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Volterra's function is differentiable everywhere just as ''f'' (as defined above) is. One can show that ''f'' ′(''x'') = 2''x'' sin(1/''x'') - cos(1/''x'') for ''x'' ≠ 0, which means that in any neighborhood of zero, there are points where ''f'' ′ takes values 1 and −1. Thus there are points where ''V'' ′ takes values 1 and −1 in every neighborhood of each of the endpoints of intervals removed in the construction of the [[Smith–Volterra–Cantor set]] ''S''. In fact, ''V'' ′ is discontinuous at every point of ''S'', even though ''V'' itself is differentiable at every point of ''S'', with derivative 0. However, ''V'' ′ is continuous on each interval removed in the construction of ''S'', so the set of discontinuities of ''V'' ′ is equal to ''S''.
Since the Smith–Volterra–Cantor set ''S'' has positive [[Lebesgue measure]], this means that ''V'' ′ is discontinuous on a set of positive measure. By [[Riemann_integral#Integrability|Lebesgue's criterion for Riemann integrability]], ''V'' ′ is not Riemann integrable. If one were to repeat the construction of Volterra's function with the ordinary measure-0 Cantor set ''C'' in place of the "fat" (positive-measure) Cantor set ''S'', one would obtain a function with many similar properties, but the derivative would then be discontinuous on the measure-0 set ''C'' instead of the positive-measure set ''S'', and so the resulting function would have
==See also==
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