# Given $E \subset \mathbb{R^n}$ with $m^*(E) < \infty$. Show that if $E$ is compact then $m^*(E) = \lim_{m \to \infty} m^*(\sigma_m)$

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.

• A simpler argument: $E$ and $\sigma_m$ are all measurable sets of finite measure and $\sigma_m$ decreases to $E$. QED – Kavi Rama Murthy Feb 12 at 6:05
• @KaviRamaMurthy, if $E$ is open, there are counterexamples of measurable sets of finite measure where $\lim_{m \to \infty} m^*(\sigma_m) \neq m^*(E)$ – clay Feb 12 at 15:49
• @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. – Kavi Rama Murthy Feb 12 at 23:11
• @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. – clay Feb 12 at 23:48
• @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 $x \in E$. – Kavi Rama Murthy Feb 12 at 23:56