0
$\begingroup$

So, here is the formal statement:

Let $m^*$ denote the Lebesgue outer measure on $\mathbb{R^n}$, and suppose $E \subset \mathbb{R^n}$ with $m^*(E) < \infty$. Let $\sigma_m = \{x\in \mathbb{R^n} : d(x,E) < \frac{1}{m}\}$. Show that if $E$ is compact, then $m^*(E) = \lim_{m \to \infty } m^*(\sigma_m)$.

I basically have come up with a proof attempt, but unfortunately, I'm highly suspicious of it because I didn't really use the fact that $E$ was compact. I was hoping that someone can point out a flaw in my argument or provide suggestions on how to more clearly incorporate this hypothesis if it turns out I'm implicitly using it.

Proof Attempt: Fix $\epsilon > 0$ and choose an open cover $Q = \bigcup_{j=1}^{\infty} Q_j$ of $E$ such that $$\sum_{j=1}^{\infty} m^*(Q_j) \leq m^*(E) + \epsilon$$

Since $m^*$ possesses countable subadditivity, it follows that $m^*(Q) \leq \sum_{j=1}^{\infty} m^*(Q_j)$. But this implies $m^*(Q) \leq m^*(E) + \epsilon$. Now, (I suspect this is where my argument begins to fail) for all $m$ large enough, we can have $\sigma_m \subset Q$, and by the monotonicity of $m^*$, it will follow that $m^*(\sigma_m) \leq m^*(Q)$ for all $m$ large enough. But then, this means that $m^*(\sigma_m) \leq m^*(E) + \epsilon \implies m^*(\sigma_m) - m^*(E) \leq \epsilon$. Since we notice that $E \subset \sigma_m$ for every $m$, it follows by monotonicity that $m^*(E) \leq m^*(\sigma_m) \implies 0 \leq m^*(\sigma_m) - m^*(E)$. Thus, for all $m$ large enough, $|m^*(\sigma_m) - m^*(E)| = m^*(\sigma_m) - m^*(E) \leq \epsilon$. This is what we wished to show.

$\endgroup$
  • 1
    $\begingroup$ A simpler argument: $E$ and $\sigma_m$ are all measurable sets of finite measure and $\sigma_m$ decreases to $E$. QED $\endgroup$ – Kavi Rama Murthy Feb 12 at 6:05
  • 2
    $\begingroup$ @KaviRamaMurthy, if $E$ is open, there are counterexamples of measurable sets of finite measure where $\lim_{m \to \infty} m^*(\sigma_m) \neq m^*(E)$ $\endgroup$ – clay Feb 12 at 15:49
  • $\begingroup$ @clay I don't understand why you are making this comment. It is given in the question that $E$ is compact and my argument uses the fact that $E$ is closed. $\endgroup$ – Kavi Rama Murthy Feb 12 at 23:11
  • $\begingroup$ @KaviRamaMurthy, I can't see how you can conclude that $\sigma_m$ would necessarily decrease to $E$. Or, more specifically that $m^*(\sigma_m) \to m^*(E)$. Yes, it is given that $E$ is closed, and I'm merely thinking of an open set counter-example to validate or invalidate your argument. $\endgroup$ – clay Feb 12 at 23:48
  • 1
    $\begingroup$ @clay The argument is very simple. If $d(x,E) <\frac 1 n$ fore all $n$ then there exist points $x_n \in E$ such that $d(x,x_n) <\frac 1 n$. This implies that $x_n \to x$. Since $E$ is closed, we get $x \in E$. $\endgroup$ – Kavi Rama Murthy Feb 12 at 23:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.