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

First we show that $F$ is well defined on $[a,b]$. Let $y\in[a,b]$. Since $f$ is continuous at every point of $[a,b]$, it is in particular continuous at every point of $[a,y]$. Hence $f|_{[a,y]}:[a,y]\to\mathbb{R}$ is continuous on $[a,y]$ in the sense of Continuity on a Closed Interval. By Continuous Functions on a Closed Interval are Riemann Integrable, the restriction $f|_{[a,y]}$ is Riemann integrable on $[a,y]$. Therefore the integral

is defined for every $y\in[a,b]$.

Now fix a point $x\in(a,b)$. We will show that $F$ is differentiable at $x$ and that $F'(x)=f(x)$.

Let $\varepsilon>0$. Since $f$ is continuous at $x$ in the sense of Continuity at a Point, there exists $\delta>0$ such that whenever $|u-x|<\delta$, one has

Let $h\in\mathbb{R}$ satisfy $0<|h|<\delta$ and $x+h\in[a,b]$. Since $F$ is already known to be well defined on $[a,b]$, both $F(x+h)$ and $F(x)$ exist. If $h>0$, then $a\le x\le x+h\le b$, so by Additivity of the Riemann Integral on Adjacent Intervals applied to $f$ on the adjacent intervals $[a,x]$ and $[x,x+h]$,

If $h<0$, then $a\le x+h\le x\le b$, so by Additivity of the Riemann Integral on Adjacent Intervals applied to $f$ on the adjacent intervals $[a,x+h]$ and $[x+h,x]$,

Writing $k=-h>0$, it follows that

when $h<0$, while

when $h>0$.

If $h>0$, then for every $t\in[x,x+h]$ one has $|t-x|\le h<\delta$, so $|f(t)-f(x)|<\varepsilon$. Therefore

If $h<0$, then for every $t\in[x+h,x]$ one has $|t-x|\le |h|=k<\delta$, so $|f(t)-f(x)|<\varepsilon$. Hence

Thus, for the number $f(x)\in\mathbb{R}$ and for every $\varepsilon>0$, there exists $\delta>0$ such that whenever $0<|h|<\delta$ and $x+h\in[a,b]$, one has

This is exactly the condition in Derivative at an Interior Point for $F$ to be differentiable at $x$ with derivative $f(x)$. Therefore $F$ is differentiable at $x$ in the sense of Derivative at an Interior Point, and $F'(x)=f(x)$.