Does the fundamental theorem of calculus require continuity of the function being integrated? The (first) fundamental theorem of calculus is typically stated as follows, assuming continuity of the given function:


*

*Suppose that f is continuous on the closed interval [a,b] and F
is defined by $F(x)= \int_a^x f(t)\ dt.$ Then  F is differentiable
on (a,b) and for all $x\in(a,b)$, we have $F'(x) = f(x)$.


One (apparent) application of this theorem is in the proof that the winding number of a closed path is an integer. One example is lemma 1.1 of Lang's textbook Complex Analysis. (A very similar proof appears on page 115 of a book with the same title by LV Ahlfors.) The key step is to form the definite integral of a certain function and immediately take the derivative, getting the same function back. But the proof includes no argument that the function in question is continuous, and indeed the function could be undefined on a finite number of points in the interval. (It involves a derivative of a piecewise differentiable path.) It is possible to show that the integral exists, however. Is this a sufficient condition for the fundamental theorem of calculus?
 A: A more general version of the fundamental theorem of calculus appears in most introductory measure theory textbooks:

Suppose that $f : [a,b] \to \mathbb{R}$ is measurable and Lebesgue integrable.  Set $F(x) = \int_a^x f(t)\,dt$ (Lebesgue integral).  Then at almost every point $x \in [a,b]$, $F$ is differentiable at $x$ and $F'(x) = f(x)$.

More precisely, there is a set $A \subset [a,b]$ whose Lebesgue measure is zero, such that for every $x \in [a,b] \setminus A$ we have that $F$ is differentiable at $x$ and $F'(x) = f(x)$.
$F$ also satisfies the stronger property of being absolutely continuous on $[a,b]$.  Indeed, a "converse" of this fundamental theorem of calculus is that every absolutely continuous function $G : [a,b] \to \mathbb{R}$ may be written $G(x) = G(a) + \int_a^x g(t)\,dt$ for some measurable and Lebesgue integrable function $g$, which is unique up to measure-zero sets.
See, for instance, Theorem 3.35 of Folland's Real Analysis.
You may of course replace the codomain $\mathbb{R}$ by $\mathbb{C}$ simply by taking real and imaginary parts.  More general vector spaces are also possible, with appropriate modifications.
