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26

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 ...


11

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 ...


11

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 ...


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 ...


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

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 ...


5

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 ...


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

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


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. ...


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, ...


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

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 ...


3

Normal Human already gave a good answer, but I came up with another one that I would like to share. It is based on the coarea formula and I tried to make it feel natural, with few arbitrary choices. As has been noted, it suffices to consider $L^1$ functions, so let $f\in L^1$. (For $f\in L^p$ we have $|f|^p\in L^1$.) This argument works for any ...


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

Let $F,U,x,y$ be as in your attempt. For every $\theta$ the projection map $\operatorname{proj}_\theta$ is continuous; hence, $\operatorname{proj}_\theta (U)$ is connected. A connected subset of a line with two distinct points has positive length. For all $\theta\in [0,\pi)$ except one (the one parallel to the line segment $xy$), ...


3

The point is simply that if $F_i = \{\overline{B}_1, \overline{B}_2, \ldots\}$, by countable additivity $$\mu(C \cap \bigcup_{i=1}^\infty \overline{B}_i) = \lim_{N\to \infty} \mu(C \cap \bigcup_{i=1}^N \overline{B}_i)$$


3

It seems to me that what you ask is a particular instance of Theorem 2.10.10 (page 176) of H. Federer "Geometric Measure Theory" (Springer 1969) (here the Zmath reference)


3

It is a standard fact that a Lebesgue measurable set can be written as the union of an $F_\sigma$ set and a set of measure zero. You can adjust this result rather easily to show that a measurable set $E$ has the form $$E = \left( \bigcup_j K_j \right) \cup N$$ where $\{K_j\}$ is a sequence of compact sets and $N$ has Lebesgue measure zero. A continuous ...


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

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 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 ...


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

$|x|^{-t}$ is locally integrable when $t\lt1$. In fact $$ \int_{-1}^1|x|^{-t}\,\mathrm{d}x=\frac2{1-t} $$ This function, being bounded, can be integrated again with no problem. We compute the integral on $[-1,1]$ since $-1\le x-y\le1$ for $x,y\in[0,1]$. In other words, $$ \begin{align} \int_0^1\int_0^1|x-y|^{-t}\,\mathrm{d}x\,\mathrm{d}y ...



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