> [!definition] > > Let $[a, b]$ be a [[Compactness|compact]] interval and $E$ be a [[Banach Space|Banach space]]. Let $f: [a, b] \to E$ be a [[Function|function]]. If there is a [[Partition|partition]] of $[a, b]$ by > $ > a = a_0 \le a_1 \le \cdots \le a_n = b > $ > and elements $\seqf{v_k} \subset E$ such that $f(t) = v_i$ for all $t \in (a_{i - 1}, a_i)$, then $f$ is a **step map** with respect to $P$. > [!definition] > > Let $f:[a, b] \to E$ be a step map with a partition $P$, then the **integral** of $f$ with respect to $P$ is > $ > I_P(f) = \sum_{i = 1}^{n}(a_i - a_{i - 1})v_i > $ > which is [[Bounded Linear Functional|bounded]] with respect to the [[Uniform Norm|uniform norm]]. > [!definition] > > Let $\Sigma$ be the collection of all step maps from $[a, b]$ to $E$. Then the [[Topological Closure|closure]] $\ol{\Sigma}$ is the space of all **regulated** maps, and there is a unique extension of the integral to $\ol{\Sigma}$ by the [[Linear Extension Theorem]]. > [!theorem] > > Let $\lambda: \cx \to \cy$ be a bounded linear map. If $f: [a, b] \to E$ is regulated, then $\lambda \circ f$ is also regulated, and > $ > \int_a^b \lambda \circ f = \lambda\braks{\int_a^b f} > $ > *Proof.* Let $\phi = \sum_{k = 1}^{n}\alpha_k\chi_{(a_{k - 1}, a_{k})}$, then > $ > \int_a^b \lambda \circ \phi = \sum_{k = 1}\lambda(a_k)\chi_{(a_{k - 1}, a_k)} = \lambda \braks{\int_a^b \phi} > $ > If $f$ is regulated, then there is a supporting sequence $\seq{\phi_n}$ of step maps such that $\phi_n \to f$ uniformly. Since $\lambda$ is bounded, $\lambda \circ \phi_n \to \lambda \circ f$ uniformly as well. Using the above case for step maps, we have > $ > \int_a^b \lambda \circ \phi_n = \lambda \braks{\int_a^b \phi_n} \to \lambda \braks{\int_a^b f} > $ > [!theorem] Fundamental Theorem of Calculus > > Let $f \in \text{Reg}([a, b])$, and suppose that $f$ is [[Continuity|continuous]] at a point $c \in [a, b]$. Then the map > $ > t \to \int_a^yf = \phi(t) > $ > is differentiable at $c$, with derivative $f(c)$. > > *Proof*. > $ > \begin{align*} > \phi(c + h) - \phi(c) &= \int_c^{c + h}f \\ > \phi(c + h) - \phi(c) - hf(c)&= \int_c^{c + h}(f - f(c))\\ > \phi(c + h) - \phi(c)&=h\underbrace{f(c)}_{\lambda} + \int_c^{c + h}(f - f(c)) > \end{align*} > $ > where > $ > \abs{\int_c^{c + h}(f - f(c))} \le \abs{h} \cdot \sup_{t \in [c, c + h]}\abs{f(t) - f(x)} > $ > Since $f$ is continuous at $c$, > $ > \lim_{h \to 0}\frac{\abs{h} \cdot \sup_{t \in [c, c + h]}\abs{f(t) - f(x)}}{\abs{h}} = \lim_{h \to 0}\sup_{t \in [c, c + h]}\abs{f(t) - f(x)} = 0 > $ > the integral remainder is $o(h)$.