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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 …

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 …

The following is not a full proof, as I skip on some calculation details, just an extension of the remark I made in a comment but I am pretty certain it is correct. The answer is no, at least not wrt. …

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, …

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 $S$ …

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 …

Since Pietro Majer definitely was right about the structure of the set but hasn't supplied a proof let me jump in with an elementary one. I think the problem is to specific to find a reference, but it …

If by not agree you include that the derivative may not exist, you can get any dimension and measure that does not contradict the almost everywhere. Consider the worst case $d=1$: Take the constructio …

I assume that by maximal you mean with respect to inclusion. Then the answer is no. Consider the following counterexample on the real line: Let $\mathcal{B}_\epsilon := \epsilon\mathbb{Z}$ and $\widet …

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_ …

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 …

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 …

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 $\phi(0) = 0$, …

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 …

If I am not missing something then the corrected statements look equivalent to me on the level of sets, no need for smoothness. Fix $\tau > 0$ and assume that 1. holds. For any $t\geq \tau$ and $y \in …