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Suppose $A$ is a given set in $\mathbb R^d$, and $O_n$ is the open set

$$O_n = \left\{x\in \mathbb R^d, d(x,A)< \frac{1}{n}\right\}.$$

show that:

  1. If $A$ is bounded, then $\lim_{n\to\infty}\lambda(O_n)=\lambda(\bar A).$ Where $\lambda$ is the Lebesgue measure and $\bar A$ is the closure of $A$.

  2. Give an example of an unbounded and closed set $A$ such that the inequality in (1) is not true.

Here is my thinking:

For 1, since $O_i$'s are decreasing in sense of inclusion, and $O_1$ has finite measure. So I have $\lim_{n\to\infty}\lambda(O_n)=\lambda(\lim_{n\to\infty}O_n).$ I only need to show that $\lim O_n = \bar A$ (or maybe differ from a set of measure $0$?) How can I prove this part?

For 3, I don't have any idea how to find this counterexample. Any thoughts?

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up vote 1 down vote accepted

For 1, your conclusion should be $\lim_{n\to\infty} \lambda(O_n) = \lambda(\bigcap_n O_n)$. What's left is to show that $\bigcap_n O_n = \overline{A}$. Notice that if $x \in \bigcap_n O_n$, then $d(x, A) = 0$.

For 2, consider a line in $\mathbb R^2$. The measure of the line is $0$, but the measure of each $O_n$ is $\infty$.

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Let $C_n = \{x\in \mathbb R^d, d(x,A) \le \frac{1}{n}\}$, note $C_n$ is closed and $A \subset C_{n+1} \subset O_n \subset C_n$. Hence $C= \cap_n C_n$ is closed, and $A \subset C$.

Furthermore, $C = \{x\in \mathbb R^d, d(x,A) =0\} = \overline{A}$. Since $A$ is bounded, $\lambda O_1 < \infty$, and so $\lim_n \lambda O_n = \lim_n \lambda C_n = \lambda C = \lambda \overline{A}$.

For a counterexample, take $A= \mathbb{Z}$. Then $\lambda O_n = \infty$, but $\lambda A = 0$.

In fact, any unbounded set of measure zero will do.

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Why do you construct the family $C_n$? The same argument can be done via $O_n$. – Ayman Hourieh Sep 23 '13 at 21:35
@AymanHourieh: To be sure, to be sure :-). My initial thought was that it would be easier to show the intersection of the $C_n$ was the closure, but as you have pointed out, it was entirely unnecessary. – copper.hat Sep 23 '13 at 21:45

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