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?

  • $\begingroup$ 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. $\endgroup$ – 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 $\bigcup_n (G_n \setminus E) \subset \bigcup_n (G_n \setminus E_n)$.

  • $\begingroup$ 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? $\endgroup$ – Tom Offer Oct 23 '14 at 21:11
  • $\begingroup$ I added an elaboration. $\endgroup$ – copper.hat Oct 23 '14 at 21:27
  • $\begingroup$ 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! $\endgroup$ – Tom Offer Oct 23 '14 at 21:32
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    $\begingroup$ Well $A \setminus B = A \cap B^c$, so $(\cup_n A) \setminus B = (\cup_n A) \cap B^c = \cup_n (A \cap B^c) = \cap_n (A \setminus B)$. $\endgroup$ – copper.hat Oct 23 '14 at 21:34
  • $\begingroup$ 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 $E = \cup_n E_n$. $\endgroup$ – copper.hat Oct 23 '14 at 21:36

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