# Characteristic function of the Smith-Volterra-Cantor set

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