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I'm studying Functional Analysis and Lebesgue Measure.

I know that the standard Cantor Set $C$ is compact because is bounded by $[0,1]$ and it's closed because is an intersection of closed sets. Moreover:

$$\mathcal{L}(C)=\mathcal{L}(\bigcap\limits_{0}^{\infty} C_k)=\lim\limits_{k \to \infty} \mathcal{L}(C_k)=\lim\limits_{k \to \infty}\frac{2^k}{3^k}=0$$

So $C$ is Lebesgue Measurable with measure 0.

I'm struggling because if it's compact I think it's Riemann measurable too, isn't it? What's its Riemann Measure? Or maybe it can't be defined on this set?

james watt
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    What does "Riemann measurable" mean? – Mees de Vries Oct 11 '18 at 11:23
  • I don't know if it is a proper term. I mean the measurable sets that can be measured with rectangulars, rectangular prisms in $\mathbb{R}^3$ or other strange volumes in $\mathbb{R}^n$. – james watt Oct 11 '18 at 11:27
  • So basically you're asking if the indicator function of the Cantor set is Riemann-integrable, correct ? – Matrefeytontias Oct 11 '18 at 11:29
  • What do you mean with indicator function of the Cantor set? – james watt Oct 11 '18 at 11:31
  • I think @Matrefeytontias's description is the closest to the original question. The indicator (characteristic) function of Cantor set is Riemann integrable; see https://math.stackexchange.com/questions/864206/characteristic-function-of-cantor-set-is-riemann-integrable. It takes a little while to mint a proof for this. – Miles Zhou Oct 11 '18 at 11:43
  • If you're thinking of rectangular bounding regions you may be thinking of Jordan Measure which is a weaker but more historically motivated measure than the lebesgue measure. – CyclotomicField Oct 11 '18 at 12:06

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