Problem 3

Exercise 10.3 (Compactess and continuity from above TODO)   Let $E\subseteq\mathbb{R}$ and define $\mathcal{O}_n=\{x\in\mathbb{R} \;\vert\; d(x,E) < 1/n\}$

1. Show that if $E$ is compact, then $\lim\mu(\mathcal{O}_n)=\mu(E)$.
2. Show that the conclusion may be false if $E$ is closed and unbounded or if $E$ is open and bounded.

Proof. If $E$ is compact, then each $\mathcal{O}_n$ is bounded, because $E$ is bounded. Moreover, we have the following inclusions by the monotonicity of $1/n$:

$$\mathcal{O}_1 \supseteq \mathcal{O}_2 \supseteq \cdots \supseteq \mathcal{O}_n \supseteq \cdots$$ (10.15)

To apply the Lebesgue measure's continuity from above, note that

$$\bigcap_{n=1}^\infty \mathcal{O}_n = E$$ (10.16)

because $E$ is closed. Therefore we are free to determine that

$$\lim_{n\to\infty} \mu(\mathcal{O}_n) = \mu\left(\bigcap_{n=1}^\infty \mathcal{O}_n\right)=\mu(E)$$ (10.17)

For a first counterexample, consider the closed and unbounded sequence $E_j=[j,\infty)\cap\mathbb{N}$ with the counting measure. Each set is countable, so the intersection is empty, so the result from continuity does not apply, because

$$\mu\left(\bigcap_{j=1}^\infty E_j\right) = 0 \neq \mu(E_1) - \lim_{j\to\infty} \mu(E_j) = \infty$$ (10.18)

In a similar way, we can exploit $\sigma$-finiteness by setting $E_j = [j,\infty)$ with the Lebesgue measure to observe an identical inequality.

$\qedsymbol$