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Consider the surface (hyperboloid) $S=\{(x,y,z)\in \mathbb{R}^3: \frac{1}{z}+\frac{1}{y}+\frac{1}{x}=0\}$. How to prove that: it is of Lebesgue measure zero?

The idea is to show that for all $\epsilon>0$, there is a countable set of Cubes$E_n$ such that $S$ is contained in the union of $\{E_n: n\in N\}$ and that $\sum_{n=1}^\infty m(E_n)<\epsilon$

Raafat
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3 Answers3

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I don't know if this helps you.

The function $f(x,y,z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is defined on $V=(\mathbb{R}^3$ \ $\{0\}) \times (\mathbb{R}^3$ \ $\{0\}) \times (\mathbb{R}^3$ \ $\{0\})$ and is continuous there.

Let $\epsilon>0$

Take the cube $U=(-\epsilon,\epsilon) \times (-\epsilon,\epsilon) \times (-\epsilon,\epsilon)$

$B=\mathbb{R}^3$ \ $U$ is closed and $B \subseteq V$

So you can write $B$ as a countable union of bounded closed cubes.

A closed cube in $\mathbb{R}^n$ is compact .

So if you write $B=\bigcup_{n=1}^{\infty}C_n$ where every $C_i$ is a compact set.

Now the restriction of $f$ to $C_n$ is continuous and it is proved that in this case the graph of $f$ has measure zero on $C_n$

The graph of $f$ on each $C_n$ has a finite area say $S_n$ in the forth dimensional space now.

So you can cover that part of the graph with cubes of length sides $\frac{\sqrt[4]{\epsilon}}{2^m}$.

The volume of each cube will be at $\frac{\epsilon}{(2^m)^4}=t_{m,n}$

So you will need at most $O(At_{m,n})$ cubes for that part where the function is defined on $C_n$

And also each area $S_n$ is strictly smaller than $t_{m,n}$

So the max volume of the cubes would sum up to $O(\epsilon)$

Now the problem is reduced to $U$.

Marios Gretsas
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To get you started: Let $I = [0,1]$ and $f:I^2 \to \mathbb R$ is continuous. Then $G=\{(x,y,f(x,y)): (x,y) \in I^2\}$ has measure $0$ in $\mathbb R^3.$

Proof: Let $\epsilon>0.$ Because $f$ is uniformly continuous on $I^2$ there exists $\delta >0$ such that $p,q\in I^2,|p-q|<\delta$ implies $|f(p)-f(q)| < \epsilon.$ Choose $n$ such that $1/n < \delta/\sqrt 2.$ Partition $I^2$ into a standard grid of $n^2$ closed squares of side length $1/n.$ Label the squares $S_1,\dots, S_{n^2}$ and choose points $p_k \in S_k, k = 1,\dots ,k^2.$ Then

$$G\subset \bigcup_{k=1}^{n^2} S_k \times [f(p_k)-\epsilon,f(p_k)+\epsilon].$$

Thus $m_3(G) \le n^2\times \dfrac{1}{n^2}\cdot 2\epsilon = 2\epsilon.$

zhw.
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  • You may mean $]0,1]$ not $[0,1]$ ! – Raafat Sep 14 '17 at 23:51
  • @Raafat No, I meant what I wrote. – zhw. Sep 15 '17 at 00:21
  • @zho But $f(x,y) $ does not continuous at $(0,0)$ !! – Raafat Sep 15 '17 at 08:29
  • @zho Also f is not uniformly continuous on $]0,1]$ !! I think that the continuity may be enough! – Raafat Sep 15 '17 at 09:19
  • @Raafat In your problem $f$ is defined on an open subset of $R^2.$ That open set is the union of countably many closed squares. On each such square, $f$ is continuous. So the result follows by applying the result I sketched (which holds for any closed square, not just $I^2.) – zhw. Sep 15 '17 at 16:12
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Here is my attempt to solve: Denote by $O_i,i=1,2,...,8$ be the eight octants in $R^3$. Let $S_i$ be the portion of the surface in the $O_i$ octant. We do the following with respect to $O_1$. We have

Let $E_n=[n,n+1]\times [n,n+1]\times[f(x,y)-\frac{\epsilon}{2^{n+4}},f(x,y)+\frac{\epsilon}{2^{n+4}}]$. Then

$\mu (E_n\cap S_1)\leq\frac{\epsilon}{8\times 2^n} $. This implies that

$\mu (S\cap (R^3/[0,1]^3)<\epsilon$.

Therefore, $S\cap (R^3/[0,1]^3)$ is of lebesgue measure zero. The proof of $\mu (S\cap ]0,1]^3)<\epsilon$ is similar and will be omitted

Marios Gretsas
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Raafat
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