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I made a post about this theorem before. But I decided to create a new post to see if I am proving this theorem correctly.

Theorem 1.19 - If $E\subset \mathbb{R}$, TFAE:

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 $\mu(N_2) = 0$

Proof a.) implies b.) - If $E\subset\mathbb{R}$ and $E\in M_{\mu}$ then by lemma 1.17 $$\mu(E) = \inf\{\sum_{1}^{\infty}\mu((a_j,b_j)):E\subset \bigcup_{1}^{\infty}(a_j,b_j)\}$$ so, $E\subset \bigcup_{1}^{\infty}(a_j,b_j)$ and $\bigcup_{1}^{\infty}(a_j,b_j)$ is open. Let $U_n = \bigcup_{1}^{\infty}(a_j,b_j)$ then clearly $U_{n+1}\subseteq U_n$ that contains $E$ such that $$\mu(E) > \mu(U_n) - \frac{1}{n}$$ Thus, $$\mu(E)\geq \mu\left(\bigcap_{1}^{\infty} U_n\right)$$ Since by monotonocity, $\mu(E)\leq \mu(U_n)$ for all $n$, this implies that $$\mu(E) = \mu\left(\bigcap_{1}^{\infty}U_n\right)$$ Set $V = \bigcap_{1}^{\infty}U_n$ and $N_1 = \left(\bigcap_{1}^{\infty}U_n\setminus E\right)$ then clearly $E = V\setminus N_1$ where $V$ is a $F_{\delta}$ set and $\mu(N_1) = 0$. Thus a.) implies b.).

Proof a.) implies c.) Following the same reasoning as above we see that taking the complement of $V$ and setting equal to $H$ we get $$H = \bigcup_{1}^{\infty}U_n^c$$ which is a $F_{\sigma}$ set. Recall that $$E = V\setminus N_1 = V\cap N_1^c$$ and $$(V\cap N_1^c)^c = H\cup N_1$$ where $N_1 = \left(\bigcap_{1}^{\infty}U_n\setminus E\right)$ and $N_1^c = \bigcup_{1}^{\infty}U_n^c\cup E$ set $N_1^c = N_2$ since $\mu(N_1) = 0$ then $\mu(N_2) = 0$ and thus a.) implies c.).

Proof c.) implies a.) not sure

I am not sure if this is correct any suggestions is greatly appreciated.

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Your proof is essentially correct for the first part. I will reword it following Follands and adding some commments and including the other parts.

Theorem 1.19 - 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 $\mu(N_2) = 0$

Proof a.) implies b.) If $E\subset\mathbb{R}$ and $E\in M_{\mu}$ then, by theorem 1.18, for each $n\in \mathbb{N}$, $n>0$, there is an open set $U_n$ such that $E\subset U_n$ and $$\mu(E)+ \frac{1}{n} \geq \mu(U_n) \geq \mu(E)$$ Since $E\subset \bigcap_{1}^{\infty} U_n$, we have, for each $n\in \mathbb{N}$, $n>0$, $$\mu(E)+ \frac{1}{n} > \mu(U_n)\geq \mu\left(\bigcap_{1}^{\infty} U_n\right) \geq \mu(E)$$ Thus, $$\mu(E)= \mu\left(\bigcap_{1}^{\infty} U_n\right)$$

Set $V = \bigcap_{1}^{\infty}U_n$. $V$ is a $G_{\delta}$, $E \subset V$ and $\mu(V)=\mu(E)$. Set $N_1 = V \setminus E$ then, since $E,V \in M_\mu$, we have $N_1 \in M_\mu$. Since $E\subset V$, we have $E=V\setminus N_1$.

If $\mu(E)<+\infty$ then $\mu(N_1)=\mu(V)-\mu(E) =0$

Remark: Folland's book stops here and leaves the general case to the reader (exercise 25). However, since the countable union of $G_\delta$ may not be a $G_\delta$, it is not straight forward to use the fact $\mu$ is $\sigma$-finite to extend the result for sets of finite measures to the general case. In fact, proving the general case is more like re-doing the proof completely than simply extending the result for sets of finite measures.

General Case:

