Discussion
Which of the following statements holds for a function \(f : [0, 1] \to \mathbf{R}\) ?
Answer: A
The statement that holds is that if \(f\) is of bounded variation, then \(f\) is Riemann integrable.
A function of bounded variation on \([0, 1]\) is the difference of two increasing functions. Increasing functions are bounded and have at most countably many discontinuities, a set of measure zero, so by Lebesgue's criterion they are Riemann integrable, and so is their difference.
The other implications fail. A step function is of bounded variation but not continuous, let alone uniformly continuous (B). The Dirichlet function, 1 on rationals and 0 elsewhere, is bounded but not of bounded variation (C). And \(x\sin(1/x)\) with value 0 at 0 is uniformly continuous on \([0, 1]\) but of unbounded variation (D).
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