# Show example $E \subset \mathbb{R}^n$ where exterior measure does not equal that of the smallest open ball cover.

Let $$m^*$$ be Lebesgue outer measure (also called exterior measure) on $$\mathbb{R}^n$$. Suppose $$E$$ is a subset of $$\mathbb{R}^n$$ with $$m^*(E) < \infty$$. Let $$\mathcal{O}_m$$ be the open set: \begin{align*} \mathcal{O}_m &= \left\{ x : d(x, E) < \frac{1}{m} \right\} \\ \end{align*} Part (a): show that if $$E$$ is compact, then: \begin{align*} m^*(E) &= \lim\limits_{m \to \infty} m^*(\mathcal{O}_m) \\ \end{align*} Part (b): Give examples to show that this property may not hold for cases where $$E$$ is closed and unbounded or $$E$$ is open and bounded.

My work:

$$\mathcal{O}_m$$ is an open ball covering of $$E$$ with ball radius $$\epsilon = 1/m$$

The precise definition of Lebesgue outer measure is:

\begin{align*} m^*(E) &= \inf \sum\limits_{j=1}^{\infty} |Q_j| \\ \end{align*}

taken over all countable coverings of closed cubes: $$E \subseteq \cup_{j=1}^\infty Q_j$$

I suspect I can work out a proof for (a) with the Vitali covering lemma.

I'm mainly looking for help on part (b). I can't think of any irregular open and bounded sets that would break this property. By definition, all open sets are Lebesgue measurable. I also can't think of any closed but unbounded set that would break this property.

Example when $$E$$ is closed:

Let $$E = [1,1+1/2] \cup [2,2+1/4] \cup [3,3+1/8] \cup \cdots = \bigcup_{k=1}^\infty [k,k+2^{-k}]$$.

Then $$m(E) = 1$$, but $$m(O_m) = \infty$$ for all $$m$$.

Example when $$E$$ is open:

Let $$\{r_k\}$$ denote an enumeration of the rational numbers in $$(0,1)$$.

Fix $$\epsilon > 0$$ and for each $$k$$ choose $$\delta_k$$ so that the interval $$I_k = (r_k - \delta_k,r_k+\delta_k)$$ satifies

• $$I_k \subset (0,1)$$, and
• $$\ell(I_k) < \dfrac{\epsilon}{2^k}$$.

One way to do this is to take $$\delta_k = \min\{ \epsilon/2^{k+1}, \mathrm{dist}(x,\{0,1\}))$$.

Let $$E = \cup_k I_k$$. Then $$E$$ is open and $$m(E) \le \sum \ell(I_k) < \epsilon$$.

On the other hand, since $$\{r_k\} \subset E$$ each set $$O_m$$ contains $$[0,1]$$ so that $$m(O_m) \ge 1$$ for all $$m$$.

• wow, thanks!!!! – clay Feb 11 '19 at 22:38