Defining an Antiderivative for Monotone Functions I'm reading through a paper and I'm having trouble following the logic of the following step.
"Suppose $\psi:(0,\infty) \to [0,\infty)$ is a non-negative, non-decreasing function. Let $\Psi$ be the primitive function of $\psi$, i.e. $\Psi' = \psi$."
Can we actually do this? I believe monotonicity of $\psi$ means that an antiderivative can indeed be defined, and would be continuous, but wouldn't it only be differentiable almost everywhere? In particular, at points of discontinuity of $\psi$ (the set of which I believe will have measure zero, again by monotonicity) we cannot say $\Psi' = \psi$?
In summary, is it instead true that $\Psi' = \psi$ a.e.?
Thank you for your time!
 A: If the author has presented no other definition of a function, then you are probably correct and it should be almost everywhere. However, it may be the case for the author's requirements it is only necessary that the function be continuous almost everywhere, in which case it is just poor writing (Off the top of my head, Vrabie has something to do with null-measure sets called a "strongly measurable function" in his book on $C_0$-semigroups; can't remember the details).
Alternatively, it may be the case that the author's definition of a function lets them get away with this statement. The construction a Cauchy sequence of "simple" functions (polynomial, piecewise-linear) to represent the functions, and taking the sequence of the antiderivatives to represent the antiderivative, with limits of sequences of functions converging in some definition of a measure on functions (sup, l-p, Lebesgue), lets one define distributional derivatives of non-differentiable things. 
If someone can tell you more definitively that this is wrong, the author would probably appreciate notification of the error. 
A: Yes you are right. The statement is in general false. Consider the monotonic function $f$ on the reals such that $f(x)=0$ for $x \leq 0$ and $f(x) = 1 + x$ for $x > 0$, if $f = F'$ for some $F$ for all $x$ then $f$ must have the intermediate value property, but it clearly doesn't.
