Let $X$ in $\mathbb{R}$ be a closed and bounded set. Define $X_\epsilon = \{x: d(x, X) < \epsilon\}$ for $\epsilon > 0$. I want to prove that $\lim_{\epsilon \to 0} m(X_{\epsilon}) = m(X)$? Suggestions?
Sequence of sets
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0Due to the modifications made to the question, it is now impossible to understand Robert's and David's mention of a second part. One should not do this. Please revert to the original question. – 2012-03-17
4 Answers
Take any sequence $\epsilon_n \searrow 0$, notice that $X_{\epsilon_n} \subset X_{\epsilon_{n-1}}$ and that $\bar{X}=X=\bigcap_{n} X_{\epsilon_{n}}$. Now use continuity from above.
Can you use the Lebesgue Dominated Convergence Theorem, or the Monotone Convergence Theorem, or the outer regularity of Lebesgue measure?
For the second part, think of a sequence that increases to $+\infty$ but where the "gaps" go to $0$.
Hints for the first part: Let $\epsilon>0$. Choose an open set $O$ containing $E$ such that $m(O\setminus E)<\epsilon$. Argue that you can take $O$ to be a finite union of disjoint open intervals (recall any open set is a countable union of disjoint open intervals). Next, argue that there is an \epsilon'>0 such that for each $x\in E$, the open ball B_{\epsilon'}(x)\subset O. Also, note $E_\epsilon=\bigcup\limits_{x\in E} B_\epsilon(x)$.
Hint for the second part: Think of the set of rational numbers.
It's not hard to see that $X_\epsilon$ is open for all $\epsilon$, for the interval $(x-\gamma, x+\gamma) \subseteq X_\epsilon$ when $\gamma < \frac12(\varepsilon - d(x,X))$. Moreover, $X = \cap X_\varepsilon$. Hence the outer measure of $X$ is $\lim_{\varepsilon\to 0} m(X_\varepsilon)$ by definition of outer measure, and since $X$ is closed it's also measurable. Thus its measure equals its outer measure.