## sets with positive Lebesgue measure boundary

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

Consider a compact subset $K$ of $R^n$ which is the closure of its interior. Does its boundary $\partial K$ have zero Lebesgue measure ?

I guess it's wrong, because the topological assumption is invariant w.r.t homeomorphism, in contrast to being of zero Lebesgue measure. But I don't see any simple counterexample.

-

## 3 Answers

Construct a Cantor set of positive measure in much the same way as you make the `standard' Cantor set but make sure the lengths of the deleted intervals add up to 1/2, say. Let $U$ be the union of the intervals that are deleted at the even-numbered steps and let $V$ be the union of the intervals deleted at the odd-numbered steps. The Cantor set is the common boundary of $U$ and $V$; their closures are as required.

-

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

http://www.jstor.org/pss/1986455 Here is constructed a Jordan Curve with positive measure. This gives an example.

-

4
Ah! You beat me to it. I'll delete my own answer, which is a duplicate of yours (and you need the rep more than I do). Let me just add that this curve (the Osgood curve) has been mentioned here on MO before. The search box will find it for you. – Harald Hanche-Olsen May 26 2010 at 12:21
It does the trick. The Jordan curve in that paper is a "thinned out" variant of the Peano curve. – Xandi Tuni May 26 2010 at 12:24

Let $D_0,D_1,\ldots$ enumerate a sequence of disjoint intervals in the unit interval with $\bigcup_n D_n$ open dense and having measure less than $1$. For example, place a very tiny interval around each rational number, so that the sum of the intervals is less than $1$. Now, let $E=\bigcup_n D_{2n}$ be the union of the even intervals and $O=\bigcup_n D_{2n+1}$, the union of the odd intervals. The entire interval is the union of $E$, $O$ and their boundaries, so one of these boundaries must have positive measure. So we may take $K$ to be the closure of $E$ or $O$.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9544548988342285, "perplexity_flag": "head"}