# Proof of Outer Regularity of Lebesgue Measure on $\mathbb{R}$

Let $$E \subseteq \mathbb{R}$$ be a measurable set, and $$\varepsilon > 0$$. Show that there exists an open set $$G \supset E$$ such that $$\mu(G \setminus E) < \epsilon$$.

By the definition of Lebesgue measure we can find a countable collection of open intervals $$(A_{n})_{n \in \mathbb{N}}$$ such that $$\sum\limits_{n=1}^{\infty} |A_n| \leq \mu(E) + \epsilon.$$

And I think setting $$G = \cup_{n \in \mathbb{N}}A_n$$ gives us what we want. But what about if $$E$$ is of infinite measure?

• Try consider the family $B_N=[-N,N]$. This is a countable collection of measurable sets having finite measure and that cover $\mathbb{R}$. Approximate $E_N=E\cap B_N$ with an open set $G_N$ and then take the union. – Giuseppe Negro Oct 23 '14 at 16:04

The key element here is that $$\mathbb{R}$$ is $$\sigma$$-finite. The rest is nitty-gritty and a $$\sum_{n=1}^\infty {1 \over 2^n} = 1$$ trick.

Suppose for each set $$E$$ of finite measure and each $$\epsilon >0$$ you can find some open set $$G$$ containing $$E$$ with $$\mu(G \setminus E) < \varepsilon$$.

Now let $$E$$ be and $$\varepsilon>0$$ be arbitrary, and let $$E_n = E \cap (n, n+1]$$. Now let choose open $$G_n$$ such that $$E_n \subset G_n \qquad \text{and} \qquad \mu(G_n \setminus E_n) < \varepsilon {1 \over 3\cdot2^{|n|}}.$$

Now let $$G = \bigcup_n G_n$$, which is open. By monotonicity, we have $$G \setminus E = \bigcup_n G_n \setminus E \subset \bigcup_n G_n \setminus E_n \implies \mu(G \setminus E) \le \sum_n \mu (G_n \setminus E_n ) < \varepsilon.$$

Elaboration: $$G \setminus E = (\bigcup_n G_n) \setminus E = \bigcup_n (G_n \setminus E)$$. Now, since $$G_n \setminus E \subset G_n \setminus E_n$$, we have $$\bigcup_n (G_n \setminus E) \subset \bigcup_n (G_n \setminus E_n)$$.

• Thanks! This looks good. However I'm not sure about the final line. We have $\cup_n G_n \setminus E \subset (\cup_n G_n) \setminus E_n$, but surely not necessarily $\cup_n G_n \setminus E \subset \cup_n (G_n \setminus E_n)$ which is what we need for the conclusion? – Tom Offer Oct 23 '14 at 21:11
• I added an elaboration. – copper.hat Oct 23 '14 at 21:27
• Oh yeah. Can Just take the set difference inside the union first... thanks! I had thought that we'd need to use some sort of covering of $\mathbb{R}$ to do this for sets of infinite measure. The $E_n$ you suggest don't even cover $\mathbb{R}^+$, but it seems to work. I didn't expect it to! – Tom Offer Oct 23 '14 at 21:32
• Well $A \setminus B = A \cap B^c$, so $(\cup_n A) \setminus B = (\cup_n A) \cap B^c = \cup_n (A \cap B^c) = \cap_n (A \setminus B)$. – copper.hat Oct 23 '14 at 21:34
• Why do you care about a cover of $\mathbb{R}^+$??? All you need is a collection of disjoint sets $E_n$ such that the measures are bounded and the union is $E$. Since $\mathbb{R} = \cup_n (n,n+1]$, you know that $E = \cup_n E_n$. – copper.hat Oct 23 '14 at 21:36