# Tag Info

34

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 ...

17

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 ...

10

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 ${\mathbb R}^... 10 Apparently you ask about metric as two-point function, not about Riemannian metric in the projective space. The answer (consistent with the Riemannian metric) is: angle between lines through the origin in$\mathbb R^{n+1}$(which represent points in$\mathbb RP^n$). The diameter of the space is$\pi/2$. Correct on both counts. Using the Euclidean metric on ... 10 A$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 $$m(S)=\int_{\mathbb R^n}\lvert\det A\rvert\,dx=0.$$ ... 10 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 $$f\in \mathrm{Lip}_{1/2}.$$... 9 The Minkowski 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$m$-... 9 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 ... 8 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 ... 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 ... 7 There are many ways to interpret your question, and even if you only need one, others might be interested in other, so I'll write about a few of them. Projective transformation of the plane You ask whether the same results you obtained by measuring the photo could also be obtained from a formula. The answer is yes, if you know the transformation. The class ... 7 Yes, this is possible. First, let me set up a bit of machinery. Define a dyadic cube to be an open cube of the form$\prod_{i=1}^n \left(\frac{a_i}{2^k},\frac{a_i+1}{2^k}\right)$where$a_1,\dots,a_n\in\mathbb{Z}$and$k\in\mathbb{N}$. Write$N\subset\mathbb{R}^n$for the set of points which have at least one coordinate that is a dyadic rational; note ... 7 This is most easily understood when$N = 1$. Suppose$A \subset \mathbb{R}^m$is a bounded domain and$f: A \rightarrow \mathbb{R}$is a smooth function whose gradient is everywhere nonzero. Its level sets are non-intersecting hypersurfaces. Suppose first that you want to compute the volume of the$A$in terms of the surface areas of the hypersurfaces. ... 6 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, ... 6 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 ...

6

Let's first clarify this definition. For an $m$-form $\omega \in \Lambda^m(V^*)$ on an $n$-dimensional vector space $V$ with inner product $\langle \cdot, \cdot \rangle$, we are defining \begin{align*} \Vert \omega \Vert & = \sup\left\{\left|\omega(v_1 \wedge \cdots \wedge v_m)\right| \colon \text{ vectors } v_1, \ldots, v_m \in V \text{ with } \left|... 6 It's false; f(A) need not be Borel. Let C be the middle-thirds Cantor set. Define \pi_1:C\times C\to C by \pi_1(x,y)=x. True Fact There exists a Borel set B\subset C\times C such that \pi_1(B) is not Borel. Hand Waving Well, that's well known with \Bbb R in place of C. There simply cannot be any difference between \Bbb R and C in this ... 6 In your estimates of s-dimensional Hausdorff measure of a set E, you have to do this: For any \epsilon > 0, choose a countable cover \{U_k\} of E with \text{diam}\; U_k < \epsilon for all k. Then your estimate is \sum_k \left(\text{diam}\; U_k\right)^s \tag{$*$} $$You take the infimum over all covers. Then the limit as \epsilon \... 5 Hint: Nearest-point projection from a sphere S containing A onto the boundary of A is a contraction. 5 In case you are still interested in an answer, I encourage you to look at Leon Simon's notes. The idea of currents, is to serve as a far reaching generalization of oriented submanifolds. They were introduced as part of the fairly long and involved program that went into solving the Plateau's problem: such a minimization problem would require three important ... 5 This is false even for curves in the plane. Consider a sequence of ellipses converging to a line segment. The length of the limit curve drops by the factor of 2. If you want an example of curves without boundary, consider a similar example in 3d space. Or consider curves which are disjoint unions of pairs of concentric circles converging to a single circle. 5 I regret having to note that the other answer is wrong. It is well known, since work of Moran in 1946, that the Hausdorff dimension of the particular set that you describe does not depend on the location of the intervals at each step. For a proof you can see the book Dimension and Recurrence in Hyperbolic Dynamics, by Barreira (see Theorem 3.1.1). 5 There are homeomorphisms (see Henning Makholm's very good answer) that take a set of zero measure to a set of positive measure (where again measure is taken w.r.t. some local charts), so without modification "measure zero" is not well-defined for topological manifolds. (This pathology is not possible, however, on smooth manifolds.) On the other hand, ... 5 Take a set C\subset\mathbb{R}^n that is not Borel. For every X\subset \mathbb{R}^n, let$$ \mu(X) = \begin{cases} 0 & \text{if } X\subset C; \\ \infty & \text{if } X\not\subset C. \\ \end{cases} $$This is a measure on the entire P(\mathbb{R}^n). In particular, all Borel sets are measuable. For every Borel set B\supset C we have \mu(B)=\... 5 We will use the fact that$$ |x_{1}-x_{2}|^{\alpha}\leq(|x_{1}-x_{3}|+|x_{2}-x_{3}|)^{\alpha}\leq |x_{1}-x_{3}|^{\alpha}+|x_{2}-x_{3}|^{\alpha}. $$Let E\subseteq\mathbb{R}^{n} and let f:E\rightarrow\mathbb{R} be such that  |f(x)-f(y)|\leq L|x-y|^{\alpha}  for all x,y\in E. Define$$ h(x):=\inf\left\{ f(y)+L|x-y|^{\alpha}:\,y\in E\right\} ,\quad ...

5

As suggested by fourierwho, perhaps the most the natural domains for which the divergence (also called Gauss-Green) theorem holds are the sets of finite perimeter, i.e. Caccioppoli sets, so let's precisely see why. Definition 1 ([1], §3.3 p. 143). Let $\Omega$ a Lebesgue measurable set in $\mathbb{R}^n$. For any open subset $G\subseteq\mathbb{R}^n$ the ...

5

The axioms of a $\sigma$-algebra itself do not guarantee that the sets can't produce the Banach-Tarski paradox. After all, the full power set of $\mathbb R^3$ is a $\sigma$-algebra and obviously allows Banach-Tarski. What prevents it is the defining feature of the Lebesgue $\sigma$-algebra (the measurable sets). For any set $E\subseteq \mathbb R^n$ to be in ...

4

Some remarks. The original question: Does a function $a(y)$ exist? is (by the Radon-Nikodym theorem) equivalent to the question Is the image measure $D_*H^m$ absolutely continuous with respect to $H^m$? (Let's suppose $A$ has sigma-finite measure.) And the answer is yes if $D$ is a non-singular matrix; the answer is sometimes no if $D$ is a ...

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