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.
 A: 
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
$$
\int_{[0,1]} \beta ~ d{\mu} = \mu(\text{SVC}) > 0.
$$
