Proof of Rules for Oriented Riemann Integrals of Real-Valued Maps: Sums, Scalar Multiples, Constants, Continuous Maps, Additivity and Evaluation by an Antiderivative

Throughout, for $u,v\in\mathbb{R}$ exactly one of $u<v$, $u=v$ and $v<u$ holds, by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vi), the order of $\mathbb{R}$ being total since $\mathbb{R}$ is an ordered field. For every $E\subseteq\mathbb{R}$, $g:E\to\mathbb{R}$ and $u\in E$, the integral $\int_{u}^{u}g$ is defined and equals $0$, by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (viii) and The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vii); and if $s<t$, $[s,t]\subseteq E$ and $g$ is integrable on $[s,t]$, then $\int_{s}^{t}g$ and $\int_{t}^{s}g$ are defined and $\int_{t}^{s}g=-\int_{s}^{t}g$, by the same clauses.

Integrals over a closed interval. Let $E\subseteq\mathbb{R}$, let $s,t\in E$ with $s<t$ and $[s,t]\subseteq E$, and let $f,g:E\to\mathbb{R}$ and $c\in\mathbb{R}$. The restrictions to $[s,t]$ of $f+g$, of $cf$ and of the constant map $E\to\mathbb{R}$ with value $c$ are $f|_{[s,t]}+g|_{[s,t]}$, $c\,f|_{[s,t]}$ and the constant map $[s,t]\to\mathbb{R}$ with value $c$: in each case both are maps $[s,t]\to\mathbb{R}$ with the same value at every point of $[s,t]$, by Pointwise Operations on Real-Valued Maps (i) and Pointwise Operations on Real-Valued Maps (ii). Hence, by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vi) and Rules for the Riemann Integral: Sums, Scalar Multiples, Monotonicity, Bounds and Constants (i), Rules for the Riemann Integral: Sums, Scalar Multiples, Monotonicity, Bounds and Constants (ii) and Rules for the Riemann Integral: Sums, Scalar Multiples, Monotonicity, Bounds and Constants (vi), applied with $s$ and $t$ for $a$ and $b$ and with $f|_{[s,t]}$ and $g|_{[s,t]}$: if $f$ and $g$ are integrable on $[s,t]$, then so are $f+g$ and $cf$, with

$$\int_{s}^{t}(f+g)=\int_{s}^{t}f+\int_{s}^{t}g,\qquad\int_{s}^{t}cf=c\int_{s}^{t}f;\qquad(1)$$

and the constant map $c$ is integrable on $[s,t]$, with $\int_{s}^{t}c=c\,(t-s)$.

Sums and scalar multiples ((i), (ii)). Let $E\subseteq\mathbb{R}$, $f,g:E\to\mathbb{R}$, $u,v\in E$ and $c\in\mathbb{R}$, with $\int_{u}^{v}f$ defined, and also $\int_{u}^{v}g$ for (i). If $u=v$, all integrals from $u$ to $v$ are defined and $0$, and $0=0+0$ and $0=c\cdot0$, by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (i). If $u<v$, then $[u,v]\subseteq E$ and $f$, and for (i) also $g$, are integrable on $[u,v]$ by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (viii), and (1) with $u$ and $v$ for $s$ and $t$ gives both clauses. If $v<u$, then $[v,u]\subseteq E$ and $f$, and for (i) also $g$, are integrable on $[v,u]$; by (1) with $v$ and $u$ for $s$ and $t$, $f+g$ and $cf$ are integrable on $[v,u]$, so $\int_{u}^{v}(f+g)$ and $\int_{u}^{v}cf$ are defined, and, by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (ii),

$$\int_{u}^{v}(f+g)=-\Bigl(\int_{v}^{u}f+\int_{v}^{u}g\Bigr)=\Bigl(-\int_{v}^{u}f\Bigr)+\Bigl(-\int_{v}^{u}g\Bigr)=\int_{u}^{v}f+\int_{u}^{v}g,$$ $$\int_{u}^{v}cf=-\Bigl(c\int_{v}^{u}f\Bigr)=c\Bigl(-\int_{v}^{u}f\Bigr)=c\int_{u}^{v}f.$$

