Skip to main content
14 votes
Accepted

What's the intuition behind the Co-Area formula?

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 ...
Deane's user avatar
  • 8,247
11 votes
Accepted

Volume form and measure on manifold

The easiest way to do it is to use the Riesz representation theorem. Given a volume form $\omega \in \Omega^m(M)$ (where $m = \dim M$), we have a linear functional $\Lambda \colon C^{\infty}(M) \...
levap's user avatar
  • 65.9k
9 votes

Can we always extend a Holder-boundary continuous function to whole domain?

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\...
Gio67's user avatar
  • 21k
9 votes
Accepted

Hausdorff measure on non separable spaces

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 $\text{diam}\; U_k < \epsilon$ ...
GEdgar's user avatar
  • 113k
8 votes
Accepted

Filling a unit cube with countable balls.

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,...
Eric Wofsey's user avatar
8 votes

Can a set of Hausdorff codimension 2 disconnect a connected open set?

$U\setminus S$ is path connected as soon as $\mathrm{dim} \,S<n-1$. Let us prove the local version, that is for $U$ equal to a ball. Suppose you want to connect $x,y\in U\setminus S$. Since $S$ is ...
Del's user avatar
  • 4,066
7 votes

$f : \mathbb{R} \to \mathbb{R}$ (Lipschitz) continuous implies $f(A)$ is Borel for all Borel $A$.

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 $\...
David C. Ullrich's user avatar
7 votes

The Hausdorff dimension of the zero set of a real analytic function

I would like to propose an elementary approach based on the implicit function theorem. Consider $F_0=f^{-1}(0)$. By the implicit function theorem if $\nabla f(x)\neq 0$ for some $x\in F_0$ then $F_0$ ...
Del's user avatar
  • 4,066
7 votes

The Lebesgue measures of open sets of points of epsilon distance from a compact set converge to the Lebesgue measure of that compact set

You first need to show that, for $\epsilon_n$ decreasing to 0, $\cap_n E_{\epsilon_n} = E$. Now, fix any $x \in \cap_n E_{\epsilon_n}$. Since $x \in E_{\epsilon_n}$, by definition there is a point $...
David Gao's user avatar
  • 7,665
6 votes
Accepted

Domains for which the divergence theorem holds

As suggested by fourierwho in their comment, 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 ...
Daniele Tampieri's user avatar
6 votes
Accepted

The Hausdorff dimension of the zero set of a real analytic function

Yes, this is true and follows, e.g. from Łojasiewicz's stratification theorem: Every real-analytic subset of $R^n$ is a locally finite (hence, countable) union of pairwise disjoint smooth real-...
Moishe Kohan's user avatar
6 votes
Accepted

Confusion on integration by parts on a Riemannian manifold

Why do we not have to take the complex conjugate as in Euclidean space? Because we’re dealing with real-valued stuff. “Does this just mean we can always treat $dV$ (or $d\mu$ in the notation of the ...
peek-a-boo's user avatar
6 votes
Accepted

Stokes' theorem in differential geometry vs measure theory

Both of your two theorems try to restrict the amount of smoothness needed for Stokes theorem to hold. I would say in practically all applications you only need a basic version for smooth compact ...
quarague's user avatar
  • 5,963
5 votes
Accepted

Image of Lipschitz map measure zero

Below are two proofs of this fact. Neither is self contained; the first is more classical, but I added the second as I find it very interesting. Let $\mu$ denote the $n$-dimensional Lebesgue measure, ...
Silvia Ghinassi's user avatar
5 votes
Accepted

Blow-up of derivative of BV function at the jump set

I think you forgot to divide by $r^{n-1}$. Even for sets of finite perimeters what you are claiming is not true. If you take $E=\{(y,t)\in\mathbb{R}% ^{n-1}\times\mathbb{R}:\,t<f(y)\}$, where $f$ ...
Gio67's user avatar
  • 21k
5 votes
Accepted

Details in a proof of coarea formula in $\mathbb{R}^n$.

No, you cannot assume $\tilde B$ to be independent of $\lambda$. Consider the example $u(x)=x_n$ and $B=B^n(0,1)\subset\mathbb R^n$. Now $\Gamma_\lambda=\{\lambda\}\times B^{n-1}(0,\sqrt{1-\lambda^2})$...
Joonas Ilmavirta's user avatar
5 votes
Accepted

Why is Hausdorff measure Borel regular?

