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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))$2012-03-02
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    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.2012-03-02
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    @chessmath: What is $$B_t(0)$? Is it an open or closed ball, or something else perhaps?2012-03-02
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    You have to choose a compact $C\subset S$ such $m(C)>\varepsilon$ then the function to be considered is $$F(B(0,t)\cap C)$$. The ball is closed since you are looking for a compact set.2012-03-02
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    @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)?2012-03-02

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