# Real Analysis, Folland Theorem 1.18 Borel measures on the real line

Background information - We fix a complete Lebesgue-Stiltjes measure $$\mu$$ on $$\mathbb{R}$$ associated to the increasing right continuous function $$F$$, and we denote by $$M_{\mu}$$, the domain of $$\mu$$.

1.17 Lemma - For any $$E\in M_{\mu}$$, $$\mu(E) = \inf\{\sum_{1}^{\infty}\mu((a_j,b_j)):E\subset \bigcup_{1}^{\infty}(a_j,b_j)$$

Theorem 1.18 - If $$E\in M_{\mu}$$, then \begin{align*} \mu(E) &= \inf\{\mu(U):E\subset U, U \ \text{open}\}\\ &=\sup\{\mu(K):E\subset K, K \ \text{compact}\}\end{align*}

Proof - By lemma 1.17, for any $$\epsilon > 0$$ there exists intervals $$(a_j,b_j)$$ such that $$E\subset \bigcup_{1}^{\infty}(a_j,b_j)$$ and $$\sum_{1}^{\infty}\mu((a_j,b_j))\leq \mu(E) + \epsilon$$. If $$U = \bigcup_{1}^{\infty}(a_j,b_j)$$ then $$U$$ is open, $$E\subset U$$, and $$\mu(U)\leq \sum_{1}^{\infty}\mu((a_j,b_j))\leq \mu(E) + \epsilon$$. Since $$E\subset U$$ then $$\mu(E)\leq \mu(U)$$ so the first part is done.

I am not sure how to prove the second part, specifically following follands proof I am confused how when we choose $$\overline{E}\setminus E\subset U$$ such that $$\mu(U)\leq \mu(\overline{E}\setminus E) + \epsilon$$. Then if we let $$K = \overline{E}\setminus U$$ this somehow makes $$K$$ compact?

• Recall that for subsets of $\mathbb R$, compact $\iff$ closed and bounded. The assumption at that part of the proof is that $E$ is bounded. Consequently, $\overline{E}$ is also bounded. Now $\overline{E}$ and $U^c$ are closed, so $K = \overline{E}\setminus U = \overline{E} \cap U^c$ is also closed. As $K$ is a subset of $\overline{E}$, it is also bounded. Since $K$ is closed and bounded, it is compact.
– user169852
Commented Jul 8, 2016 at 2:41
• I see thanks for that. Commented Jul 8, 2016 at 2:42
• Oops, sorry, I didn't notice that there was already an answer below, saying the same thing.
– user169852
Commented Jul 8, 2016 at 2:44
• It's ok don't worry about it. As an aside why is $E_j = E\cap [-j,j]$ where $j\in \mathbb{N}$ bounded? Commented Jul 8, 2016 at 2:46
• Well, $[-j,j]$ is bounded, and $E \cap [-j,j]$ is a subset of $[-j,j]$. Any subset of a bounded set is bounded.
– user169852
Commented Jul 8, 2016 at 2:52

Proof of part 1 is fine. I have just copied it to add few more details. The proof of part 2 I have also written it in detail.

If $$E\in M_{\mu}$$, then \begin{align*} \mu(E) &= \inf\{\mu(U):E\subset U, U \ \text{open}\}\\ &=\sup\{\mu(K):E\subset K, K \ \text{compact}\}\end{align*}

Proof -

(Part 1) Let us prove $$\mu(E) = \inf\{\mu(U):E\subset U, U \ \text{open}\}$$.

By lemma 1.17, for any $$\epsilon > 0$$ there exists intervals $$(a_j,b_j)$$ such that $$E\subset \bigcup_{1}^{\infty}(a_j,b_j)$$ and $$\sum_{1}^{\infty}(a_j,b_j)\leq \mu(E) + \epsilon$$. If $$U = \bigcup_{1}^{\infty}(a_j,b_j)$$ then $$U$$ is open, $$E\subset U$$, and $$\mu(U)\leq \sum_{1}^{\infty}(a_j,b_j)\leq \mu(E) + \epsilon$$. Since $$E\subset U$$ then $$\mu(E)\leq \mu(U)$$, we have that $$\mu(E)\leq \mu(U) \leq \mu(E) + \epsilon$$ So, we have proved that, for any $$\epsilon > 0$$, there $$U$$ such that $$U\supset E$$, $$U$$ is open and
$$\mu(E)\leq \mu(U) \leq \mu(E) + \epsilon$$ so the first part is done.

(Part 2) Let us prove $$\mu(E) =\sup\{\mu(K):K\subset E, K \ \text{compact}\}$$.

