# Real Analysis, Folland Theorem 1.19 Borel Measures

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.

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 $$E=H \cup N_2\in M_n$$

• 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. Jun 15 '16 at 19:34
• @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 $\mathbb{R}$. Jun 15 '16 at 22:14
• 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 Jun 15 '16 at 22:18
• @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. Jun 16 '16 at 1:17
• Ok, good to know thanks Jun 16 '16 at 2:21