I have a sequence of signed, bounded measures $\{\mu_n\}_{n\in \mathbb N}$ and an open, bounded set $\Omega \subset \mathbb R^N$. I know that $\vert \mu_n \vert(\Omega) \to \vert \mu \vert (\Omega)$ (as real numbers) for a suitable measure $\mu$. In particular, the sequence $\mu_n$ is uniformly norm-bounded (being the norm on the space of measures the total variation) hence by compactness there exists a measure $\lambda$ such that (up to subsequences) $\mu_n \rightharpoonup \lambda$ weakly.

Can I conclude that $\lambda \ll \mu$? What can I say on the relationship between $\lambda, \mu$? I apologize for the confusion, but I am puzzled.


This question arises essentially from reading this paper. It seems to me that, mutatis mutandis, the authors claim what I am asking at page 20. Let me present a simplified and self-contained version of the argument: let $B_1 \subset \mathbb R^N$ be the unit ball in $\mathbb R^N$. The authors define $u^\varepsilon(x) = \min \left\{\frac{d(x, \mathbb R^N \setminus B_1)}{\varepsilon}, 1 \right\}$ and they assume (see eq. (17) in the paper) that $$ \vert Du^\varepsilon \vert(\mathbb R^N) \to P(B_1) = \mathcal H^{N-1} \llcorner_{\partial B_1}(\mathbb R^N) =: \mu(\mathbb R^N) $$ where $\mathcal H^{N-1}$ is the Hausdorff measure (the last equation is the one right under (18) in the paper). From this convergence they conclude that the vector valued measures $(Du^\varepsilon \mathcal L^N)$ are compact and they converge - up to subsequences - to a vector valued measure $\mu^\infty$ which they prove is a.c. w.r.t. $\mathcal H^{N-1}\llcorner_{\partial B_1}$... What have I misunderstood? Am I missing something?


I suspect that when you wrote $|\mu_n|(\Omega)\to|\mu|(\Omega)$ maybe that's not exactly what you meant. Replying to the question as stated:

You assume so little about $\mu$ that it really has nothing to do with the rest of what's going on; there's no relationship between $\lambda$ and $\mu$. One doesn't need to "construct" a counterexample, any almost random choice of $\mu_n$ and $\mu$ works.

Say $\mu_n=\delta_{1/n}$, a point mass at $1/n$. So $\mu_n\to\delta_0$. Let $\Omega=(-1,1)$ and let $\mu$ be any positive measure with $\mu(\Omega)=1$.

Or maybe to better illustrate how $\mu$ really has nothing to do with $\lambda$, let $\Omega=(2,3)$ and let $\mu$ be any measure with $\mu(\Omega)=0$. Or $\Omega=\emptyset$ and $\mu=0$.

Heh, let $\mu_n$ be any norm-bounded sequence, $\Omega=\emptyset$, $\mu=0$.

Edit: Regarding the edit made to the question, and the comment asking whether there's any significant difference: Well of course there's a huge difference, since in the new version we have a specific "sequence" of measures! In particular, for example, if I'm reading things correctly $\mu_\epsilon$ is supported in the annulus $A_\epsilon=\{1-\epsilon\le|x|\le1\}$.

If I have the picture right it seems clear that the gradient of $u_\epsilon$ is $\nabla u_\epsilon(x)=-\frac1\epsilon\frac x{|x|}$ or something like that in $A_\epsilon$, $0$ elsewhere. So it seems clear that $\mu_\epsilon\to\lambda$, where $d\lambda=-n(x)\,dH$ (where $n$ is the outward unit normal on the sphere and $H$ is surface area on the sphere), which certainly appears to be ac wrt $H$.

Of course that could be all wrong, it's just my first impression. But note that the various things I say "seem clear" seem clear to me based on my picture of what $\mu_\epsilon$ actually is, not because of any general principle analogous to what you ask about in the original version of the question!

As a general rule, if they assert P and you don't see why P holds you might be better off actually stating what they actually assert and asking why it holds, instead of sort of guessing that they seem to be saying that P follows from Q and asking whether Q implies P.

