Connection between absolute continuity of function and measure

[psie]

TL;DR Summary

I'm stuck with a small detail in the proposition below. I assume the reader is familiar with the space $BV$, the functions $F:\mathbb R\to\mathbb C$ of bounded variation. We also have $NBV$, where $N$ stands for normalized; this is the space of $BV$ functions which are right continuous and $F(-\infty)=0$.

Let $m$ be Lebesgue measure. It is another proposition that the functions $NBV$ are in one-to-one correspondence between complex Borel measures, e.g. $F\in NBV$ induces a complex measure $\mu_F$ such that $F(x)=\mu_F((-\infty,x])$. Then in Folland's real analysis text,

3.32 Proposition. If $F\in NBV$, then $F$ is absolutely continuous iff $\mu_F\ll m$.

I'll omit the $\impliedby$ direction. In the $\implies$ direction, we suppose that $m(E)=0$ for a Borel set $E$ and we want to show $\mu_F(E)=0$. If $\delta$ and $\epsilon$ are as in the definition of absolute continuity of $F$ (see here), we can find open sets $U_1\supset U_2\supset\cdots\supset E$ such that $m(U_1)<\delta$. This is possible by so-called regularity of $m$. To be exact, we have $m(E)=\inf\{m(U):U\supset E,U\text{ open}\}$, and then by definition of infimum we can find such a sequence. However, it is also claimed that $\mu_F(U_j)\to \mu_F(E)$. Why is this true?

Some observations; $\mu_F$ is a complex measure which is regular. This means the positive measure $|\mu_F|$ is regular. So $|\mu_F|(K_j)\to|\mu_F|(E)$ for some sequence $\{K_j\}$ of open sets. But it is not clear to me that we can simply choose $K_j=U_j$ (and why we'd get convergence for $\mu_F$ rather than $|\mu_F|$).

 

[psie]

Here’s one motivation I’ve worked out. I will omit some details, but just so that this receives an answer.

We can decompose the real and imaginary parts of $F$ into differences of bounded and increasing functions. Each of these functions induce finite measures who decompose $\mu_F$. Now we apply regularity to all these four measures plus Lebesgue measure (and for Lebesgue measure we do it in a way so that the measure of all the sets is smaller than $\delta$). So we will obtain five sequences of sets. Then we just take the intersection of the first entries of all the five sequences, then the intersection of the second entries, and so on. Then we obtain a sequence of sets for which the Lebesgue measure is smaller than $\delta$ and the $\mu_F$ measure of the sets converges to $\mu_F(E)$. Finally, to make them decreasing, define a new sequence of sets whose $n$th entry is the intersection of the first $n$ entries of our sequence of sets.