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Let $\mu$ be a finite, nonatomic, signed Radon measure on $([0,1],\mathcal{B}([0,1]))$. For all $x \in ]0,1[$ the limit $$\lim_{\varepsilon \to 0} {\frac{\mu(]x-\varepsilon,x+\varepsilon[)}{2\varepsilon}}$$ exists. Does this imply that $\mu$ is absolutely continuous with respect to Lebesgue measure on $[0,1]$?

If $\mu$ was a positive measure, this would follow from a version of Besicovitch's Differentiation Theorem, see Theorem 2.22 in Functions of Bounded Variation and Free Discontinuity Problems by Ambrosio, Fusco, Pallara. In the general case I could imagine that two positive, singular continuous measures could be used to give a counterexample, but I haven't been able to construct them.

Edit: I was thinking about a middle-half Cantor probability distribution as positive part and as negative part a probability measure constructed in the following way: At step $n$ of the construction of the Cantor set we have $2^n$ intervals of length $4^{-n}$ left; consider the two intervals next to each of these intervals, also of length $4^{-n}$ and uniformly distribute weight on them in order to get a probability measure. The weak*-limit of those measures shall constitute the negative part. - I'm rather convinced that the resulting signed measure has pointwise Radon-Nikodým derivative 0, but I fear that in the end it's the zero measure, so both constructions (of positive and negative part) give the same measure.

Edit no. 2: It is indeed the zero measure. :(

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What about considering the decomposition of positive measures, $\mu=\mu_+ - \mu_-$? –  Berci Nov 12 '12 at 15:27
    
That's what I had in mind when I was talking about the two positive singular continuous measures. But it doesn't seem to help for a proof of absolute continuity. Negative part and positive part could both have infinite density at some points, even at points of a Lebesgue-nullset. –  Thomas Nov 13 '12 at 9:53
    
When $\lim_{\epsilon\to 0}\frac{\mu(]x-\epsilon,x+\epsilon[)}{2\epsilon}$ is integrable or even continuous, do you have any conclusion? –  23rd Dec 6 '12 at 21:03
    
Thanks for the hint, @richard. After I read your comment, I had another look at my sources and stumbled upon the de la Vallée Poussin theorem, see my answer below. –  Thomas Dec 10 '12 at 13:23
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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:

  1. 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 $\nu(A) \geq q \lambda (A)$.
  2. 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.

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Provided that $D\nu$ is finite everywhere, does part 2 of your lemma imply $\lambda(B)=0\Rightarrow |\nu|(B)=0$ and hence $\nu\ll\lambda$ immediately? –  23rd Dec 10 '12 at 14:03
    
Another question: why is Vitali covering theorem applicable to $|\nu|$? –  23rd Dec 10 '12 at 14:48
    
Hmm, I don't think so. $\nu(B) = 0$ does not imply $|\nu|(B) = 0$. And I don't see a good connection between $D\nu$ and $\overline{D}|\nu|, \underline{D}|\nu|$. - And the Vitali-Besicovitch theorem only needs a positive Radon measure, which $|\nu|$ should be. –  Thomas Dec 10 '12 at 15:00
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$\nu(B)=0$ does not imply $|\nu|(B)=0$, but your lemma implies that if $D\nu$ is finite everywhere, then $\lambda(B)=0\Rightarrow \lambda(C)=0,\forall C\subset B\Rightarrow \nu(C)=0,\forall C\subset B\Rightarrow |\nu|(B)=0$. –  23rd Dec 10 '12 at 15:17
    
Ah, good point! It's still a nice theorem, so I'll leave it there as reference. –  Thomas Dec 10 '12 at 15:50
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Edit: There is a problem with the below argument: just because $f(x) = \lim_{\epsilon \to 0} \frac{1}{2\epsilon} (F(x+\epsilon)-F(x-\epsilon))$ exists does not let us conclude that $F$ is differentiable (consider $F(x)=x^{1/3}$ at $x=0$). So more work is needed.

Set $f(x) = \lim_{\epsilon \to 0} \frac{1}{2\epsilon} \mu(]x+\epsilon, x-\epsilon[)$. It is shown in Theorem 3.22 of Folland's Real Analysis that $f$ is the density of the absolutely continuous part of $\mu$; in particular, $f$ is integrable (with respect to Lebesgue measure).

If $F(x) = \mu([0,x))$, then $f = F'$ (Edit: not yet proved) (note that $F(x+\epsilon) - F(x-\epsilon) = \mu([x-\epsilon, x+\epsilon[)$ instead of $\mu(]x-\epsilon, x+\epsilon[)$, but $\mu$ has no atoms so they are actually equal). Now Corollary 3.22 of A First Course in Sobolev Spaces by Giovanni Leoni (free preview of relevant chapter) says that any everywhere differentiable function with integrable derivative is absolutely continuous. That means that $\mu$ itself is an absolutely continuous measure.

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In general, $F'$ only exists Lebesgue almost everywhere, so how can you conclude that $F'=f$ everywhere? –  23rd Dec 7 '12 at 21:04
    
@richard: I was thinking that $F'$ was precisely the limit that we are assuming to exist, but I've noted that this is not right. –  Nate Eldredge Dec 7 '12 at 22:22
    
It is too tricky for me to make use of the existence of the limit. Maybe as suggested by Thomas in the question, $\mu$ could be singular? –  23rd Dec 7 '12 at 22:49
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