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I have a two-dimensional Lebesgue integral over $\partial D(0,1)$. I think it should be zero because the two-dimensional measure of the unit circle is $0$.

But I'm not sure how to justify this. My idea is to cover the unit circle by countably many boxes of area $\epsilon/2^n$ but I'm not sure how to go about this. Can you please help?

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3 Answers 3

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For a very nice region, the Lebesgue measure in the two-dimensional space measures the area of the region.

Consider the region consisting on the points $S_\epsilon$ between the circle of radius $1-\epsilon$ and $1+\epsilon$. The area of this region is

$$\begin{align*} S_\epsilon&=\pi\cdot (1+\epsilon)^2 - \pi\cdot (1-\epsilon)^2\\ &=\pi\cdot [(1+\epsilon)+(1-\epsilon]\cdot [(1+\epsilon)-(1-\epsilon)]\\ &=4\pi\cdot \epsilon \end{align*}$$ Since the circle of radius $1$ (which is usually denoted by $S^1$) is contained in each $S_\epsilon$, we know that

$$0\leq \lambda(S^1)\leq \lim_{\epsilon\to 0}\lambda (S_\epsilon)=\lim_{\epsilon\to 0}4\pi\cdot \epsilon=0.$$ Thus, $\lambda(S^1)=0$.

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Cover the circle by annuli of arbitrarily small width.

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  • $\begingroup$ Do you know that the Lebesgue measure coincides with the $n$-dimensional volume? $\endgroup$ Nov 2, 2016 at 16:21
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You can approximate the circle with a sequence of inscribed regular n-gons and another sequence of regular n-gons such that the circle is inscribed into them. The limiting measures for the two sequences will be the same, thus the measure of the circle itself will be 0.

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