Proof of Fundamental Theorem of Calculus, Part I: the Indefinite Integral is Lipschitz, Continuous, and Differentiable at Points of Continuity

Proof of (i). $[a,b]$ is an interval containing the two distinct points $a$ and $b$, by Basic Facts about Intervals of the Real Line (v), $a\in[a,b]$, and $f$ is integrable on $[u,v]$ for all $u,v\in[a,b]$ with $u<v$, by Additivity of the Riemann Integral over Adjacent Intervals, Subintervals and Oriented Integrals, and Integrals of Restrictions (ii). So A Predicative Formula for the Oriented Riemann Integral, and Maps Given by Integrals (ii), applied with $K=[a,b]$, $c=a$ and $g=f$, gives exactly one map $F:[a,b]\to\mathbb{R}$ with $F(x)=\int_{a}^{x}f$ for every $x\in[a,b]$.

Step 1: increments of $F$. Let $p,q\in[a,b]$ with $p<q$. Then $[p,q]\subseteq[a,b]$ by Basic Facts about Intervals of the Real Line (v) and Basic Facts about Intervals of the Real Line (ii), and $f$ is integrable on $[p,q]$ by Additivity of the Riemann Integral over Adjacent Intervals, Subintervals and Oriented Integrals, and Integrals of Restrictions (ii). By Additivity of the Riemann Integral over Adjacent Intervals, Subintervals and Oriented Integrals, and Integrals of Restrictions (iii), applied with $u=a$, $v=p$ and $w=q$, we have $\int_{a}^{q}f=\int_{a}^{p}f+\int_{p}^{q}f$, that is, by (i),

$$F(q)-F(p)=\int_{p}^{q}f.\qquad\text{(1)}$$

Suppose moreover that $c,C\in\mathbb{R}$ satisfy $c\le f(s)\le C$ for every $s\in[p,q]$. The restriction $f|_{[p,q]}:[p,q]\to\mathbb{R}$ has the values $f(s)$ for $s\in[p,q]$, so it satisfies the same bounds, and it is bounded and integrable on $[p,q]$. By Rules for the Riemann Integral: Sums, Scalar Multiples, Monotonicity, Bounds and Constants (iv), applied to $f|_{[p,q]}$ on $[p,q]$, and as $\int_{p}^{q}f=\int_{p}^{q}f|_{[p,q]}$ by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vi), (1) gives

$$c\,(q-p)\le F(q)-F(p)\le C\,(q-p).\qquad\text{(2)}$$

Proof of (ii). Let $B\in\mathbb{R}$ with $|f(t)|\le B$ for every $t\in[a,b]$, and let $x,y\in[a,b]$. If $x=y$, both sides of the claimed inequality are $0$. Otherwise let $p$ be the smaller and $q$ the larger of $x$ and $y$, so $p<q$, $|F(y)-F(x)|=|F(q)-F(p)|$ and $|y-x|=q-p$ (Difference and Quotient Identities in ℝ (i)). Every $s\in[p,q]$ lies in $[a,b]$ by Step 1, so $-B\le f(s)\le B$. By (2) with $c=-B$ and $C=B$, $-B\,(q-p)\le F(q)-F(p)\le B\,(q-p)$, that is, $|F(y)-F(x)|=|F(q)-F(p)|\le B\,(q-p)=B\,|y-x|$.

Proof of (iii). As $f$ is bounded, there is $B\in\mathbb{R}$ with $|f(t)|\le B$ for every $t\in[a,b]$; then $B\ge|f(a)|\ge0$. Let $x\in[a,b]$ and let $\varepsilon>0$ be real, and put $\delta=\varepsilon/(B+1)$, a real number with $\delta>0$. Every $y\in[a,b]$ with $|y-x|<\delta$ satisfies, by (ii),

$$|F(y)-F(x)|\le B\,|y-x|\le B\,\delta<(B+1)\,\delta=\varepsilon .$$

By The Real Line: Standing Notation and Background for Calculus on Intervals (vi), $F$ is therefore continuous at $x$; as $x\in[a,b]$ was arbitrary, $F$ is continuous on $[a,b]$, again by The Real Line: Standing Notation and Background for Calculus on Intervals (vi).

Proof of (iv). Let $x_{0}\in[a,b]$ be a point at which $f$ is continuous, and let $\varepsilon>0$ be real. By The Real Line: Standing Notation and Background for Calculus on Intervals (vi), applied with $\varepsilon/2$, there is a real $\delta>0$ such that every $s\in[a,b]$ with $|s-x_{0}|<\delta$ satisfies $|f(s)-f(x_{0})|<\varepsilon/2$. Let $t\in[a,b]$ with $0<|t-x_{0}|<\delta$, and let $p$ be the smaller and $q$ the larger of $t$ and $x_{0}$, so $p<q$ and $q-p=|t-x_{0}|$. Every $s\in[p,q]$ satisfies $s\in[a,b]$ and $|s-x_{0}|\le q-p<\delta$, as $x_{0}$ is one of $p$ and $q$; hence $f(x_{0})-\varepsilon/2\le f(s)\le f(x_{0})+\varepsilon/2$. By (2) with $c=f(x_{0})-\varepsilon/2$ and $C=f(x_{0})+\varepsilon/2$, and dividing by $q-p>0$,

$$f(x_{0})-\frac{\varepsilon}{2}\le\frac{F(q)-F(p)}{q-p}\le f(x_{0})+\frac{\varepsilon}{2}.$$

If $t=q$ the quotient here is $\frac{F(t)-F(x_{0})}{t-x_{0}}$; if $t=p$, numerator and denominator of $\frac{F(t)-F(x_{0})}{t-x_{0}}$ are the negatives of those of $\frac{F(q)-F(p)}{q-p}$ (Difference and Quotient Identities in ℝ (i)), so the two quotients agree again. Hence

$$\Bigl|\frac{F(t)-F(x_{0})}{t-x_{0}}-f(x_{0})\Bigr|\le\frac{\varepsilon}{2}<\varepsilon .$$

As $\varepsilon$ was arbitrary and $\delta$ was chosen after $\varepsilon$, $f(x_{0})$ is a derivative of $F$ at $x_{0}$. So $F$ is differentiable at $x_{0}$ by Derivative of a Real Function on an Interval (ii), and since this derivative is unique by Derivative of a Real Function on an Interval (ii) and Uniqueness of the Derivative of a Real Function on an Interval (i), $F'(x_{0})=f(x_{0})$.