this. I remember when I took measure theory this was an occasionally useful one to solve homework problems, but I don't remember it being discussed in the text I used (Rudin's text).

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Thanks for the clarification Shawn, it really helped! –  Malikah Mar 2 '12 at 23:59