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Volterra's function is differentiable everywhere just as ''f''(''x'') (defined above) is. One can show that ''f''′(''x'') = 2''x'' sin(1/''x'') - sin(1/''x'') for ''x'' ≠ 0, which means that in any neighborhood of zero, there are points where ''f''′(''x'') takes 1 and −1. Thus there are points where ''V'' ′(''x'') 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]] ''C<sub>SV</sub>''. In fact, ''V'' ′ is discontinuous at every point of ''C<sub>SV</sub>'', even though ''V'' itself is differentiable at every point of ''C<sub>SV</sub>'', with derivative 0. The set of points where ''V'' ′ is the set ''C<sub>SV</sub>''.
Since the Smith–Volterra–Cantor set ''C<sub>SV</sub>'' 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 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 ''C<sub>SV</sub>'', 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 ''C<sub>SV</sub>'', and so the resulting function would have an integrable derivative.
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