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

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$, so that $d_{\mathbb{R}}(s,t)=|s-t|$ for all real $s,t$ by The Absolute Value Metric on the Real Line.

Claim 1. Let $x\in(a,b]$. 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 $B=[a,x]$, the restriction $f|_{[a,x]}$ is continuous on $[a,x]$. Since $a<x$, claim 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval, applied on the interval $[a,x]$ to the function $f|_{[a,x]}$, shows that $f|_{[a,x]}$ is Riemann integrable on $[a,x]$. Hence $\int_a^x f(t)\,dt$ is defined for every $x\in(a,b]$, and $F$ is well defined on $[a,b]$, with $F(a)=0$ by the degenerate-interval convention of Mean-Square Riemann Integral of a Family of Random Variables.

Claim 2. By claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals, applied to $\varphi=f$ on $[a,b]$, the function $t\mapsto\int_a^t f(u)\,du$ on $[a,b]$ is continuous on all of $[a,b]$. That function is exactly $F$, both being defined with the same degenerate-interval convention at $t=a$, so $F$ is continuous on $[a,b]$.

Claim 3. Fix $x\in(a,b)$ and a real $\varepsilon>0$. Since $f$ is continuous at $x$ relative to $[a,b]$, there is a real $\delta_0>0$ such that every $u\in[a,b]$ with $|u-x|=d_{\mathbb{R}}(x,u)<\delta_0$ satisfies $|f(u)-f(x)|<\varepsilon/2$, that is,

$$f(x)-\varepsilon/2<f(u)<f(x)+\varepsilon/2 .$$

Set $\delta=\min\{\delta_0,\,x-a,\,b-x\}$, which is positive since $a<x<b$. Let $h$ be real with $0<|h|<\delta$ and $x+h\in[a,b]$.

Case $h>0$. From $h<\delta\le b-x$ we get $a\le x<x+h\le b$, and every $u$ with $x\le u\le x+h$ lies in $[a,b]$. By claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals, applied to $\varphi=f$ with $s=x$ and $t=x+h$,

$$F(x+h)-F(x)=\int_x^{x+h}f(u)\,du .$$

By claim 1 of Restriction Stability of Continuity and of the Derivative the restriction of $f$ to $[x,x+h]$ is continuous on $[x,x+h]$, hence Riemann integrable on $[x,x+h]$ by claim 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval. Every $u\in[x,x+h]$ satisfies $|u-x|=u-x\le h<\delta\le\delta_0$, so the two-sided bound above gives $f(x)-\varepsilon/2\le f(u)\le f(x)+\varepsilon/2$ on $[x,x+h]$. By claim 2 of Uniform Partitions and Order Bounds for the Riemann Integral, applied to this restriction on $[x,x+h]$ with $m=f(x)-\varepsilon/2$ and $M=f(x)+\varepsilon/2$,

$$\bigl(f(x)-\varepsilon/2\bigr)\,h\le F(x+h)-F(x)\le\bigl(f(x)+\varepsilon/2\bigr)\,h .$$

Dividing by $h>0$,

$$\left|\frac{F(x+h)-F(x)}{h}-f(x)\right|\le\varepsilon/2<\varepsilon .$$

Case $h<0$. From $|h|=-h<\delta\le x-a$ we get $a\le x+h<x\le b$, and every $u$ with $x+h\le u\le x$ lies in $[a,b]$. By claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals, applied with $s=x+h$ and $t=x$,

$$F(x)-F(x+h)=\int_{x+h}^{x}f(u)\,du .$$

As before, the restriction of $f$ to $[x+h,x]$ is continuous on $[x+h,x]$, hence Riemann integrable, and every $u\in[x+h,x]$ satisfies $|u-x|=x-u\le -h<\delta_0$, so $f(x)-\varepsilon/2\le f(u)\le f(x)+\varepsilon/2$ on $[x+h,x]$. By claim 2 of Uniform Partitions and Order Bounds for the Riemann Integral, applied on $[x+h,x]$, whose length is $x-(x+h)=-h>0$,

$$\bigl(f(x)-\varepsilon/2\bigr)(-h)\le F(x)-F(x+h)\le\bigl(f(x)+\varepsilon/2\bigr)(-h) .$$

Dividing by $-h>0$ and noting that

$$\frac{F(x+h)-F(x)}{h}=\frac{F(x)-F(x+h)}{-h},$$

we again obtain

$$\left|\frac{F(x+h)-F(x)}{h}-f(x)\right|\le\varepsilon/2<\varepsilon .$$

In both cases, whenever $0<|h|<\delta$ and $x+h\in[a,b]$, the difference quotient of $F$ at $x$ lies within $\varepsilon$ of $f(x)$. Since $\varepsilon>0$ was arbitrary and $x$ is an interior point of the interval $[a,b]$, the function $F$ is differentiable at $x$ with $F'(x)=f(x)$. $\blacksquare$