Volterra's function is differentiable everywhere just as ''f''(''x'') (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''′(''x'') takes values 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>S''. In fact, ''V''′ is discontinuous at every point of ''C<sub>SV</sub>S'', even though ''V'' itself is differentiable at every point of ''C<sub>SV</sub>S'', with derivative 0. TheHowever, ''V''′ is continuous on each interval removed in the construction of ''S'', so the set of pointsdiscontinuities whereof ''V''′ is theequal setto ''C<sub>SV</sub>S''.
Since the Smith–Volterra–Cantor set ''C<sub>SV</sub>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 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>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 ''C<sub>SV</sub>S'', and so the resulting function would have an integrable derivative.
