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?
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 $\bigcup_n (G_n \setminus E) \subset \bigcup_n (G_n \setminus E_n)$.
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