Htykuut

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 $dmu_F= frac{1}{1+x^2} chi_{x>0} dm$. (dm meaning respect to Lebesgue measure).
   Then how do I get $mu_F$?
   I'm honestly not sure why the expression for $dmu_F$ is true. Now I'm very anxious why I'm not understanding these formulations. Any help would be much appreciated.

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*}