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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 ...

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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 remaining....

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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 ...

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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 ${\mathbb R}^... 8 We want to mimic$f(x) = x^{-1/2}$on$[0,1]$, but then cut$[0,1]$into lots of intervals, and then on each interval remake the function so that its integral over that interval remains the same, but the support of the function is much smaller. So let$x_n \to 0$be a decreasing sequence with$x_0 = 1$. Write$x_{n} = (1+\epsilon_n) x_{n+1}$, and suppose ... 8 No. Just for good old$\mathbb R$, there is a homeomorphism$[0,1]\to[0,1]$which takes the middle-thirds Cantor set (of measure$0$) to the "fat" Smith-Volterra-Cantor set (of measure$1/2$). (The excluded intervals in the two constructions, including their endpoints, correspond naturally to each other -- simply map each of them to its counterpart ... 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 ... 7 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 ... 6 Algebraic stacks are a far-reaching generalization of algebraic varieties. If an algebraic variety is considered as a stack, then its dimension as stack is the same as its dimension as variety. However there are many stacks that do not correspond to varieties, and some of these have negative dimension. Specifically, if$V$is a variety and$G$is an ... 6 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 non-... 6 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 $$m(S)=\int_{\mathbb R^n}\lvert\det A\rvert\,dx=0.$$ ... 6 First, notation/terminology: Talking about Lipschitz maps with respect to that nonstandard metric is a bad idea, simply because there exists a simple and much more standard way to say the same thing. Your function is Lipschitz with respect to that funny metric if $$|f(x)-f(y)|\le c|x-y|^{1/2}.$$ This is exactly the definition of $$f\in \mathrm{Lip}_{1/2}.$$... 5 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 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 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,$\mathbb{...

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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 ...

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I will concentrate on comparing (3) and (4). The definition (1) is meant for finite signed measures, whereas all the other definitions are meant for arbitrary positive measures; (1) is equivalent to (4) in the case of finite positive measures. (2) appears to be equivalent to (4) ["locally finite" can mean "finite on compact sets", although it is sometimes ...

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

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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, ...

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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 $m$-...

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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.

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I regret having to note that the other answer is wrong. It is well known, since work of Moran in 1946, that the Hausdorff dimension of the particular set that you describe does not depend on the location of the intervals at each step. For a proof you can see the book Dimension and Recurrence in Hyperbolic Dynamics, by Barreira (see Theorem 3.1.1).

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The two measures are within a constant factor of each other. In K. Falconer, Fractal Geometry: Mathematical Foundations and Applications, see the material on "net measure". In the 2nd edition, that is page 36

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As felipeh noted, the problem reduces to $p=1$ by replacing $f$ with $g = |f|^p$. (The case $p=\infty$ should be treated separately.) Also, the annulus can be replaced by the sphere $S^{n-1}$. Indeed, define a function $h$ on the unit sphere by $h(\xi) = \int_r^1 g(t\xi)\,dt$. Then $h\in L^1(S^{n-1})$, with a norm comparable to $\|g\|_{L^1(A)}$ because ...

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Assume we have a Jordan curve of positive area measure whose complement has connected components $U,V$ as in the Jordan Curve theorem. Let $U$ be the bounded component. Every point in $\partial U = \partial V$ is the limit of a sequence of points in $\overline U$ and the limit of a sequence of points in $V= \overline U ^c.$ Thus the boundary of $\overline ... 4 The answer is negative. Let's consider a Cantor-type set$K\subset [0,1]$built by removing the$c_j$proportion of each interval at$j$th step ($c_j<1$,$j=1,2,\dots$). It's a compact set with empty interior, so$\partial K=K$. The measure of$K$is zero provided that $$\prod_{j=1}^\infty (1-c_j)=0$$ Hence,$K$is Jordan measurable in this case. On ... 4 No. As shown in Lebesgue measure of the graph of a function, there exists a function$f : \mathbb{R} \to \mathbb{R}$whose graph$G = \{(x,f(x)) : x \in \mathbb{R}\}$is a Lebesgue non-measurable subset of$\mathbb{R}^2$. But if we let$\pi(x,y) = (x,0)$be projection onto the$x$axis, then$\pi|_G : G \to \mathbb{R}$is bijective. If you can first show ... 4 Take a set$C\subset\mathbb{R}^n$that is not Borel. For every$X\subset \mathbb{R}^n$, let $$\mu(X) = \begin{cases} 0 & \text{if } X\subset C; \\ \infty & \text{if } X\not\subset C. \\ \end{cases}$$ This is a measure on the entire$P(\mathbb{R}^n)$. In particular, all Borel sets are measuable. For every Borel set$B\supset C$we have$\mu(B)=\...

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