In dimension $\mathbb R^N$, we say a set $S$ is $N-1$-rectifiable if there exists a countably many $C^1$ hyper surface $\Gamma_i$ so that $$ \mathcal H^{N-1}(S_u\setminus \bigcup\Gamma_i)=0 $$

Now I am reading a paper it use the term compact rectifiable set without define it, and I can't find a reliable reference where it has the definition of this term. Also, the paper uses 'compact rectifiable set' and 'closed rectifiable set' together, so it looks to me they are the same thing.

Can any body point out a good reference which has the definition of them?

Also, if the set $S$ so that $\mathcal H^{N-1}(S)<\infty$. Then by properties of measure, I can always have a compact set $K\subset S$ so that $\mathcal H^{N-1}(S\setminus K)<\epsilon$ for arbitrary $\epsilon>0$. Then, can I say the set $K$ is compact rectifiable?

Thank you!

  • 2
    $\begingroup$ I don't know anything about rectifiable sets, but I'm guessing "compact rectifiable set" refers to a set that is both compact and rectifiable. Do you know what a compact set is? EDIT: Or a closed set? $\endgroup$ Nov 25, 2015 at 21:48
  • $\begingroup$ @AkivaWeinberger Yes yes of course I know what is compact and closed means. I am just afraid there are something special for this special term... $\endgroup$
    – spatially
    Nov 25, 2015 at 21:53
  • 1
    $\begingroup$ @AkivaWeinberger I know something about rectifiable sets and yours is still my best guess... $\endgroup$ Nov 25, 2015 at 23:51
  • $\begingroup$ To add more weight to Akiva Weinberger's comment, I am virtually certain that "compact rectifiable" means a rectifiable set that is also compact, and similarly for "closed rectifiable". $\endgroup$ Jan 13, 2016 at 17:02

1 Answer 1


Definition from Munkres : Let $S$ be a bounded set in $\mathbb R^n$. If the indicator function $\mathbb 1_S$ is integrable. We say that $S$ is rectifiable and we define the ($n$-dimensional) volume of $S$ by the equation $$V(S) = \int_{\mathbb R^n} {\mathbb 1_S}.$$

Furthermore a subset $S$ of $\mathbb R^n$ is rectifiable if and only if $S$ is bounded and $\partial S$ has measure zero. So basically its a set whose Jordan content is defined i.e it is a Jordan-measurable set.


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