Theorem 11.5.4 (Fundamental Theorem of Calculus for Riemann Integrals). Let $[a, b] \subset \real$ and $E$ be a separated locally convex space, then:

  1. For any Riemann integrable function $f: [a, b] \to E$ and $x \in (a, b)$ such that $f$ is continuous at $x$, the function

    \[F: [a, b] \to E \quad y \mapsto \int_{a}^{y} f(t)dt\]

    is differentiable at $x$ with $DF(x) = f(x)$.

  2. For $F \in C^{1}([a, b]; E)$,

    \[F(b) - F(a) = \int_{a}^{b} DF(t)dt\]

Proof. (1): Let $x_{0} \in (a, b)$ such that $f$ is continuous at $x$, and $\delta > 0$ such that $(x_{0} - \delta, x_{0} + \delta) \subset (a, b)$, then for any $h > 0$,

\begin{align*}\frac{1}{h}\braks{\int_a^{x+h} f(t)dt - \int_a^{x} f(t)dt}&= \frac{1}{h}\int_{x}^{x+h}f(t)dt \in \overline{\text{Conv}(f([t, t +h)))}\\ -\frac{1}{h}\braks{\int_a^{x-h} f(t)dt - \int_a^{x} f(t)dt}&= \frac{1}{h}\int_{x-h}^{x}f(t)dt \in \overline{\text{Conv}(f([t, t +h)))}\end{align*}

As $E$ is locally convex and separated, and $f$ is continuous at $x$, this implies that $[F(x + h) - F(x)]/h \to f(x)$.

(2): Let $G(x) = \int_{a}^{x} DF(t)dt + F(a)$, then $G - F$ has derivative $0$. By the Mean Value Theorem, $G - F$ is constant. As $G(a) - F(a) = 0$, $G = F$.$\square$