Fubini’s theorem for Hausdorff measures

$Bsubset mathbb{R}^2$ is a Borel set. Define the slices $B_x:= {y in mathbb{R}: (x,y) in B }$.
If $lambda$ denotes the Lebesgue measure on $mathbb{R}$, presentations of Fubini’s theorem often include that fact that the function $lambda(B_x)$ is measurable.

Question: If $H^s$ denotes the $s$th Hausdorff measure, how do I see that the function $H^s(B_x)$ is measurable? $(star)$

I came across this while looking at Marstrand’s slice theorem in a book. The authors suggest to use a Monotone class argument wherein one would have to show the following

  • If $B=Utimes V$ then, $H^s(B_x)= mathbb{1}_U(x)cdot H^s(V)$ which is measurable.
  • If $B$ is a finite union of disjoint rectangles then again $H^s(B_x)$ is measurable.
  • If $B_n$ is an increasing family of sets each of which satisfies $(star)$ then $H^sleft((cup B_n)_xright) = H^s(cup (B_n)_x) = lim H^s((B_n)_x)$ which is again measurable.
  • If $B_n$ is a decreasing family of sets each of which satisfies $(star)$, one would like to show the same for $H^s((cap B_n)_x)$. However, this is equal to $H^s(cap (B_n)_x)$. This in general won’t be $lim H^s((B_n)_x)$ since we don’t know if any of the terms has finite $H^s$ measure.

I don’t see how to prove the last point in the absence of $sigma$-finiteness. Am I missing something easy?

Also posted on mse.

at.algebraic topology – Relationship Between Hausdorff Convergence of Sets and Indicator Functions

Let ${K_n}_n$ be a sequence of compact subsets of a metric space $X$, and $Ksubset X$ be compact. If $K_n$ Hausdorff converges to $K$, i.e.:
limlimits_{ntoinfty} d_{mathrm H}(K_n,K) = maxleft{,sup_{x in K_n} d(x,K),, sup_{y in K} d(K_n,y) ,right} = 0

then what can be said of the convergence of $I_{K_n}$ to $I_K$?

What I know:
Indeed, since Hausdorff convergence strictly implies convergence of the upper Kuratowski limits then this post we know that $I_{K_n}$ converges to $I_K$ in $C(X,{0,1})$ for the compact-open topology if ${0,1}$ has the Sipersinki topology.

What I expect: Since the Hausdorff topology is strictly finer than the Sierpinski topology then, is there a topology $tau$ on $C(X,{0,1})$ for which:
I_{K_n} overset{tau}{to} I_K Leftrightarrow K_n overset{d_H}{to} K.

calculus – prove that a Paraboloid is a hausdorff measurable subset

I’m trying to study for my exam and the question is:


A={ ( x,y,z) R^3 : || (x,y )|| , z=1/2 ||( x,y )||^2}

Show that A is a H^2 – measurable subset and calculate the surface area.

I don’t have a probleme computing the surface area but i don’t know how to show the first part.

I would be really thankful if someone could give me a hint.

reference request – Hausdorff Dimension e Surface Measure

Could someone please indicate me some reference that contains the proof of the following theorem?

Below $mathcal{H}^n$ denotes the $n$-dimensional Hausdorff outer measure in $mathbb{R}^n$.

Theorem: Let $Msubset mathbb{R}^N$ be a $k$-dimensional manifold of class $C^1$, $1leq kleq N$.

  1. Let $varphi$ be a local chart, that is, $varphi:Ato M$ is a function of class $C^1$ for some open set $A subset mathbb{R}^k$ such that $nabla varphi $ has maximum rank $k$ in $A$. Define $g_{ij}:=frac{partial varphi }{partial y_i}!cdot !frac{partial varphi }{partial y_j}$, where $cdot$ is the inner product in $mathbb{R}^N$. Then $varphi (A)$ has Hausdorff dimension $k$ and

$$mathcal{H}^k(varphi(A))=int_A sqrt{det g_{ij}(y)}dy$$

  1. $M$ has Hausdorff dimension $k$ and that $mathcal{H}^k(M)$ is the standard surface measure of $M$.

I found this theorem in the file “Measure and Integration” (pg 9).

I searched for some reference that contains the proof of the above theorem but couldn’t find it.

I am posting this request here since a similar question on Mathematics Stack Exchange didn’t receive an answer!

reference request – On the Hausdorff dimension of a Cantor set

In what follows I refer to this paper by Orevkov.

I am writing a paper on this, so if somebody is interested we could consider to write a joint paper.

Consider a sequence $R={R_n}_n$ of strictly increasing positive real number such that $R_ntoinfty$.

Take any $a={a_n}_nsubsetBbb C$ such that $R_{n-1}<|a_n|<R_n$ and define the following polydiscs in $Bbb C^2$:
B_n={|z|le R_n}^2.


f_n(x,y)=left(x,y+g_n(x)right);;mbox{n odd}\
f_n(x,y)=left(x+g_n(y),yright);;mbox{n even}

for $(x,y)inoverline{Bbb C^2}$ which is the one point compactification of $Bbb C^2$.

