0
$\begingroup$

Is every compact set rectifiable? The set is rectifiable iff it is compact and the boundary is of measure $0$ (This is stated as a theorem). Can I infer from this that every compact set is rectifiable?

Edit Definition of a rectifiable set: A set is rectifiable if the constant function $1$ is integrable over that set.

Thanks for your help!

$\endgroup$
13
  • $\begingroup$ what is the definition of rectifiable ?' $\endgroup$
    – Learnmore
    Jan 20, 2015 at 5:06
  • $\begingroup$ I added it @learnmore $\endgroup$ Jan 20, 2015 at 5:08
  • $\begingroup$ There are two definition of rectifiability stated. Which one is the correct one? $\endgroup$
    – user99914
    Jan 20, 2015 at 5:37
  • $\begingroup$ Where did you find this definition of rectifiable? What it actually is is a definition of "finite measure." $\endgroup$ Jan 20, 2015 at 5:37
  • 1
    $\begingroup$ I've got it: rectifiable = Jordan measurable with finite measure $\endgroup$ Jan 20, 2015 at 6:11

2 Answers 2

4
$\begingroup$

Construct a modified Cantor set as follows. Start with $A_0 = [0,1]$. Then take out the open subinterval constituting the middle $1/4$. Let $A_1$ be the union of two intervals you have left. Then let $A_2$ be what you have left when you take out the middle $1/9$ of each of those intervals. And so on. Then let $A = \bigcap A_n$.

$A$ is compact with empty interior, so $A$ is its own boundary. However, its measure is $\prod_{n \geq 2} (1 - 1/n^2) > 0$. It might take a little work to show that it's not of measure zero without using measure theory.

$\endgroup$
2
  • $\begingroup$ "This might take a little work to show that it's not of measure zero without using measure theory." Presumably they use measure theory if they know what measure zero means? $\endgroup$
    – Squirtle
    Jan 20, 2015 at 6:22
  • 2
    $\begingroup$ No, it's easy to define a set $S$ of measure zero. It means that for any $\varepsilon > 0$, you can cover $S$ by a countable collection of intervals the sum of whose lengths is $< \varepsilon$ $\endgroup$
    – user208259
    Jan 20, 2015 at 6:49
0
$\begingroup$

Here's my idea. Take any set that you can perform a one-point compactification on it where the original set is not rectifiable, then the compactified set won't be either.

Example, take $\mathbb{N}$ with the modified counting measure (such that it is zero on the integer zero, i.e. $require$ that $\mu(0)=0$) then $\mu(\mathbb{N})=\infty$, and the one-point compactification of $\mathbb{N}$ is $\{\frac{1}{n}:n\in \mathbb{N}\}\cup \{0\}$ and the boundary $\{0\}$ has measure zero because we defined our measure this way. This is clearly compact and not rectifiable, because $\sum_{n=1}^\infty \frac{1}{n}=\infty$.

$\endgroup$
6
  • $\begingroup$ Notice that under my modified counting measure that $\mu(\{0,1\}) = \mu(\{0\}\cup \{1\})=\mu(\{0\})+\mu(\{1\})=0+1$ NOT $2$. $\endgroup$
    – Squirtle
    Jan 20, 2015 at 6:02
  • $\begingroup$ So to answer your question.... the answer is no. $\endgroup$
    – Squirtle
    Jan 20, 2015 at 6:04
  • $\begingroup$ I'm not sure the question intends you to take an arbitrary measure. $\endgroup$
    – user208259
    Jan 20, 2015 at 6:06
  • $\begingroup$ Well.... The question was asked in pretty wide generality so I gave a general sort of answer. $\endgroup$
    – Squirtle
    Jan 20, 2015 at 6:08
  • $\begingroup$ @Squirtle: Actually, what is the definition of rectifiability? $\endgroup$
    – user99914
    Jan 20, 2015 at 6:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .