Complex Borel Measure and Bounded Variation Functions

Question

3.29 Theorem in Folland states that: If $\mu$ is complex Borel measure on $\mathbb{R}$ and $F(x)=\mu(-\infty,x])$, then $F \in NBV$. Conversely if $F \in NBV$, there is a unique complex Borel measure $\mu_F$ such that $F(x)=\mu(-\infty,x])$. NBV={F is of Bounded Variation: F is right-continuous and $F(-\infty)=0$}.

Now all, of sudden I am confused/completely lost. My professor writes that:

1. If $F(x)= \chi_{[a, b)}$, then it is in BV and right continuous so it is in NBV? and $u_F= \delta_a-\delta_b$. I have no idea how $u_F= \delta_a -\delta_b$. How was this formulated/come up with? I have no idea how $u_F$ was determined Also I'm not exactly sure what is meant by $\delta_a$ and -$\delta_b$

Does Dirac measure in this case mean, $\delta_a$ mean$$ \left\{ \begin{array}{c@{,\quad}l} 1 & x=a \\ 0 & x \neq a. \end{array}\right. $$ $\delta_b$ mean$$ \left\{ \begin{array}{c@{,\quad}l} 1 & x=b \\ 0 & x \neq b. \end{array}\right. $$ ? Or is this wrong.

Why then is $\mu_F=\delta_a- \delta_b$? I'm very anxious why I don't understand how this was derived/ why it is true.

2. Like wise, my professor writes that if $F(x)= \arctan(x)$, for $x>0$, $0$ otherwise then $F \in$ NBV. and $d\mu_F= \frac{1}{1+x^2} \chi_{x>0} dm$. (dm meaning respect to Lebesgue measure). Then how do I get $\mu_F$? I'm honestly not sure why the expression for $d\mu_F$ is true. Now I'm very anxious why I'm not understanding these formulations. Any help would be much appreciated.

Answer 1

1. Yes, this $F$ is NBV. The $\delta$'s are indeed Dirac measures, however, as such, they take measurable sets as values, rather than function values.

$$\delta_a(A) := \begin{cases}1, &A \ni a\\ 0, & \text{otherwise.} \end{cases} \qquad (A \in \mathcal{B}\mathbb{R})$$

$\mu_F$ is now the Lebesgue-Stieltjes measure based on NBV function $F$. In order to show this, we need to show that $$\mu_F((c,d]) = \delta_a((c,d]) - \delta_b((c,d]) = F(d)-F(c) \qquad (c, d \in \mathbb{R}, c < d)$$ It is probably best to consider different cases:

(i) $c < d < a < b \Rightarrow \mu_F((c,d]) = 0 - 0 = F(d)-F(c)$

(ii) $c < a \leq d < b \Rightarrow \mu_F((c,d]) = 1 - 0 = F(d)-F(c)$

(iii) $a \leq c < d < b \Rightarrow \mu_F((c,d]) = 1 - 1 = F(d)-F(c)$

(iv) $a \leq c < b \leq d \Rightarrow \mu_F((c,d]) = 0 - 1 = F(d)-F(c)$

(v) $a < b \leq c < d \Rightarrow \mu_F((c,d]) = 0 - 0 = F(d)-F(c)$

As the set of intervals $\{(c,d] : c, d \in \mathbb{R}\}$ is a semiring, we can now apply Carathéodory's extension theorem and know that the $$\mu_F(A) = \delta_a(A) - \delta_b(A) \qquad (A \in \mathcal{B}\mathbb{R}).$$

Maybe one hint to give you an intuition, also towards the second question: If you have an NBV function, you can compute a something like a generalised "derivative". "Derivative" refers to the measure-theoretic fundamental theorem of calculus. Like in the case above, this "derivative" may be represented by a measure rather than a function - especially in cases like in 1., where $F$ is clearly not differentiable. It actually is a function, if $F$ is absolutely continuous (with respect to the Lebesgue measure).

2. The latter is the case here. In particular, we have $$\mu_F(A) = \int_{A} F'(x)\mathrm{d}x \qquad (x \in \mathcal{B}\mathbb{R})$$ Also, $\arctan'(x) = 1/(1+x^2)$, (x > 0)

\begin{align*}F(x) &:= \begin{cases}\arctan(x), &x > 0\\ 0, & \text{otherwise.} \end{cases} \qquad (x \in \mathbb{R})\\ F'(x) &= \begin{cases}\arctan'(x), &x > 0\\ 0, & \text{otherwise.} \end{cases} \\ &= \begin{cases}1/(x^2 +1), &x > 0\\ 0, & \text{otherwise.} \end{cases} \qquad (x \in \mathbb{R}) \qquad (x \in \mathbb{R}) \end{align*}