Is the Lebesgue measure of the boundary of a simply connected domain in $\mathbb{R}^n$ necessarily 0?
Acturally, I want to know the sufficient condition to guarantee the measure of the boundary of a domain is $0$.
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$\begingroup$ might be useful math.stackexchange.com/questions/1421934/… $\endgroup$– Tim kinsellaJan 6, 2018 at 9:24
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2 Answers
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Take a fat Cantor set, $C\subset \mathbb{R}$, i.e. a closed set with empty interior and nonzero measure. Then consider the set $$\{(x,y)\in \mathbb{R}^2: y <0\} \cup \{(x,y)\in \mathbb{R}^2: y\geq 0, x \notin C\}.$$
This set seems open, simply connected and with boundary $C \times [0,\infty)$.
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Tim kinsella's answer is really fine, here is another suggestion (way more complicated).
Just take an Osgood curve (= Jordan arc of positive area)
This is an arc (continuous image of $[0,1]$ into the plane) therefore simply connected, and closed.
It is also a non self intersecting curve (an arc is the image of an injective function), so it has empty interior (see this or this, the proof is easy)
Thus the boundary of the Osgood curve is the curve itself, so it has positive measure.
See for instance this step by step construction of an Osgood curve (by Knopp, on Wolfram Demonstrations).
