If $\mathcal{H}^{n-1}(K)=0$ then $\mathcal{H}^n(K\times \mathbb{R})=0$

Question

I am reading a paper Simon and Wickramasekera - A Frequency Function and Singular Set Bounds for Branched Minimal Immersions where the authors seem to claim that if $K\subset\mathbb{R}^n$ is a compact set with $\mathcal{H}^{n-1}(K)=0$ in $\mathbb{R}^n$ ($\mathcal{H}^{n-1}$ is the $(n-1)$-dimensional Hausdorff measure), then $\mathcal{H}^n(K\times \mathbb{R})=0$ in $\mathbb{R}^{n+1}$.

I want to know how to prove this.

For example, I would be able to prove it if I knew how to prove the following: If $K\subset \mathbb{R}^n$ is a compact set such that $\mathcal{H}^s(K)=0$ ($\mathcal{H}^s$ is the $s$-dimensional Hausdorff measure) for some $s<n$, then for all $\epsilon>0$ $K$ can be covered by finitely many $B_r(x_i)$, $i=1,...,N$ (same radius) such that $N r^s<\epsilon$.

Any hints are appreciated.

Answer 1

$\newcommand\de\delta\newcommand\ep\varepsilon\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Saúl RM proved the desired result by an application of Frostman's lemma.

Here is an elementary proof:

Take any $\de$ and $\ep$ in $(0,1)$. Then, by the condition $H^{n-1}(K)=0$ and the definition of the Hausdorff measure, there is a set $\{Q_i\colon i\in\N\}$ of closed $n$-cubes in $\R^n$ (that are product sets) such that $\bigcup_{i\in\N}Q_i\supseteq K$, $0<l(Q_i)<\de$ for all $i\in\N$, and \begin{equation*} \sum_{i\in\N}l(Q_i)^{n-1}<\ep/2, \tag{1}\label{1} \end{equation*} where $l(Q_i)$ is the edge length of the $n$-cube $Q_i$.

For each $i\in\N$, let $N_i:=\lceil 1/l(Q_i)\rceil$, so that \begin{equation*} \frac1{l(Q_i)}\le N_i\le1+\frac1{l(Q_i)}<\frac 2{l(Q_i)}. \tag{2}\label{2} \end{equation*} For each $i\in\N$ and each $j\in[N_i]:=\{1,\dots,N_i\}$, consider the $(n+1)$-cube \begin{equation*} R_{i,j}:=Q_i\times[(j-1)l(Q_i),jl(Q_i)], \end{equation*} with edge length $l(R_{i,j})=l(Q_i)<\de$. Then $\bigcup_{i\in\N}\bigcup_{j\in[N_i]}R_{i,j}\supseteq K\times[0,1]$ and \begin{equation*} \sum_{i\in\N}\sum_{j\in[N_i]}l(R_{i,j})^n =\sum_{i\in\N}\sum_{j\in[N_i]}l(Q_i)^n =\sum_{i\in\N}N_i l(Q_i)^n \\ <\sum_{i\in\N}\frac 2{l(Q_i)}l(Q_i)^n =2\sum_{i\in\N}l(Q_i)^{n-1}<\ep, \end{equation*} by \eqref{2} and \eqref{1}.

So, again by the definition of the Hausdorff measure, \begin{equation} H^n(K\times[0,1])=0. \end{equation} So/similarly, $H^n(K\times[k,k+1])=0$ for all integers $k$.

So, by the subadditivity of the Hausdorff measure, $H^n(K\times\R)=0$. $\quad\Box$

(The condition that $K$ is compact was not needed or used here.)

Answer 2

Frostman's lemma seems to work for this problem.

Suppose that $H^n(K\times\mathbb{R})>0$. Then $H^n(K\times[0,1])>0$, so there is a measure $\mu$ in $\mathbb{R}^{n+1}$ with $\mu(K\times[0,1])>0$ and $\mu(B(x,r))\leq r^n$ for all $x\in\mathbb{R}^{n+1}$ and $r>0$.

Now consider the measure $\nu$ in $\mathbb{R}^n$ given by $\nu(A)=\mu(A\times[0,1])$ for Borel sets $A$. Then $\nu(K)>0$, and for any ball $B(x,r)$ in $\mathbb{R}^n$ we can cover the set $B(x,r)\times[0,1]\subseteq\mathbb{R}^{n+1}$ by $\frac{N_n}{r}$ balls of radius $r$, where $N_n$ is some big constant dependent only on $n$.

So we have $\nu(B(x,r))\leq\mu(B(x,r)\times[0,1])\leq\frac{N_n}{r}r^n=N_nr^{n-1}$ for all $x\in\mathbb{R}^n,r>0$, which by the Frostman lemma implies $H^{n-1}(K)>0$.