Second edit: Two things.

(i) A conjecture regarding the sort of "soft" or "abstract" argument the authors might have had in mind: It's easy tp see that $||\mu_\epsilon||$ is bounded. And $\mu_\epsilon$ has a certain sort of rotational symmetry (which I'm not going to try to define precisely; this isn't my argument after all); hence any weak limit $\lambda$ must have the same symmetry. It's clear that $\lambda$ must be supported on $S=\{|x|=1\}$, since the support of $\mu_\epsilon$ shrinks to $S$, and the only vector-valued measure on $S$ with that symmetry is $cn\,dH$.

That's pretty vague; I'm not going to try to make it more precise, since it's just my guess regarding sort of what the authors may have had in mind. But:

(ii) Why it seems clear to me that $\mu_\epsilon$ simply does converge to what I say it does:

First, in general if $u$ is a radial function, $$u(x)=\phi(|x|),$$then $$\nabla u(x)=\phi'(|x|)\frac x{|x|}.$$("Advanced calculus": $\nabla u$ is the directional derivative in the direction of greatest increase, which is to say in a direction orthogonal to the level sets of $u$...) Hence $\nabla u_\epsilon$ is what I say it is above.

Now assume $f\in C_c(\Bbb R^n)$ and integrate in polar coordinates:

$$\int_{\Bbb R^n}f(x)\nabla u_\epsilon(x)\,dx=\int_S\frac{-1}{\epsilon}\int_{1-\epsilon}^1f(r\xi)\xi r^{n-1}\,drdH(\xi).$$And since $f$ is continuous it's clear that $$\lim_{\epsilon\to0}\frac{-1}{\epsilon}\int_{1-\epsilon}^1f(r\xi)\xi r^{n-1}\,dr=-f(\xi)\xi,$$uniformly over $\xi\in S$.

("Polar coordinates": In general $$\int_{\Bbb R^n}g(x)\,dx=c_n\int_S\int_0^\infty g(r\xi)r^{n-1}\,drdH(\xi).$$Note that if $H$ is actual "surface area" on $S$, in particular not normalized to be a probability measure as is sometimes done in that formula, then $c_n=1$. See Folland Real Analysis or various other places.)

  • $\begingroup$ Thank you very much for your kind reply. I see your point and I totally agree with you, the question is essentially ill-posed. However, I am still in troubles, see the edit to my question. It seems to me that the authors kind of claim what I asked. Is there any significative difference in their "particular" case? Have I misunderstood anything? Thanks for your help. $\endgroup$ – Romeo Jan 20 at 10:13
  • $\begingroup$ Understood, you gave me a "lesson of method" and I thank you for it. Yet, I believe that in the aforementioned paper the peculiar form of the sequence seems to be not so relevant (at least this was my impression). In some sense, reading your edit, one might see the picture simply as: well, we have $(\mu_\epsilon)_\epsilon$ which has a specific form and this sequence converges to the surface measure. But this is not the way the authors propose the arguments. Last question, I promise: Could you please justify why it seems clear to you that $\mu_\epsilon \to \lambda$ (in your notation)? Thanks. $\endgroup$ – Romeo Jan 20 at 15:10
  • $\begingroup$ Cux you just write down what $\int f\,d\mu_\epsilon$ is and let $\epsilon\to0$. See edit.... $\endgroup$ – David C. Ullrich Jan 20 at 15:43
  • $\begingroup$ "Cux" $\to$ "Cuz" $\endgroup$ – David C. Ullrich Jan 20 at 15:50
  • $\begingroup$ @Romeo Hmm. A way one could deduce that $\lambda<<H$ without saying anything about what $\lambda$ is: If $E\subset S$ say $V(E)=\{r\xi:r>0,\xi\in E\}$. Suppose there exists a function $\phi$ with $\lim_{t\to0}\phi(t)=0$ such that $\mu_n(V(E)\le\phi(H(E))$ for all $n$.. If $\mu_n\to\lambda$ weakly and $\lambda$ is supported on $S$ then $\lambda<<H$. $\endgroup$ – David C. Ullrich Jan 20 at 17:04

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