# Is every compact set rectifiable? example

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.

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

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.

• "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? Jan 20, 2015 at 6:22
• 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 $< \varepsilon$ Jan 20, 2015 at 6:49

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 $\sum_{n=1}^\infty \frac{1}{n}=\infty$.

• Notice that under my modified counting measure that $\mu(\{0,1\}) = \mu(\{0\}\cup \{1\})=\mu(\{0\})+\mu(\{1\})=0+1$ NOT $2$. Jan 20, 2015 at 6:02