$\DeclareMathOperator{\diam}{diam}$First off, we can replace the $C_j$ with either open or closed sets for the reasons described in your second image. Since you seem to be a little confused on that ...
Xander Henderson's user avatar
5 votes
Accepted

How did we resolve the Banach-Taraski paradox?

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 ...
eyeballfrog's user avatar
  • 22.8k
5 votes

Linear map of borel set is a measurable set?

If by measurable set you mean a Borel set then the answer is NO. There exist Borel sets in $\mathbb R^{2}$ whose images under the projection $(x,y) \to x$ are not Borel. See A Borel set whose ...
Kavi Rama Murthy's user avatar
5 votes

Help with the proof that $E\subset \mathbb{R}$ with finite perimeter and area has to be equal to the finite union of bounded intervals

In case you fancy Lebesgue density point there is a fairly simple argument that goes as follows: Suppose $a<b$ are Lebesgue density points of $E$ and $E^c$, respectively. Then by Lebesgue density ...
H. H. Rugh's user avatar
  • 35.3k
5 votes

Investigating a $1/2$-dimensional sphere

Since no one has answered this question yet, let me write what I think about this interesting, but ill posed (in my opinion) question. First of all, In mathematics there are multiple meaning of ...
Bumblebee's user avatar
  • 18.3k
5 votes
Accepted

Understanding the notion of approximate tangent space, with examples

I think part of the confusion comes from the fact that the approximate tangent space is fundamentally about measures, whereas you are asking about the approximate tangent space of sets. Now there is a ...
felipeh's user avatar
  • 3,810
5 votes
Accepted

Error in Falconer's proof of Hausdorff dimension of Cantor set

When you choose the "largest such intervals", I don't believe it's necessary for them to be net intervals (although that certainly seems to be what Falconer is taking them to be). First note ...
lonza leggiera's user avatar
5 votes
Accepted

If there is a zero measure set with positive measure pre-image, is there a level set with positive measure?

There is a $C^\infty$ counterexample. For $x \in [0,1]$, let $$\psi_0(x)=\int_0^x\exp\left(- \, \frac1{t(1-t)}\right)\,dt \,.$$ Then the strictly increasing function $\psi \in C^\infty [0,1]$ ...
Yuval Peres's user avatar
  • 22.2k
4 votes

Change of Variables for Hausdorff Measure

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 ...
GEdgar's user avatar
  • 113k
4 votes
Accepted

Using small spherical balls to fill a cube and also find volume of the cube

Your problem is the same as filling balls of unit radius into cubes of larger and larger side length and taking the limit as the side length goes to $+\infty$. This converts your problem into the ...
Lee Mosher's user avatar
  • 123k
4 votes
Accepted

Suitable non-countable union of sets with measure zero is still a set of measure zero?

Let us denote your set by $E$, Lebesgue measure on $\mathbb{R}$ by $\mu_1$, and Lebesgue measure on $\mathbb{R}^2$ by $\mu_2$. If $E$ is measurable, then by Fubini's theorem applied to the ...
Eric Wofsey's user avatar
4 votes

What is the definition of $L^{p}(M)$, $M$ is a Riemannian manifold?

For (1): No, that approach no longer works if $f$ no longer has compact support. Remember that the convenience of compactness resides in the fact that you may cover a compact set by finitely many ...
Alex M.'s user avatar
  • 35.3k
4 votes
Accepted

Are immersions injective outside a set of measure zero?

(Updated to take into account $\partial M \ne \emptyset$) I'll write my answer using the Lebesgue measure class of a smooth volume form on $M$. The answer is that $A$ exists if and only if the ...
Lee Mosher's user avatar
  • 123k
4 votes
Accepted

Other references for Marstrand's density theorem

What a curious coincidence! I was reading Marstrand's theorem from De Lellis' notes just a couple of weeks ago (after reading it a year ago without checking all the details) and I noticed the same ...
Del's user avatar
  • 4,066

Only top scored, non community-wiki answers of a minimum length are eligible