First suppose $$E$$ is bounded. Then there is $$n\in \mathbb{N}$$ such that $$E\subset [-n,n]$$. Then, $$[-n,n]\setminus E \in M_{\mu}$$ and then, by part 1, $$\mu([-n,n]\setminus E) = \inf\{\mu(U):([-n,n]\setminus E)\subset U, U \ \text{open}\}$$ So , for any $$\varepsilon>0$$, there is $$U$$ such that $$([-n,n]\setminus E)\subset U$$, $$U$$ is open and $$\mu([-n,n]\setminus E)\leq \mu(U) \leq \mu([-n,n]\setminus E) + \epsilon$$ So we have, for any $$\varepsilon>0$$, there is $$U$$ such that $$([-n,n]\setminus U)\subset E$$, $$U$$ is open and $$\mu(E)=\mu([-n,n])-\mu([-n,n]\setminus E)\geq \mu([-n,n])-\mu(U) \geq \\ \geq \mu([-n,n])-\mu([-n,n]\setminus E) -\varepsilon=\mu(E)-\varepsilon$$ Note that $$([-n,n]\setminus U)$$ is closed and bounded, so it is compact and $$\mu([-n,n]\setminus U)=\mu([-n,n])-\mu(U)$$. So we proved that for any $$\varepsilon>0$$, there is $$K$$ (take $$K=[-n,n]\setminus U$$) such that $$K\subset E$$, $$K$$ is compact and $$\mu(E)\geq \mu(K) \geq \mu(E)-\varepsilon$$ So we have proved that, if $$E\in M_{\mu}$$ is bounded then $$\mu(E) =\sup\{\mu(K):K\subset E, K \ \text{compact}\}$$

If $$E$$ is not bounded, then define, for $$j\in \mathbb{Z}$$, $$Ej=E\cap (j,j+1]$$. Note that for each $$j$$, $$E_j\in M_{\mu}$$, $$E_j$$ is bounded. Moreover, $$\{E_j\}_{j \in \mathbb{Z}}$$ is a family of disjoint sets. Now, given any $$\varepsilon>0$$, since each $$E_j$$ is in $$M_{\mu}$$ and is bounded, there is $$K_j$$ compact, such that $$K_j\subset E_j$$ and $$\mu(E_j)\geq \mu(K_j) \geq \mu(E_j)-\frac{\varepsilon}{2^{|j|+2}}$$ Let $$H_n=\bigcup_{-n}^{n-1} K_j$$. We have that $$H_n$$ is compact and $$H_n \subset E$$. We also have \begin{align*}\mu(E) \geq \mu(H_n)=\sum_{j=-n}^{n-1}\mu(K_j) & \geq \sum_{j=-n}^{n-1}\mu(E_j)-\sum_{j=-n}^{n-1}\frac{\varepsilon}{2^{|j|+2}}= \\ & =\mu(E\cap (-n,n])-\sum_{j=-n}^{n-1}\frac{\varepsilon}{2^{|j|+2}}\geq \\ & \geq \mu(E\cap (-n,n])-\frac{3}{4}\varepsilon \end{align*} Since $$(E\cap (-n,n])_{n\in\mathbb{N}}$$ is a non decreasing sequece of sets and $$E=\bigcup_n (E\cap (-n,n])$$ we have that $$\lim_{n\to\infty}\mu(E\cap (-n,n])=\mu(E)$$. So take $$n_0$$ such that $$\mu(E\cap (-n_0,n_0])\geq\mu(E)-\frac{\varepsilon}{4}$$ So we get \begin{align*}\mu(E) \geq \mu(H_n)\geq \mu(E\cap (-n,n])-\frac{3}{4}\varepsilon \geq\mu(E)-\varepsilon\end{align*} So we have proved that, if $$E\in M_{\mu}$$ (bounded or not) then $$\mu(E) =\sup\{\mu(K):K\subset E, K \ \text{compact}\}$$

Remark: An alternative proof for the case where $$E$$ is not bounded in (Part 2) is as follows:

If $$E$$ is not bounded, since for all $$K\subset E$$, $$\mu(K)\leq \mu(E)$$, we $$\sup\{\mu(K):K\subset E, K \ \text{compact}\} \leq \mu(E)$$

Define, for $$j\in \mathbb{N}$$,$$j>1$$ $$E_j=E\cap [-j,j]$$. Note that $$\{E_j\}_j$$ is a monotone non-decreasing sequence of measurable sets and $$E=\bigcup_jE_j$$ So, since $$\mu$$ is continuous from below, we have $$\lim_{j\to \infty}\mu(E_j)=\mu(E)$$.

Now, given any $$\varepsilon>0$$, since each $$E_j$$ is in $$\mathcal{L}^n$$ and is bounded, there is $$K_j$$ compact, such that $$K_j\subset E_j$$ and $$\mu(E_j)-\frac{\varepsilon}{j}\leq \mu(K_j) \leq \mu(E_j)$$ So we have $$\lim_{j\to \infty}\mu(K_j)=\mu(E)$$. So we have $$\mu(E) =\sup\{\mu(K):K\subset E, K \ \text{compact}\}$$

• Isn't $\sum_{j=-n}^n\mu(E_j) = \mu(E\cap (-n,n + 1])$ since $Ej=E\cap (j,j+1]$??
– Abel
Commented Jul 23, 2022 at 14:34
• @Abel , You are right. Thanks for pointing it. I have edited my answer, to correct this. The proof itself remains the same. I preferred to use $\sum_{j=-n}^{n-1}\mu(E_j)=\mu(E\cap (-n,n])$. Commented Jul 24, 2022 at 4:53

In the part of the proof you're referring to, $E$ is assumed to be bounded. Therefore $K=\overline{E}\setminus U$ is also bounded, and since $K=\overline{E}\cap U^c$ is closed it follows that $K$ is compact.