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25

Suppose $x_0\in A$. Let $B := \{ d(x,K) < 1\}$. Observe that since $K$ is compact, there exists $y_0\in K$ such that $d(x_0,y_0) = 1$. By definition $B_1(y_0) = \{x: d(x,y_0) < 1\}$ is a subset of $B$. This implies that for all $\epsilon < 1/2$, we have that $$ \mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0)) $$ where $\mu$ is ...


9

Negative dimension is actually much easier to talk about than complex dimension. Super vector spaces are a natural collection of objects that can have negative dimension; given a super vector space $(V_0, V_1)$ we can define its dimension to be $\dim V_0 - \dim V_1$, and this definition has many nice properties; see this blog post, for example. More ...


9

A thin version of the usual Cantor middle third set works. The idea is just that you need to omit more than just a third of the remaining intervals as the construction proceeds, enough so as to force the Hausdorff dimension to $0$. Specifically, we construct the set in stages. At each stage, we've omitted a "middle third" from each finite interval ...


8

I was going to make this a comment but it occurred to me there might be sufficient interest that perhaps I should not bury it in a comment. At the beginning of the paper below Erdős gives a short proof (that he attributes to Tibor Radó) making use of the Lebesgue density theorem that $E_r$ has Lebesgue measure zero, where $E$ is a closed set in ${\mathbb ...


7

You only need to consider the case $\mathfrak{h}_{s}^\ast(A) \lt \infty$, but you need to be a bit careful in choosing the outer approximations since swapping $\inf$ and $\sup$ certainly isn't allowed without some thinking. If you knew that you can always take the same set $E$ in the $\inf$ (which I will show in \eqref{eqn:ast} below) then you'd be ...


6

I just checked the book and I found out that the author defines the $m$-dimensional density of a set $E\subset \mathbb{R}^n$ near a point $x$ for $m\le n$ by $$\lim_{r\to 0} \frac{\mathcal{H}^m(E\cap B^n(x,r))}{|B^m(0,r)|}$$ Note that in the numerator, the $n$-dimensional ball appears, whereas in the denominator it's the $m$-dimensional ball! For $m=n$ the ...


4

As Henning says, whether or not you have a Banach-Tarski paradox depends on what symmetries of the Cantor set you allow. My understanding is that you want to allow all measure preserving continuous bijections. With this understanding, the Cantor has a Banach-Tarski decomposition. Note that we can identify the Cantor set with the $2$-adic integers, ...


4

It is, at least partially, a chicken-and-egg problem. We'd like to define the Lebesgue integral which has nicer convergence properties when we integrate sequences of functions, and take limits of the functions and integrals -- but the definition of the Lebesgue integral requires that we already have a well-behaved measure (i.e., "area") concept to build it ...


4

A good anotated list of textbooks on geometric measure theory can be found in this blog post. Besides comments on Federer and Mattila it has several more examples. As my personal favorite I found, while lecturing geometric measure theory, "Measure Theory and Fine Properties of Functions" by Evans and Gariepy. It is short and crisp (often you have to build ...


4

A $k$-dimensional subspace $S$ of $\mathbb R^n$, with $k<n$, is the range of linear transformation of rank $k$. That means that there exists a matrix $A\in\mathbb{R}^{n\times n}$, with $\mathrm{rank}(A)=k<n$, such that $\mathrm{Ran}(A)=S$. But then $\det A=0$. The theorem of change of variables $$ m(S)=\int_{\mathbb R^n}\lvert\det A\rvert\,dx=0. $$ ...


4

This is false even for curves in the plane. Consider a sequence of ellipses converging to a line segment. The length of the limit curve drops by the factor of 2. If you want an example of curves without boundary, consider a similar example in 3d space. Or consider curves which are disjoint unions of pairs of concentric circles converging to a single circle. ...


3

I think a good thing for you to think about is upper and lower Riemann sums. If you have an over-approximation of the area (via the assumption of finite additivity of area) and an under-approximation of the area (again via this assumption), and the over-approximation and under-approximation converge to the same thing then we can safely call/define this the ...


3

Anon beat me to the point in the comments above :) In measure theory, there is a notion of regular measures of which the Lebesgue measure is an example. Being regular essentially says that (measurable) sets can be approximated in this way. Now if you're thinking of things like integrals, remember that integrals are based upon measures, and that a function ...


