Folland, Real Analysis Theorem 1.19

Question

Theorem: If $E\subset\mathbb{R}$, the following are equivalent

a.) $E\in M_\mu$

b.) $E = V\setminus N_1$ where $V$ is a $G_\delta$ set and $\mu(N_1) = 0$

c.) $E = H\cup N_2$ where $H$ is a $F_\sigma$ set and $\mu(N_2) = 0$

Background information:

$M_\mu$ denotes the domain of $\mu$ where $\mu$ is a lebesgue-stiltjes measure on $\mathbb{R}$

$G_\delta$ are countable intersections of open sets and $F_\sigma$ are countable unions of closed sets.

Attempted proof: Suppose $E\in M_\mu$ and let $\mu(E) < \infty$. Let, $V$ and $H$ be a $G_\delta$ and $F_\sigma$ set respectively, so $V,H\subset E$. Lets define a set $\mathcal{N} = \{n\in M_\mu:\mu(N) = 0\}$ (note may want to refer to theorem 1.9 in book) and set $$V = \bigcap_{1}^{\infty}V_j \ \ \text{and} \ \ H = \bigcup_{1}^{\infty}H_j$$ I believe we will have to use theorem 1.18 (and I don't really feel like writing all of it down) so from theorem 1.18 we can choose an open $V_j\supset E$ and a compact $H_j\subset E$ such that for $j\in\mathbb{N}$ $$\mu(V_j) - 2^{-j} \leq \mu(E) \leq \mu(H_j) + 2^{-j}$$

I am not sure where to go from here the author provides an unfinished proof of this but I would like to do this on my own, any suggestions is greatly appreciated.

Answer 1

If you are assuming $\mu (E)<\infty $, then I think this will work:

Let $n\in \mathbb N$. Then using the definition of the outer measure, there is are open sets $U_{n+1}\subseteq U_n$ containing $E$ such that

$\mu^* (E)=\mu(E)>\mu (U_n)-1/n$. Thus,

$\tag1\mu (E)\geq \mu\left ( \bigcap _{n\in \mathbb N} U_n \right )$

(because $U_n\subseteq \cdots \subseteq U_1$ implies that $\mu \left ( \bigcap _{n\in \mathbb N} U_n\right )=\lim _{n\to \infty }\mu (U_n)$)

and since clearly $\mu (E)\leq U_n\quad \forall n\in \mathbb N$, this implies now that

$\tag2\mu (E)= \mu\left ( \bigcap _{n\in \mathbb N} U_n \right )$

Now, $E= \bigcap _{n\in \mathbb N}U_n\setminus \left ( \bigcap _{n\in \mathbb N}U_n\setminus E \right )$ so as $\mu (E)<\infty $ we conclude that

$\tag3\mu \left ( \bigcap _{n\in \mathbb N}U_n\setminus E \right )=0$

We may now take $V=\bigcap _{n\in \mathbb N}U_n$ and $N_1=\left ( \bigcap _{n\in \mathbb N}U_n\setminus E \right )$ to see that $a)\Rightarrow b)$.

Now, Lebesgue measure is regular, so there are closed sets $F_n\subseteq F_{n+1}$ such that

$\mu^* (E)=\mu(E)<\mu (F_n)+1/n$. Then, arguing as above we have

$\tag4E=\left ( E\setminus \bigcup _{n\in \mathbb N}F_n \right )\cup \left ( \bigcup _{n\in \mathbb N}F_n \right ) \text {with}\ \mu \left ( E\setminus \bigcup _{n\in \mathbb N}F_n \right )=0$

(we used the fact that $\mu \left ( \bigcup _{n\in \mathbb N} F_n\right )=\lim _{n\to \infty }\mu (F_n)$).

We may now take $H= \bigcup _{n\in \mathbb N}F_n$ and $N_2=\left ( E\setminus \bigcup _{n\in \mathbb N}F_n \right )$so that $b)\Rightarrow c).$

To show that $c)\Rightarrow a)$, it is enough to observe that all sets of measure zero are $\mu $-measureable, for, if $A,B\in \mathbb R$, such that $\mu (B)=0$ then

$\tag5 \mu ^*(A)\geq \mu ^*(A\setminus B)=\mu ^*(A\setminus B)+\mu ^*(A\cap B)$

using the fact that $A\cap B\subseteq B$ and the outer measure is subadditive.

Answer 2

I re-worked my proof but I am not sure it is entirely correct, please provide comments or suggestions if needed

Suppose $E\in M_\mu$, and $\mu(E) < \infty$, by lemma 1.17 $$\mu(E) = \inf\left\lbrace \sum_{1}^{\infty}\mu((a_j,b_j)): E\subset \bigcup_{1}^{\infty} (a_j,b_j)\right\rbrace$$ Since, $E\subset \bigcup_{j}(a_j,b_j) \Rightarrow E\subset \bigcup_{j}\left[\bigcup_{n}[a + 1/n, b - 1/n]\right]$, then $E\in F_\sigma$ and by De'Morgan's law $E\in G_\delta$ as well. By, Theorem 1.18, for $j\in\mathbb{N}$ we can choose an open $O_j\supset E$ and a compact $K_j\subset E$ such that $$\mu(O_j) - 2^{-j} \leq \mu(E) \leq \mu(K_j) + 2^{-j}$$ Let $V = \bigcap_{1}^{\infty}O_j$ and $H = \bigcup_{1}^{\infty}K_j \Rightarrow H\subset E\subset V$. For any j, we have $$\mu(V\setminus E) \leq \mu(O_j\setminus E) < \frac{1}{2^j}$$ So we must have $\mu(V\setminus E) = 0$. Hence, we have $$E = V\setminus (V\setminus E) = V\setminus N_1$$ where $V$ is a $G_\delta$ set and $N_1 = V\setminus E$ is a null set, hence (a)$\Rightarrow$(b). Now, to show (a)$\Rightarrow$(c). Since $K_j$ is closed, $H$ is a $F_\sigma$ set, and $H\subset E$. For each $j$, $$E\setminus H\subset E\setminus K_j$$ so, $$\mu(E\setminus H) \leq \mu(E\setminus K_j) < \frac{1}{2^j}$$ thus, $$\mu(E\setminus H) = 0$$ Thus we have $E = H \cup N_2$, where $N_2 = E\setminus H$ is a null set. Hence, (a)$\Rightarrow$(c). Next we prove (b)$\Rightarrow$(a). Suppose $E = V\setminus N_1$ where $V$ is a $G_\delta$ set and $N_1$ is a null set. We know that $M_\mu$ contains the Borel set, and hence the $G_\delta$ set. Since the $\sigma$-algebra $M_\mu$ is closed under complements and finite intersections $E\in M_\mu$.The proof that (c)$\Rightarrow$(a) is similar. Suppose $E = H \cup N_2$ where $H$ is an $F_\sigma$ set and $N_2$ is a null set. Then $N_2\in M_\mu$ and $H\in M_\mu$. Since $M_\mu$ contains the Borel set and hence the $F_\sigma$ sets and is also closed under finite unions $E\in M_\mu$