Pointwise existence of Radon-Nikodým derivative sufficient for absolute continuity?
Solution 1:
I had another look at A. Bruckner's »Differentiation of Integrals« (The American Mathematical Monthly, Vol. 78, No. 9, Part 2: Differentiation of Integrals (Nov., 1971), pp. i-iii+1-51) and found a mention of the de la Vallée Poussin decomposition theorem, which is a refinement of Lebesgue's decomposition theorem. The expository paper cites Saks' »Theory of the Integral«, where you can find the general theorem as Theorem (15.7) on page 155. A modern treatment can be found as Theorem 8.21 in »Real Analysis« by A. Bruckner, J. Bruckner and Thompson (free download here). The problem with both sources is that they prove the theorem only for differentiation with respect to net structures, which is not applicable here.
But I think you can modify their approach to make it work here where the differentiation base consists of balls. I will sketch how:
The following lemma corresponds to the growth lemma, Lemma 8.20 in »Real Analysis«:
Lemma: Let $\nu$ be a (signed) finite Radon measure on a cube $X$ in $\mathbb{R}^n$. Then:
- If $A \subseteq X$ is a Borel set, $q \in \mathbb{R}$, and $\overline{D} \nu(x) := \limsup \limits_{r \downarrow 0} {\frac{\nu(\overline{B}_r(x))}{\lambda(\overline{B}_r(x))}} \geq q$ for all $x \in A$, then $\nu(A) \geq q \lambda (A)$.
- If $B \subseteq X$ is a Borel set, $\lambda(B)=0$, and $\nu$ does not have an infinite derivative at any point of $B$, then $\nu(B) = 0$.
Proof: Assume $q=0$; the general case can then be proved by considering $\tilde{\nu} := \nu - q\lambda$. Let $\varepsilon > 0$. Since $|\nu|$ is outer regular, there exists an open set $G \supseteq A$ such that $\lambda(G) < \infty$ and $|\nu|(E) < \varepsilon$ for every Borel set $E \subseteq G \setminus A$. Consider $$\mathcal{F} := \{\overline{B}_r(x) \: | \: x \in A,\, \overline{B}_r(x) \subseteq G,\, \nu(\overline{B}_r(x)) > -\varepsilon\, \lambda(\overline{B}_r(x))\}.$$ By the Vitali-Besicovitch covering theorem (Theorem 2.19 in »Functions of Bounded Variation and Free Discontinuity Problems«) there exists a countable disjoint subfamily $\mathcal{F}' \subseteq \mathcal{F}$ such that $|\nu|\left(A \setminus \bigcup \mathcal{F}'\right) = 0$. Now $\bigcup \mathcal{F}' \subseteq G$ and $$\left|\nu \left(G \setminus \bigcup \mathcal{F}'\right)\right| \leq |\nu| \left(A \setminus \bigcup \mathcal{F}'\right) + |\nu| \left(G \setminus \left(A \cup \bigcup \mathcal{F}'\right)\right) < 0 + \varepsilon = \varepsilon.$$ Now the rest of the proof of Lemma 8.20 goes through without changing anything but the notation. $\Box$
Edit: As richard remarked in the comments, part 2 of the lemma already suffices to answer my question: Take $n=1$, $X= [0,1]$. If $B \subseteq [0,1]$ is a Borel set with $\lambda(B) = 0$, then $\lambda(C) = 0$ for all Borel sets $C \subseteq B$. The lemma gives $\nu(C) = 0$ for all such $C$, so $|\nu|(B) =0$, since $D\nu$ exists everywhere by hypothesis (and is finite). This proves $\nu \ll \lambda$.
Theorem 8.21 in »Real Analysis« and its proof can be copied verbatim for the differentiation using balls, applying the lemma above instead of Lemma 8.20.
Theorem: Let $\nu$ be a finite Radon measure on a cube $X$ in $\mathbb{R}^n$. Then $D\nu(x)$ (which is defined as $\overline{D}\nu(x)$ in case $\overline{D}\nu(x) = \underline{D}\nu(x) := \liminf \limits_{r \downarrow 0} {\frac{\nu(\overline{B}_r(x))}{\lambda(\overline{B}_r(x))}}$) exists for a. e. $x \in X$ and is integrable on $X$. Furthermore, $$ \nu(E) = \int_E {D\nu}\, \mathcal{d}\lambda + \nu(E \cap B_\infty) + \nu(E \cap B_{-\infty})$$ for every Borel set $E \subseteq X$, where $$ B_{\pm \infty} := \{x \in X \: | \: D\nu(x) = \pm\infty\}.$$
For my problem, choose $n=1$ and $X = [0,1]$. By hypothesis, $D\nu$ exists everywhere, so $B_{\pm \infty} = \emptyset$ and the theorem shows that $\nu$ is absolutely continuous with respect to Lebesgue measure.