Define then
gamma_n(z)=f_ncircdotscirc f_1(z,0);.

The sequence $epsilon={epsilon_n}_n$ is defined to go to $0$ fast enough such that $gamma_n$ converges.

Set $A_n=gamma_n(overline{Bbb C})=f_n(A_{n-1})$ and consider the following object
Delta_n=gamma_n^{-1}(A_nsetminus B_n).

Now $A_nsetminus B_nsubsetoverline{Bbb CtimesBbb C}$ has $b_n$ “branches” and some of them go to infinity on the first component, some other on the second. We refer to them as horizontal and vertical; correspondingly we write $b_n=h_n+v_n$.
Creating $Delta_n$, these $b_n$ components become $b_n$ (disjoint) open sets of the Riemann Sphere $overline{Bbb C}$; we write them as

Creating $A_{n+1}=f_{n+1}(A_n)$, what happens, is that the singularity of $f_{n+1}$ is inside the first component for odd $n$, thus it depends on $y$ and thus it creates as many new horizontal branches as the singularity $a_{n+1}$ is met in the vertical components of $A_n$, that is, one for every $v_n$ components.
Far from this singularity, $f_{n+1}$ behaves like the identity, thus just slightly moves the remaining components of $A_n$ (i.e. the $h_n$ horizontal ones)
Hence $b_{n+1}$ is $h_n+2v_n$ and writing $b_{n+1}=v_{n+1}+h_{n+1}$, one has $h_{n+1}=h_n+v_n$ and $v_{n+1}=v_n$.

Call $c_j^{(n)}$ the diameter of $U_j^{(n)}$.

The sequence ${Delta_n}_n$ is decreasing and it can be proved that $C:=bigcap_nDelta_n$ is a Cantor set.

Now I am investigating the Hausdorff dimension of such a Cantor set. I proved that for $R$ going to infinity sufficiently fast, $dim_H(C)$ can be taken arbitrarily small.
It can’t be zero, since its topological dimension is $0$ and fractals are defined to be topological spaces whose Hausdorff dimension is strictly bigger than their topological dimension.

I am searching now for an upperbound.

If $T$ is another diverging sequence and we write $R<T$ to say $R_n<T_n;;forall n$, it’s clear that

(we just made explicit the dependence of the Cantor set by the sequence).

Clearly we need something stronger than $R_n<T_n$ for the above definition, like $R<T$ if $R_n=o(T_n)$ for example; I would say, something such that the inequality $dim_H(C_T)<dim_H(C_R)$ is strict.

But this is still NOT enough to spot an upperbound.

It seems we should find some relation between the sequence $R_n$ and the size $c_j^{(n)}$.

Any hint?

Edit: As pointed out by Gerald Edgar, this Cantor set can have Hausdorff measure zero.

Maybe, since for every $jinBbb N;;exists R^{(j)}={R_n^{(j)}}_n$ such that $dim_H(C_{R^{(j)}})lefrac1j$, then setting $tilde{R_n}:=R_n^{(n)}$ it could be $dim_H(C_{tilde R})=0$.

But then the question is: is it possible to find some $R$ such that $dim_H(C_{R})=alpha>0$? What is an upper bound for $alpha$?

It’s clear that $alphale2$; but for example, is it possible, for every $epsilon>0$ to find $R^{(epsilon)}$ such that $dim_H(C_{R^{(epsilon)}})>2-epsilon$?

reference request – If $(X,tau _X)$ is a locally compact and second countable Hausdorff space, then a Borel locally finite measure is regular

Could somebody please give me one reference that contains the proof of the following theorem?

Let $(X,tau _X)$ be a locally compact and second countable Hausdorff space. Suppose that $mu _X:mathfrak{B}_Xto overline{mathbb{R}}$ is a locally finite measure in which $mathfrak{B}_X$ is the Borel $sigma$-algebra of $(X,tau _X)$. Then the following propositions are true

  1. For all $Binmathfrak{B}_X$ we have $mu_X(B)=inf big{mu_X (A):Bsubseteq Awedge Ain tau_X big}$
  2. For all $Binmathfrak{B}_X$ we have $mu_X (B)=supbig{mu_X (K):Ksubseteq Bwedge (K text{ é compact})big}$
  3. Given any $Bin mathfrak{B}_X$ and $varepsilon in (0,infty)$, there exists $Aintau_X$ and a closed set $F$ such that $Fsubseteq B subseteq A$ and $mu_X(Asetminus F)<varepsilon $.

I know that the above theorem is true because of the theorems that are in the book “Notes on Measure Theory” (Theorem 5.7, pg 64) and the book “Manifolds, Sheaves, and Cohomology” (Proposition 1.10, pg 4)