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Can There be a domain in $\mathbb{R^n}$, for any $n$ such that some domain has non zero boundry volume? I.E. volume of boundry is non zero?

Motivation:

In some theorems, it is specified that volume of boundary is non zero. But I cannot think of domains where volume of boundry is non zero.

EDIT

If domain by definition is expected to be open subset of $\mathbb{R^n}$, then I would be looking for such open subsets.

Thank You.

  • Maybe I'm not understanding the problem and oversimplifying it, but what about in ${(x,y,z)|x^2+y^2+z^2=1}$? The volume is $\frac{4}{3}\pi$. – rurouniwallace May 11 '13 at 15:33
  • that is volume of function, not boundry. –  May 11 '13 at 15:38
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    What exactly do you mean by domain? Arbitrary subsets? Open subsets? – fgp May 11 '13 at 15:41
  • I believe domain would any subset. Is a domain defined as open? Then I overlooked some fine detail. –  May 11 '13 at 15:44
  • If open subset is standard definition then that is what I should be looking for. –  May 11 '13 at 15:45
  • The word "domain" has many different meanings: In complex analysis it usually means a connected open subset, some topologists (Engelking) take it to be a regular open subset (closure = (closure of interior)), etc. – Martin May 11 '13 at 15:48
  • @Martin, Thank you for the info. Would it be different in real analysis? –  May 11 '13 at 15:51
  • I like this question a lot, and I'm curious whether there exists a more "geometric" example, i.e. an example for which the open set is the interior of its closure. I have posted this as a new question. – Jim Belk May 11 '13 at 17:38

2 Answers2

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Let $(0,1) \cap \mathbb Q = \{ q_1,q_2, \dots\}$. We define $A = (0,1) \cap\bigcup_k (q_k-4^{-k}, q_k+4^{-k})$. By construction we have $\bar A = [0,1]$ but an easy estimate shows that $\lambda( A) < 1$ where $\lambda$ is the Lebesgue measure.

If you want a connected example, you can take $\left(A \times (0,1)\right) \cup \left((0,1) \times (0, 1/2)\right) \subset \mathbb R^2$. This one is actually contractible.

Dark Malthorp
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Alexander Thumm
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  • Can you elaborate on why $\big(A × (0,1)\big) ∪ \big((0,1) ×(0,1/2)\big)$ is contractible? – k.stm May 11 '13 at 18:04
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    That space looks like a comb. We can contract the 'teeth' first by 'pushing them into the shaft' ($h_t(x,y) = (x, 1/4t + y(1-t))$) and then contract the 'shaft' ($k_t(x,1/4) = (1/2t + x(1-t),1/4)$). In this way we obtain a homotopy from $id_A$ to the map that sends $A$ to $(1/2,1/4)$. – Alexander Thumm May 11 '13 at 18:23
  • Thanks. I had it, but then lost it again. – k.stm May 11 '13 at 18:32
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Let $ν \colon ℕ → [0,1]∩ℚ$ be a enumeration of the rationals there. Then the set $M := \bigcup_{k∈ℕ} B_{r(k)}\big(ν(k)\big)$ with $r(k) = \tfrac{1}{3}·\big(\tfrac{1}{2}\big)^k$ has measure less than $\sum_{k=1}^∞ 2r(k) = \tfrac{1}{3}·1 < 1$ (as $B_{r(k)}$ has measure $2r(k)$), so it’s not all $[0,1]$. But $M$ lies dense in $[0,1]$ since it contains all the rationals. Therefore, the boundary of $M$ has to be $[0,1]\setminus{M}$, having nonzero measure, viz measure $\tfrac{2}{3}$. Also $M$ is open because it’s the union of open balls.

k.stm
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