So I am reading Folland Chapter 3 and I'm try to understand the bigger picture. So there are positive measures, signed measures, and complex measures. Positive measures are signed measures, although signed measures can also take on negative values.
For signed measures either $\nu+$ or $\nu-$ must be finite.
Section 3.3 Complex measures
However, for complex measures, the measure can take on complex values. Infinite values are not allowed. Positive measures are complex measures if they are finite.
- Can complex measures also be signed? Can they take on negative values?
Folland does state that if $\nu$ is a complex measure then $\nu_r$ and $\nu_i$ are signed measures that do not assume the values positive $\infty$ or negative $\infty$, and are thus finite. Thus $\nu$ can also also be signed?
2.Folland states that if $\mu$ is positive measure and $f\in L^1(\mu)$, then $f d\mu$ is a complex measure.
Is $f d\mu$ represented by $\int f d\mu$? What is meant by $f d\mu$ and why can it take complex values?
The total variation of a complex measure $\nu$ is the positive measure $|\nu|$ determined by property that if $d\nu=fd\mu$ where $\mu$ is a positive measure, then $d\nu=|f| d\mu$. Meaning: $|\nu(E)| =\int_E |f| d\mu$.
3.5 Functions of Bounded Variation:
Folland writes that as positive measures on $\mathbb{R}$ are related to increasing functions, complex measures on $\mathbb{R}$ are related to functions of bounded variation. What does it mean to be a complex measure on $\mathbb{R}$? Isn't a complex measure defined on $\mathbb{C}$? Is the only difference that a complex measure on $\mathbb{R}$ vs a positive measure that a complex one can take negative values?
Folland defines total variation $T_{F}(x) = \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}$ over all partitions and n. If the limit of $T_F(x)$ as x approaches infinity is finite then a function F is said to be of bounded variation or BV. Note that the representation $F=1/2(T_F+F)-1/2(T_F-F)$ of a real-valued $F \in$ BV is called the Jordan decomposition of F, and $1/2(T_F+F)$ and $1/2(T_F-F)$ are called the positive and negative variations of F. Folland defines: $NBV$={$F\in$ BV: F is right continuous and $F(-\infty)=0$}
Folland states there is a connection between BV and the space of complex Borel measures on $\mathbb{R}$. (Again what is meant by a complex Borel measure on $\mathbb{R}$?).
If $\mu$ is a complex Borel measure on $\mathbb{R}$ and $F(x)=\mu((-\infty,x])$ then $F\in$ NBV. Conversely if $F \in$ NBV, there is an unique complex Borel measure $\mu_F$ such that $F(x)=\mu(F)((-\infty,x]))$; moreover $|\mu_F|$=$\mu_{T_F}$ Now, I have no idea what is meant by $|\mu_F|$ and $\mu_{T_F}$. Is $|\mu_F|$ like total variation or is it the absolute value of the Borel measure, $\mu_F$ stated? $\mu_{T_F}$ a measure that is induced by $T_F$ because $T_F$ is increasing and if F is right continuous, then so is $T_F$? Folland says for the proof of $|\mu_F|$=$\mu_{T_F}$ see exercise 28.
Exercise 28 in Folland states that: If $F \in NBV$, let $G(x)=|\mu_F|(-\infty,x])$. Prove that $|\mu_F|$=$\mu_{T_F}$ by showing that $G=T_F$ via the following steps. What is meant by $G(x)=|\mu_F|(-\infty,x])$, again confused by the $|\mu_F|$ part
Part a of exercise. From the definition of $T_F, prove: T_F \leq G$
Solution given:If $x\in\mathbb{R}^n$ then \begin{align*} T_F(x) &= \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}\\ &= \sup\{\sum_{1}^{n}|\mu_F((-\infty,x_j]) - \mu_F((-\infty,x_{j-1}])|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}\\ &= \sup\{\sum_{1}^{n}|\mu_F((x_{j-1},x_j])|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}\\ &\leq \sup\{\sum_{1}^{n}|\mu_F(E_j)|:\{E_j\}_{1}^{n}\in\mathbb{B}_{\mathbb{R}} \ \text{disjoint such that} \ \bigcup_{1}^{n}E_j \subset (-\infty,x]\}\\ \\ &\leq \sup\{\sum_{1}^{n}|\mu_F(E_j)|:\{E_j\}_{1}^{n}\in\mathbb{B}_{\mathbb{R}} \ \text{disjoint such that} \ \bigcup_{1}^{n}E_j = (-\infty,x]\}\\ & = |\mu_F|((-\infty,x])\\ &= G(x) \end{align*}
I honestly do not understand the last 3 steps of this proof with the inequalities, but in particular do not understand why\begin{align*} & \sup\{\sum_{1}^{n}|\mu_F(E_j)|:\{E_j\}_{1}^{n}\in\mathbb{B}_{\mathbb{R}} \ \text{disjoint such that} \ \bigcup_{1}^{n}E_j = (-\infty,x]\}\\ & = |\mu_F|((-\infty,x])\\ \end{align*}.
Where does this fact come from? I don't see a definition related to anything like it in Folland. Thank you very much. Sorry for the long post, but really need to understand/am very worried. thanks. if I can understand this I will be happy.