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Let the characteristic function of the SVC set be denoted by $ \beta $. Does the Riemann integral $ \displaystyle \int_{0}^{1} \beta ~ d{x} $ exist? I think it does since $ \beta $ is bounded, but I cannot quite see how the set of discontinuities has measure zero.

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Proposition The Riemann-integral $ \displaystyle \int_{0}^{1} \beta ~ d{x} $ does not exist.

Proof: The Smith-Volterra-Cantor set, henceforth denoted by $ \text{SVC} $, is nowhere dense in $ [0,1] $ and has positive measure. Then $ [0,1] \setminus \text{SVC} $ is dense, which means that for every $ x \in \text{SVC} $, we can find a sequence $ (x_{n})_{n \in \mathbb{N}} $ in $ [0,1] \setminus \text{SVC} $ such that $ \displaystyle \lim_{n \to \infty} x_{n} = x $. We thus have

  • $ \beta(x) = 1 $ and

  • $ \displaystyle \lim_{n \to \infty} \beta(x_{n}) = \lim_{n \to \infty} 0 = 0 $.

It follows readily that every point of $ \text{SVC} $ is a point of discontinuity of $ \beta $, so the set of discontinuities of $ \beta $ has positive measure. By Lebesgue’s theorem on the necessary and sufficient conditions for Riemann-integrability, we conclude that $ \beta $ is not Riemann-integrable. $ \quad \spadesuit $

However, $ \beta $ is Lebesgue-integrable and $$ \int_{[0,1]} \beta ~ d{\mu} = \mu(\text{SVC}) > 0. $$

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I like it, but a few clarification questions. Why is it important that we construct the sequence and show its limit as $n$ approaches $\inf$ is $0$? We know that the characteristic function is $1$ when $x$ is in the set and $0$ when it is not. –  user43901 Feb 8 '13 at 6:59
    
@user43901: Consider, instead, the situation in which the characteristic function of SVC is replaced with the characteristic function of the interval $[\frac{1}{3},\;\frac{1}{2}].$ There are continuum many points where the characteristic function of $[\frac{1}{3},\;\frac{1}{2}]$ is equal to $1,$ but this function has only two points of discontinuity. –  Dave L. Renfro Feb 8 '13 at 17:18
    
@user43901: The only way that I can show that $ \beta $ is discontinuous at a given point $ x \in \text{SVC} $ is to exhibit a sequence $ (x_{n})_{n \in \mathbb{N}} $ in $ [0,1] $ such that $ \displaystyle \lim_{n \to \infty} x_{n} = x $ but $ \displaystyle \lim_{n \to \infty} \beta(x_{n}) = \beta(x) = 1 $ does not hold. –  Haskell Curry Feb 8 '13 at 18:12
    
@user43901: If you’re a little uncomfortable with what I just did, then you can replace it with the following explanation: For any $ x \in \text{SVC} $, there exist points that are arbitrarily close to $ x $ belonging to the complement of $ \text{SVC} $. At these points, $ \beta $ assumes the value $ 0 $, but at $ x $ itself, $ \beta $ jumps to the value $ 1 $. Therefore, $ \beta $ is discontinuous at $ x $. For the ‘jump’ to happen, you need the condition ‘arbitrarily close’. –  Haskell Curry Feb 8 '13 at 20:14
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