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I think the obvious way is indeed the only way. First of all $n_j=1$ for all $j$, as boundaries do not have higher multiplicity. Secondly, the $l$ rays of your varifold split $\mathbb{R}^2$ into $l$ s …

Building on Pietro Majer's answer to you previous question for a change, consider the following: Let $g: [0,\infty) \to \mathbb{R}$ be the unique continuous function such that $g(0)=0$, $g'(x) = 1$ fo …

Checking this in perfect detail might be a bit technical, but the following looks like it could work for any dimension up to n: For 1d: Take an S curve $\phi: [0,1] \to [0,1]$ such that $\phi(0) = 0$, …

If the $f_n$ are absolutely continuous, then $f'_n$ would also be the distributional derivative and $g=f'$ follows from the continuity of that. So if there is a counterexample it should probably invol …

I am not the greatest expert on the details of this stuff, but since nobody else tried so far, let me have an attempt: Prelude: Measures Since you mention measures, I start with that, though this is m …

Leo Moos gave a very good answer, but here is another way to think about this: Essentially equality does not hold when it is cheaper to have a the boundary of a hole than filling that hole. So if $S$ …

To me your definition seems to be the right one, you just need to prove that it is well defined when approximating Lipschitz with $C^1$-functions. For that you probably need the distributional diverge …

I think the following might be an example, though it will require a bit of work if you want to make it more precise: Take an immersion of a sphere, which is injective except for one cap at each pole, …

Your intuition is right. The key is in the paper you cite, in that they consider uniqueness in the class $L^1_{\text{loc}}$, which does not allow for concentrations. If you add to this, that the Lagra …

This is not a full answer, since I do not know the counterexample Lin refers to, but I can offer some explanations and guesses which are too long for a comment: You can define a first variation for cu …

I might be missing something, but if you require $\Omega$ to be precompact in $\mathbb{H}^n \setminus K$, then $K$ makes no difference for the purpose of determining the measures and you can use the o …

One can use your infinite-density example, but replace the outer lines with very sparse dotted lines: $$S = (\{0\} \times \mathbb{R}) \cup \bigcup_{i=1}^\infty \{i^{-1},-i^{-1}\} \times \left[\bigcup_ …

The general prerequisites are almost the same as for currents, mainly a strong understanding of measure theory and a bit of geometrical intuition. There is an aspect of multilinear algebra and some fu …

I think one of the classic counterexamples works here, to show that this is false: Let $\{q_i\}_{i\in\mathbb{N}}$ dense in $[0,1]^n$, $\delta >0$ and construct $$E = \bigcup_{i\in\mathbb{N}} B_{\delta …

I don't think your left hand side is well defined for the class of $u$ you are considering, I can change each $u(.,t)$ to a large value on the zero-set $S_{|t|}$, which will result in the supremum pic …