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 $m$.

  • $\begingroup$ wow, thanks!!!! $\endgroup$ – clay Feb 11 at 22:38

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