Weak convergence of measures and of total variations 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. 

EDIT 
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? 
 A: 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.)
