Discussion
Let \(f : [0, 1] \to \mathbf{R}\) be a continuous function. Then \(f\) is absolutely continuous, if :
Answer: D
\(f\) is absolutely continuous if \(f'\) exists and is continuous.
If \(f'\) is continuous on \([0, 1]\), it is bounded, say \(|f'| \le M\), and by the mean value theorem \(|f(x) - f(y)| \le M|x - y|\). A Lipschitz function is absolutely continuous: given \(\varepsilon\), take \(\delta = \varepsilon/M\). Equivalently, \(f(x) = f(0) + \int_{0}^{x} f'\), the defining property.
The other conditions are not enough. The Cantor function is continuous, hence uniformly continuous on \([0, 1]\), of bounded variation and Riemann integrable, yet it is not absolutely continuous, because it rises from 0 to 1 while its derivative is 0 almost everywhere.
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