3

In case you are still interested in an answer, I encourage you to look at Leon Simon's notes. The idea of currents, is to serve as a far reaching generalization of oriented submanifolds. They were introduced as part of the fairly long and involved program that went into solving the Plateau's problem: such a minimization problem would require three important ...


3

I would read about the local dimension of a measure here.


3

Hint: Nearest-point projection from a sphere $S$ containing $A$ onto the boundary of $A$ is a contraction.


3

One notion of complex dimension that has been used extensively has to do with self-similar sets. A $t$-neighborhood (i.e. points within distance $t$) of such a set may have volume $v(t)$ bounded above and below by constant multiples of $t^d$, where $d$ is the dimension of the boundary and $t$ is small, but such that $t^{-d} v(t)$ is oscillatory and ...


3

Of course most of these don't transform $A$ into a ball at once, but can be used to get there in the limit. I would add spherical symmetrization to the list; slicing $A$ by concentric spheres $S_r$, then rearranging each $A\cap S_r$ into a spherical cap (say, centered at North pole) of the same spherical measure. This is a generalization of classical ...


3

In practice, I prefer the (equivalent) following definition of upper Minkowski dimension: $$\overline{\dim}(E) = \limsup_{\delta \to 0} \frac{\log N(E,\delta)}{\log(1/\delta)}. $$ $N(E, \delta)$ is the number of $\delta$-balls needed to cover the set $E$. (We need $E$ to be compact). (Throughout, we write $X \approx Y$ if there is a constant $c\geq 1$ ...


3

The set $[0,1]\cup \{2\}$, with the metric induced from $\mathbb R$, does not admit a nontrivial Hausdorff measure. To obtain a connected, and less trivial example, stick a line segment into a square in $\mathbb R^2$. No matter what $\varphi$ is, either the line segment gets zero measure, or the square (containing infinitely many segments) gets infinite ...


3

The Minkoswki content is a rather simplistic way to define an $m$-dimensional measure of an object immersed in the euclidean space (however mathematically speaking Minkowski content is not a measure). If $\mathcal A$ is a regular $m$-dimensional surface in $\mathbb R^n$ it is easy to understand that the Minkowski content (both lower and upper) gives the ...


3

Be advised that Federer's book is not for newcomers to the subject (and it's not an online reference anyway). An accessible introduction to isoperimetric and related inequalities is The Brunn-Minkowski inequality by Gardner. It focuses on the volume of subsets of $\mathbb R^n$, thus avoiding most of the GMT machinery. A somewhat dated, but still good, ...


3

Roughly speaking, the answer is "because you have a (bi)cone". Any smooth embedded manifold (in fact any $C^2$ manifold) will have the property that the appropriately defined density is precisely equal to $1$. But if you have something that is not smooth, you can have other values. If you have just a single cone $z = \sqrt{x^2 + y^2}$, near the origin ...


3

I'll answer in greater generality. A common way to construct a measure is to take a nonnegative locally integrable function $w$ and define $\mu(E)=\int_E w(x)\,dx$. This does not give all measures (only those that are absolutely continuous with respect to $dx$) but for many examples that's enough. In terms of $w$, the desired condition translates to ...


3

I think you misunderstand the basic concepts of the subject. The expression $\nabla \chi_{E_c}$ is not "a function $v:\partial E_c\to\mathbb R^2$", but a distribution. If your sublevel set $E_c$ happens to be a Caccioppoli set (which is not guaranteed; the sublevel set of a Lipschitz function can be any closed set) then $\nabla \chi_{E_c}$ is a ...


2

Hint: Since $X$ is locally compact and separable, there exists a countable basis $\{U_i\}_{i=1}^{\infty}$ for the topology such that $\overline{U}_i$ is compact for every $i$...


2

Let $v=(v_i)$ be a vector field. If $c=(c_i)$ is the curl of $v$, as you did with the divergence, you can write: $$ c_i(x) = \lim_{A_i\to \{x\}}\frac{1}{\mu(A_i)} \int_{A_i}(\nabla\times v)_i\,d\mu = \lim_{A_i\to \{x\}}\frac{1}{\mu(A_i)} \int_{\partial A_i} (v\times dx)_i, $$ where: $$ (v\times dx)_i = \sum_{j,k}\epsilon_{ijk} v_j\,dx_k $$ are the ...



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