I'm trying to solve this question, but I'm not sure about part(b):
Suppose $F:[0,1] \rightarrow \mathbb{C}$ and $L>0$ satisfy $|F(x)-F(y)| \leq L|x-y|$ for all $x, y \in[0,1]$. (a) Show that $F$ is absolutely continuous on $[0,1]$.
(b) Let $\mu_F$ be the regular complex Borel measure on $[0,1]$ satisfying $\mu_F((a, b])=F(b)-F(a)$. (You do not need to show this measure exists.) Show that $\left|\mu_F(E)\right| \leq \operatorname{Lm}(E)$ for all Borel sets $E \subset[0,1]$.
My attempt: Since by (a) and FTC of Lebesgue integrals, we have F is differentiable a.e. on $[0,1]$, $F'\in L^1([0,1],m)$, and $F(x)-F(a)=\int_{a}^{x} F'(t) dx$. Then, we have the following estimation:
WLOG, $E=(a, b]$ in $[0,1]$ $$ \begin{aligned} \left|\mu_F(E)\right| & =|F(b)-F(a)| \\ & =|F(b)-F(x)+F(x)-F(a)| \\ & =\left|-\int_b^x F^{\prime}(t) d t+\int_a^x F^{\prime}(t) d t\right| \\ & =\left|\int_a^b F^{\prime}(t) d t\right| \\ & \leq L \left|\int_a^b d t\right| \\ & =\operatorname{Lm}(E) . \end{aligned} $$
The inequality holds by F is differentiable a.e., hence by Lipschitz condition, we have the estimation.
I also noticed that we can decompose complex measures into three parts, discrete, absolutely continuous, and singular continuous parts, but I don't know how to play around with this. Here, m denotes the Lebesgue measure.
I appreciate any help! Happy New Year to those who may see my questions!