Proof of Fundamental Theorem of Calculus, Part II, on a Closed Real Interval

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$. Write $R=\int_a^b f(t)\,dt$ for the Riemann integral of $f$ over $[a,b]$, and let $\varepsilon>0$ be real. By the definition of Riemann integrability there exists a real $\delta>0$ with the following property: whenever $P$ is a partition of $[a,b]$ with mesh $|P|<\delta$ and one chooses a tagged partition of $[a,b]$ relative to $P$, the corresponding Riemann sum $S$ of $f$ satisfies $|S-R|<\varepsilon$.

By claim 1 of Uniform Partitions and Order Bounds for the Riemann Integral there is a partition $P=(x_0,x_1,\dots,x_n)$ of $[a,b]$ with $|P|<\delta$.

Fix $i\in\{1,\dots,n\}$ and consider the restriction $F_i=F|_{[x_{i-1},x_i]}$ of $F$ to the closed interval $[x_{i-1},x_i]\subseteq[a,b]$, where $x_{i-1}<x_i$. By claim 1 of Restriction Stability of Continuity and of the Derivative, applied with both metric spaces equal to the real line, $A=[a,b]$ and $B=[x_{i-1},x_i]$, the restriction $F_i$ is continuous on $[x_{i-1},x_i]$. Let $u\in\mathbb{R}$ with $x_{i-1}<u<x_i$. Then $u$ is an interior point of $[x_{i-1},x_i]$, since $x_{i-1},x_i\in[x_{i-1},x_i]$, and $a=x_0\le x_{i-1}<u<x_i\le x_n=b$, by the ordering of the partition points, shows $u\in(a,b)$, so by hypothesis $F$ is differentiable at $u$ with $F'(u)=f(u)$. By claim 2 of Restriction Stability of Continuity and of the Derivative, applied with the interval $[a,b]$ in the role of $I$ and $[x_{i-1},x_i]$ in the role of $J$, the restriction $F_i$ is differentiable at $u$ with $F_i'(u)=F'(u)=f(u)$.

By Mean Value Theorem on a Closed Real Interval, applied to $F_i$ on $[x_{i-1},x_i]$, there exists $t_i$ with $x_{i-1}<t_i<x_i$ such that

using $F_i'(t_i)=f(t_i)$ and $F_i(x_{i-1})=F(x_{i-1})$, $F_i(x_i)=F(x_i)$.

The points $t_i\in[x_{i-1},x_i]$ ($1\le i\le n$) form a tagged partition of $[a,b]$ relative to $P$, and the corresponding Riemann sum of $f$ is

by cancellation of the telescoping sum (finite induction on $n$). Since $|P|<\delta$, the choice of $\delta$ gives

As $\varepsilon>0$ was arbitrary, $F(b)-F(a)=R$: otherwise $\varepsilon_0=|F(b)-F(a)-R|$ would be a positive real number, and taking $\varepsilon=\varepsilon_0$ above would give $\varepsilon_0<\varepsilon_0$, which is impossible. Hence