The converse of your assertion is not true. Even if you exclude sets of zero and infinite measure, all the sets you construct are rectifiable and for $k<n$ there are measurable, purely unrectifiable sets of nonzero measure.

Okay, thinking about it, the notion of "bigger" does indeed not work except for maybe something small enough like countable $X$. However $C([0,1])$ is separable and embeds isometrically into $W(C([0,1]))$, but as this is also separable, the Banach-Mazur theorem tells us that it in turn embeds isometrically back into $C([0,1])$. So they are of the same size for all intents and purposes.

There are the two trivial answers to the first question, the empty space and the space consisting of a single point. It feels like everything after that can probably be ruled out by a cardinality argument, though maybe for extremely large spaces there might not be enough ways to construct measures anymore.

@D.R. They are densities and absolute forms, which is easy to show by just following the axioms. If I am not completely mistaken, they should not be pseudoforms, but any restriction of them to a 1 or 2-dimensional submanifold respectively is a pseudoform on that submanifold.

I think that works well if the colleague was someone who likes to work linearly, i.e. in the case where the first n chapters are more or less done and chapters (n+1)ff only exist as an outline. In case of someone who likes to work backwards by starting with the main theorems and then works backwards from them to the introductory materials, I fear this would be rather unreadable.

People have tried similar things, e.g.: arxiv.org/abs/2006.07859 But simulating a physical string means discretizing it and with that it is very easy to accidentally jump in topology. In addition, seeing that this process seems to work in examples is easy. But showing that it always finishes the job in polynomial time seems more or less impossible.

$F_p$ is just the derivative of $F$ by $p$. Since $F$ is scalar, it has the same number of components as $p$. So you can represent it as an $n \times N$-matrix.

Something like $\int \sin^n dx$ for large n should result in lots and lots of terms when calculating it in the standard way of repeated partial integrations, but I have no idea how to prove that there is no nicer closed form.

@FeiCao $\psi_t$ depends on $\rho$ and $\nu$. Thus changing $\nu$ changes the right hand side in (4).

Well, the details of that are precisely why I did not do the full calculation. But note that $\chi_E$ and $\chi_{E^C}$ sum up to one, so adding their sum to the integrand just adds a sphere area, resp. two half spheres. So the integral is equal to $2 \int_{\partial B_r} \chi_{E^c} - \chi_{H} d\sigma$, where $H$ is the half-space "outside" at $0$. So the integral measures the signed area on $\partial B_r$ between the intersection with $\partial E$ and the equator. Now if $r$ is small, this deflection is tiny and the sides of the sphere are vertical, so we get the integral of $h$ on the equator.

Yes, the sphere is one dimension less. The integral is actually quite easy using symmetries. All you have in the Taylor-expansion are terms of the form $\partial_{ij} h(0) x_i x_j$, so you only need to integrate $x_i x_j$. For $i \neq j$ this is an odd function in $x_i$, so since the domain is symmetric, the integral of that term is $0$. For $i=j$ it is even and positive a.e., so the integral will be some positive constant, which for symmetry is independent of $i$. Summing up then gives you $c \Delta h(0)$.