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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.

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  • $\begingroup$ The proof seems to be backward. You should explain how the sets $V_j$, $H_j$ are chosen before trying to take unions and intersections of them. $\endgroup$ – Nate Eldredge Sep 3 '15 at 3:05
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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.

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  • $\begingroup$ What do you mean that Lebesgue measure is regular? $\endgroup$ – Wolfy Sep 3 '15 at 15:45
  • $\begingroup$ @MorganWeiss How much measure theory have you done? do you know what inner and outer measure are? Id not, I will be happy to supply a more detailed proof of the above or give you a reference. Folland is a great, but challenging text. $\endgroup$ – Matematleta Sep 4 '15 at 5:00
  • $\begingroup$ This is my first measure theory course, anything you think would help that would be great I am re-doing my proof and I almost have (a)$\Rightarrow$ (b) $\endgroup$ – Wolfy Sep 4 '15 at 18:33
  • $\begingroup$ @MorganWeiss: OK. Did you see the construction of $\mu $ from $\mu ^*$? If so, the the proof I provided above for $a\Rightarrow b$ just uses the definiton of outer measure. By the way it works for the infinite case also, by considering the intervals $[-n,n]$. For $b\Rightarrow c$ if you have not seen inner measure you can do it directly by applying the the previous argument to the measureable set $E^c$ $\endgroup$ – Matematleta Sep 5 '15 at 0:49
  • $\begingroup$ Yes, I believe I understand it. I think your proof is more clean then mine I will just have to spend sometime re-doing it over and over again $\endgroup$ – Wolfy Sep 5 '15 at 22:19
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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 $E\in M_\mu$

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  • $\begingroup$ you seem to be claiming $E$ is both a $G_{\delta }$ and $F_{\sigma }$ which is not necessarily true, in fact it's not in generalThe point is, $E$ is "close" to being one in terms of the measure. Notice also you do not need to be explicit about the intervals. You only need the definition of outer measure and to know what an open set is in $\mathbb R$with usual metric. When you pass to the infinite case you will use the fact that $\mathbb R$ with the usual metric is $\sigma$ compact and then you will use the intervals $[-n,n]$. For $b\Rightarrow c$ do the same proof with $E^c$ in place of $E$. $\endgroup$ – Matematleta Sep 5 '15 at 0:59

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