The measure of a sequence of open sets

Question

Another fun qualifying exam practice question!

Suppose $E$ is a given set and $O_n$ is the open set defined as $O_n = \left\{x: \operatorname{dist}(x,E) < \frac1n\right\}$

a) Show that if $E$ is compact then $m(E) = \lim_{n \to \infty} m(O_n)$

What I have so far: Define $B_k = \bigcap^k_{i=0}O_n = O_k$

Then $B_{k+1} \subseteq B_k$

Also $\lim_{n \rightarrow \infty}B_k = \bigcap^\infty_{i=0}O_n = E$ (Since $E$ is closed and bounded? I'm not really sure why this is true!)

So $B_k$ converges downward to $E$. Since measures are monotone, $m(B_k)$ converges downward to $m(E)$

So $\lim_{k \rightarrow \infty}m(B_k) =\lim_{k \rightarrow \infty}m(\bigcap^k_{n=1}O_n) =\lim_{k \rightarrow \infty}m(O_k)=m(E)$

Basically I'm wondering how compactness plays into this argument.

b) Show that the conclusion may be false for $E$ closed and unbounded or open and bounded.

Since I'm not sure how compactness plays into the above argument I'm having a hard time coming up with a counter example.

Answer 1

If $x\in\bigcap_{n=1}^{\infty}O_n$ then $d(x,E)<\frac1n$ for every positive integer $n$ and this can only be true iff $d(x,E)=0$ or equivalently $x\in\overline E$.

$E$ is compact hence closed so that $E=\overline E$, so we have $E=\bigcap_{n=1}^{\infty}O_n$.

Actually here we only need that $E$ is closed, but the condition of boundedness comes in to ensure that $m(O_n)<\infty$ for some $n$.

Under that extra condition your solution works.

If we would e.g. take $E=\mathbb R\times\{0\}\subseteq\mathbb R^2$ then $m(O_n)=\infty$ for every $n$.

Answer 2

You need closedness for $\bigcap_n O_n = E$ to hold: if $x$ lies in the intersection it means that $d(x,E)=0$ which is equivalent to $x \in \overline{E}$, so $x \in E$ if $E$ is closed. So $\bigcap_n O_n = \overline{E}$ holds in general. We do not need boundedness of $E$ at all, that I can see. The continuity argument for $m$ works when we have finite measure for $O_1$ (And hence all $O_n$) and it could fail for infinite measures.

The $B_k$ are unnecessary as the $O_n$ are already decreasing themselves: $x \in O_{n+1}$ or $d(x, E) < \frac{1}{n+1}$ trivially implies $d(x,E) < \frac{1}{n}$ or $x \in O_n$, as $\frac{1}{n+1} < \frac{1}{n}$, so $O_{n+1} \subseteq O_n$. It's the standard way to write a closed set as the intersection of countably many open sets in a metric space.

Answer 3

So far none of the answers address part (b) of your question fully, so I will give you two examples below:

1) To see why being bounded is crucial to the argument, consider $E=\mathbb{Z}\subset\mathbb{R}$.

2) To see why being closed is crucial consider an enumeration of the rationals inside the unit interval, say $(q_k)$, and place an open interval on top of each $q_k$ of length $\epsilon\cdot2^{-k}$ in order to construct your set $E$.

Answer 4

I'm assuming you mean that $E\subseteq\mathbb{R}$.

(a) Note that $\bigcap_{n=1}^\infty O_n=\{x:d(x,E)=0\}$, which is just the closure of $E$, which since $E$ is compact is just $E$ itself. By continuity of Lebesgue measure together with $O_n\supseteq O_{n+1}$ we now have $$m(E)=m\left(\bigcap_{n=1}^\infty O_n\right)=\lim_{n\to\infty}m(O_n).$$ (Note that the second equality holds if and only if the measures are finite.)

(b) For closed and unbounded $E$, take $E=\mathbb{N}$. For open and bounded $E$, take the complement of the fat Cantor set in $[0,1]$.

Good luck with the Quals! That brings back memories.