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I'm aware that Lebesgue-measurability captures $n-$dimensional volume in $\mathbb{R}^n$, so curves for example have Lebesgue measure 0.

My basic question is: what captures the 1-dimensional volume of lines and curves in $\mathbb{R}^2$? Is this simply done by "flattening" the curve and analyzing its measurability in $\mathbb{R}$? I'm very new to this stuff so forgive me if I'm very off the mark. I ultimately want to describe what subsets of the unit circle have a defined length.

CSSTUDENT
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  • That's something called Hausdorff measure which measure lower dimensional objects. – Arctic Char Mar 27 '21 at 18:41
  • I'll have to look into that; is that the simplest way to characterize what I'm talking about or is it overkill? – CSSTUDENT Mar 27 '21 at 18:44
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    For curves it's somewhat special (and simple): Just take the supremum of all inscribed polygonal paths. That concept doesn't work in greater dimensions, see https://en.wikipedia.org/wiki/Schwarz_lantern found by Hermann Amandus Schwarz in 1833. – Michael Hoppe Mar 27 '21 at 18:54

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This depends for starters on what you mean by "curve". If by curve you mean a continuously differentiable map $\gamma:[0,1]\to\mathbb R^n$ then yes, the Lebesgue measure is zero when $n>1$. But in such case you can calculate the length by $$ L(\gamma)=\int_0^1\|\gamma'(t)\|\,dt. $$

If you allow "curves" that are continuous maps $\gamma:[0,1]\to\mathbb R^n$, then it is not true that they necessarily have zero Lebesgue measure. For $n=2$ there is the famous Hilbert Curve, where $m(\gamma([0,1]))=1$.

With the unit circle, the approach is that as long as you can remove a point, it is homeomorphic (diffeomorphic too) with a segment. So you can use the usual Lebesgue measure from the line.

Martin Argerami
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  • I'm not entirely sure I've understood your last paragraph: could you be a little more explicit? Subsets of the unit circle have a length iff their image in R (under the diffeomorphism) is lebesque measurable in R? – CSSTUDENT Mar 28 '21 at 04:52
  • Consider a sector in the unit circle, say from $\theta_1$ radians to $\theta_2$ radians. As a curve, this is $e^{it}$, $\theta_1\leq t\leq\theta_2$. The length of such sector is $$L=\int_{\theta_1}^{\theta_2}|ie^{it}|,dt=\theta_2-\theta_1. $$ That allows you to define Lebesgue measure exactly as in the real line, where now the "intervals" are the sectors ${e^{it}:\ \theta_1\leq t\leq\theta_2}$ with length $\theta_2-\theta_1$. What you get is exactly the Lebesgue measure as if you cut the circle and stretch it over the line. – Martin Argerami Mar 29 '21 at 04:08