If $E\subset\mathbb{R}$ and $E\in M_{\mu}$, then since $\mu$ is $\sigma$-finite, we have that there are $\{E_k\}_1^\infty \subset M_\mu$ such that $\{E_k\}_1^\infty$ is a family of disjoint sets, $\mu(E_k) <+\infty$, for all $k$ and $E=\bigcup_1^\infty E_k$. Then, by theorem 1.18, we have, for each $k$, for each $n\in \mathbb{N}$, $n>0$, there is an open set $U_{k,n}$ such that $E_k\subset U_{k,n}$ and $$\mu(E_k)+ \frac{1}{n}\frac{1}{2^k} \geq \mu(U_{k,n}) \geq \mu(E_k) \tag{1}$$ Let $U_n=\bigcup_1^\infty U_{k,n}$ then $U_n$ is open, $E\subset U_n$ and from $(1)$ we have $$\mu(E)+ \frac{1}{n} =\sum_1^\infty \mu(E_k)+ \sum_{k=1}^\infty\frac{1}{n}\frac{1}{2^k} \geq \sum_{k=1}^\infty\mu(U_{k,n})\geq \mu(U_n) \geq \mu(E) \tag{2}$$ Note also that, using that $\mu(E_k) <+\infty$, for each $k$, we have $$\mu(U_n \setminus E)\leq \sum_{k=1}^\infty\mu(U_{k,n}\setminus E) \leq \sum_{k=1}^\infty\mu(U_{k,n}\setminus E_k)= \sum_{k=1}^\infty \left (\mu(U_{k,n})-\mu( E_k) \right)\leq \sum_{k=1}^\infty\frac{1}{n}\frac{1}{2^k} =\frac{1}{n} \tag{3} $$ Set $V = \bigcap_{1}^{\infty}U_n$. $V$ is a $G_{\delta}$ and $E \subset V$. From $(2)$, we have, for each $n\in \mathbb{N}$, $n>0$,
$$\mu(E)+ \frac{1}{n} \geq \mu(U_n)\geq \mu(V) \geq \mu(E)$$ So $\mu(V)=\mu(E)$. Since $E,V \in M_\mu$, we have $V \setminus E \in M_\mu$ and, from $(3)$, we have, for each $n\in \mathbb{N}$, $n>0$, $$\mu(V\setminus E) \leq \mu(U_n \setminus E)\leq \frac{1}{n} $$ So $\mu(V\setminus E)=0$. Set $N_1 = V \setminus E$ then, since $E\subset V$, we have $E=V\setminus N_1$ (where $V$ is a $G_\delta$ and $\mu(N_1)=0$).

Thus a.) implies b.).

Proof a.) implies c.) It can be proved in a way similar to the way we proved that a.) implies b.), by using the second part of theorem 1.18.

However there is another elegant way to prove it:

If $E\in M_{\mu}$, then $E^c \in M_{\mu}$, so by the previous item [a.) implies b.)] we know that $E^c=V\setminus N_1$, where $V$ is a $G_\delta$ and $\mu(N_1)=0$.

So, since $E^c=V\setminus N_1=V\cap N_1^c$, then, by taking complement, we have $E=V^c\cup N_1$.

Now, note that the complement of a $G_\delta$ is an $F_\sigma$, so $V^c$ is an $F_\sigma$. Just take $H=V^c$ and $N_2=N_1$.

Proof b.) implies a.)

Suppose $E\subset\mathbb{R}$ and $E=V\setminus N_1$, where $V$ is a $G_\delta$ and $\mu(N_1)=0$. Since $V$ is a $G_\delta$, we have that $V \in M_\mu$ and, since $\mu$ is complete, we also have $N_1\in M_\mu$. So $$E=V\setminus N_1\in M_n$$

Proof c.) implies a.)

Suppose $E\subset\mathbb{R}$ and $E=H \cup N_2$, where $H$ is a $F_\sigma$ and $\mu(N_2)=0$. Since $F$ is a $F_\sigma$, we have that $F \in M_\mu$ and, since $\mu$ is complete, we also have $N_2\in M_\mu$. So $$E=H \cup N_2\in M_n$$

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  • $\begingroup$ The general case is a bit tricky for me. How do we know $\mu$ is $\sigma$-finite. Looking at the definition of $\mu$ being $\sigma$-finite it is not obvious to me that this is the case for this theorem. $\endgroup$
    – Wolfy
    Jun 15 '16 at 19:34
  • $\begingroup$ @Wolfy The section 1.5 of Folland's book is about Borel-Stieltjes and Lebesgue-Stieltjes measures on $\mathbb{R}$, defined by an incresing and right-continuous function $F:\mathbb{R} \to \mathbb{R}$. All such measures are $\sigma$-finite. Please note that any result in section 1.5 is not necessarily true for an arbitrary measure on $\mathbb{R}$. For instance, Lemma 1.17 and Theorem 1.18 are not true for the counting measure on $\mathbb{R}$. $\endgroup$
    – Ramiro
    Jun 15 '16 at 22:14
  • $\begingroup$ I see, is there somewhere in section 1.5 where I missed that all measures are $\sigma$-finite? I don't recall that Folland explicitly states this. If he doesn't I do not see how I could ascertain that $\mu$ is $\sigma$-finite $\endgroup$
    – Wolfy
    Jun 15 '16 at 22:18
  • $\begingroup$ @Wolfy You are right Folland does not explicitly state this. However, for instance, in the remarks after 1.16, Folland writes that $\overline{\mu_F}$ is just the completion of $\mu_F$ and refer this result to exercise 22a. The condition in exercise 22a requires the measure to be $\sigma$-finite. So, at this point, Folland is using that all measures $\mu_F$ are $\sigma$-finite. In fact, it is not difficult to prove that all Borel-Stieltjes measures on $\mathbb{R}$, defined by an increasing and right-continuous function $F:\mathbb{R} \to \mathbb{R}$ are $$\sigma$-finite. $\endgroup$
    – Ramiro
    Jun 16 '16 at 1:17
  • $\begingroup$ Ok, good to know thanks $\endgroup$
    – Wolfy
    Jun 16 '16 at 2:21

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