Volterra's function is differentiable everywhere just as ''f''(''x'') (defined above) is. TheOne derivativecan show that ''Vf''′(''x'') is= discontinuous2''x'' atsin(1/''x'') the- endpointssin(1/''x'') offor every''x'' interval≠ removed0, inwhich themeans constructionthat ofin theany [[Smith–Volterra–Cantorneighborhood set|SVC]]of zero, butthere theare functionpoints iswhere differentiable''f''′(''x'') attakes these1 points with valueand 0−1. Furthermore, in any neighbourhood of such a pointThus there are points where ''V'' ′(''x'') takes values 1 and −1. in Itevery followsneighborhood thatof iteach isof notthe possible,endpoints forof everyintervals εremoved >in 0,the to find a partitionconstruction of the real[[Smith–Volterra–Cantor lineset]] such that |''V'' ′(''x''C<sub>2SV</sub>)''. −In fact, ''V'' ′(''x''<sub>1</sub>)|<is εdiscontinuous onat every intervalpoint of [''x''C<sub>1SV</sub>'', even though ''xV'' itself is differentiable at every point of ''C<sub>2SV</sub>]'', ofwith thederivative partition0. The Therefore,set theof derivativepoints where ''V'' ′(''x'') is notthe Riemannset integrable''C<sub>SV</sub>''.
Since the Smith–Volterra–Cantor set ''C<sub>SV</sub>'' has positive 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.
A real-valued function is Riemann integrable if and only if it is bounded and continuous almost-everywhere (''i.e.'' everywhere except a set of [[measure theory|measure]] 0). Since ''V'' ′(''x'') is bounded, it follows that it must be discontinuous on a set of positive measure, so in particular the derivative of ''V''(''x'') is discontinuous at uncountably many points.
