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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?

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    Due 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 4

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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.

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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 $0$.

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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.

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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.