Assume $F:(0,1] \to \mathbb{R}$ is given by $F(x) = \int_0^xf(u) du$ where $f:(0,1) \to \mathbb{R}$ is a nonincreasing (not necessarily continuous) function (note that F is real-valued so it is assumed implicitly the integral is well-defined)
Is it true that $F$ is differentiable almost everywhere and its derivative is equal to $f$ almost everywhere (if $f$ were continuous, then this would hold everywhere by the fundamental theorem of calculus)?
Idea: $f$ is monotone so has countable number of discontinuities...