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.

 

[thomsj4]

Define a sequence of nested open sets that approximate $E$ from the outside, and then exploit a summation-by-parts style argument with an auxiliary function to pass to the limit.

First, write the complex measure $\mu_F$ as $\mu_F = \mu_1 + i\mu_2$, where $\mu_1$ and $\mu_2$ are real finite Borel measures. Then each of $\mu_1$ and $\mu_2$ can be further decomposed using Jordan decomposition into finite positive measures. It thus suffices to argue for a single finite positive measure $\nu$, because if we prove

$$
\nu(U_j) \to \nu(E)
$$

for the nested open sets $\{U_j\}_{j=1}^\infty$ with $\bigcap_{j=1}^\infty U_j = E$, it will imply the same limit for any signed or complex combination of such measures.

Construct a function $\varphi$ with the following properties:

1. $\varphi$ is nonnegative and continuous.
2. $\varphi(x) = 1$ for $x \in E$.
3. $\varphi(x) \to 0$ sufficiently fast as $x$ moves away from $E$.

To ensure existence of $\varphi$, notice that for each $x \notin E$, we can find an open interval (possibly very small) around $x$ disjoint from $E$, and define $\varphi$ to vanish there while smoothly transitioning to 1 on $E$. One can, for instance, construct $\varphi$ via a Urysohn-type lemma argument (the set $E$ being closed in the relative topology of $\bigcap_j U_j$, though it is not necessarily closed in $\mathbb{R}$; one just arranges $\varphi=1$ on $E$ and $\varphi=0$ outside some union of slightly bigger open sets).

Define for each $j$,
$$
\Phi_j := \int_{U_j} \varphi \, d\nu,
$$

that is, the integral of $\varphi$ over $U_j$ with respect to $\nu$. Because $U_{j+1} \subset U_j$ and $\bigcap_j U_j = E$, we have $U_{j+1} \subset U_j$ implies

$$
\Phi_{j+1} = \int_{U_{j+1}} \varphi \, d\nu \le \int_{U_j} \varphi \, d\nu = \Phi_j.
$$

So $\{\Phi_j\}$ is a decreasing sequence of real numbers. Also,

$$
\Phi_1 \ge \Phi_2 \ge \cdots \ge 0.
$$

By completeness of the real numbers, there is a limit
$$
\lim_{j \to \infty} \Phi_j =: L \ge 0.
$$

On the other hand, for each $x \in E$, $\varphi(x)=1$. Hence intuitively, as the open sets $U_j$ shrink down onto $E$, we expect $\Phi_j$ to converge to

$$
\int_E \varphi \, d\nu = \nu(E),
$$

because $\varphi = 1$ on $E$. Formally, for the complement $U_j \setminus E$, we have $\varphi \le 1$ but arranged to be strictly less than 1 and vanish outside a neighborhood of $E$. Thus for large $j$, most of the mass of $\nu$ in $U_j$ is forced to lie near $E$, making

$$
\int_{U_j} \varphi\,d\nu \quad \text{arbitrarily close to} \quad \nu(E).
$$

Hence

$$
\lim_{j \to \infty} \int_{U_j} \varphi \, d\nu = \int_E \varphi \, d\nu = \nu(E).
$$

Thus $L = \nu(E)$.

Now repeat this procedure with appropriate choices of continuous auxiliary functions $\varphi$ that “pick out” the imaginary part, or the positive/negative parts from the Jordan decomposition; that gives, for each component, the analogous conclusion

$$
\lim_{j \to \infty} \mu_F(U_j) = \mu_F(E).
$$

Hence we deduce $\mu_F(U_j) \to \mu_F(E)$ as $j \to \infty$.