Proof of Fundamental Theorem of Calculus, Part II in One Dimension

Let $\varepsilon>0$. By Continuous Functions on a Closed Interval are Riemann Integrable, the function $f$ is Riemann integrable on $[a,b]$. Hence, by Riemann Integrability on a Closed Interval, there exists $\delta>0$ such that whenever $P=(x_0,\dots,x_n)$ is a partition of $[a,b]$ with $|P|<\delta$ and one chooses a tagged partition of $[a,b]$ relative to $P$, the corresponding Riemann sum $S$ satisfies

Now let $P=(x_0,\dots,x_n)$ be any partition of $[a,b]$ with $|P|<\delta$. For each $i=1,\dots,n$, apply Mean Value Theorem in One Dimension to $F$ on $[x_{i-1},x_i]$. Since $F$ is an antiderivative of $f$ on $I$ in the sense of Antiderivative on an Interval, there exists $\xi_i\in(x_{i-1},x_i)$ such that

The points $\xi_i\in[x_{i-1},x_i]$ determine a tagged partition of $[a,b]$ relative to $P$, and the corresponding Riemann sum is

Therefore

Since $\varepsilon>0$ was arbitrary, it follows that

This is equivalent to the stated identity.