Constants ((iii)). If $u=v$, then $\int_{u}^{v}c$ is defined and equals $0=0\cdot c=(v-u)\,c$, by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (i) and since $v-u=u-u=u+(-u)=0$ by Negatives, Differences, Reciprocals and Quotients (i) and Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (vi). If $u<v$, then $[u,v]\subseteq E$, and by the paragraph on closed intervals the constant map $c$ is integrable on $[u,v]$, so $\int_{u}^{v}c$ is defined and equals $c\,(v-u)=(v-u)\,c$. If $v<u$, then $[v,u]\subseteq E$, the constant map is integrable on $[v,u]$ with $\int_{v}^{u}c=(u-v)\,c$, so $\int_{u}^{v}c$ is defined, and by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (ii)

$$\int_{u}^{v}c=-\bigl((u-v)\,c\bigr)=\bigl(-(u-v)\bigr)\,c=(v-u)\,c.$$

Continuous maps ((iv)). Let $I$ and $f$ be as there, $f$ continuous on $I$; the same data are used for (v). We first show: $(\ast)$ for $s,t\in I$ with $s<t$, $[s,t]\subseteq I$ and $f|_{[s,t]}$ is bounded and integrable on $[s,t]$, so that $f$ is integrable on $[s,t]$. Let $s,t\in I$ with $s<t$. Then $[s,t]\subseteq I$ by Basic Facts about Intervals of the Real Line (ii), and $[s,t]$ is an interval containing two distinct points by Basic Facts about Intervals of the Real Line (v). As $f$ is continuous at every point of $[s,t]$ relative to $I$, $f|_{[s,t]}$ is continuous at every point of $[s,t]$ relative to $[s,t]$, by Restriction Stability of Continuity and of the Derivative (i), that is, continuous on $[s,t]$. So $f|_{[s,t]}$ is bounded and integrable on $[s,t]$ by A Continuous Function on a Closed Interval is Riemann Integrable (i), and $f$ is integrable on $[s,t]$ by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vi). This proves $(\ast)$.

Now let $u,v\in I$. If $u=v$, $\int_{u}^{v}f$ is defined. If $u<v$, $(\ast)$ with $u$ and $v$ for $s$ and $t$ shows that $[u,v]\subseteq I$ and $f$ is integrable on $[u,v]$; if $v<u$, $(\ast)$ with $v$ and $u$ shows this for $[v,u]$. In each case $\int_{u}^{v}f$ is defined, by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (viii).

Additivity ((v)). Let $u,v,w\in I$. By (iv), the three integrals are defined. Since the order of $\mathbb{R}$ is total and transitive (Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (viii)), one of $u,v,w$, call it $a$, satisfies $a\le u$, $a\le v$ and $a\le w$ (take the smaller of $u$ and $v$, then the smaller of it and $w$), and one, call it $b$, satisfies $u\le b$, $v\le b$ and $w\le b$ (likewise with the larger). If $a=b$, then each $x\in\{u,v,w\}$ satisfies $a\le x\le b=a$, so $x=a$ by antisymmetry (Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (viii)); thus $u=v=w$, and both sides equal $0=0+0$. Let $a<b$, and let $K=[a,b]$, so that $u,v,w\in K$. By $(\ast)$, $K\subseteq I$ and $f|_{K}$ is bounded and integrable on $[a,b]$. By Additivity of the Riemann Integral over Adjacent Intervals, Subintervals and Oriented Integrals, and Integrals of Restrictions (iii), applied to $f|_{K}$,

$$\int_{u}^{w}f|_{K}=\int_{u}^{v}f|_{K}+\int_{v}^{w}f|_{K}.$$

For $s,t\in K$ with $s<t$, $[s,t]\subseteq K$ by Basic Facts about Intervals of the Real Line (ii), $K$ being an interval by Basic Facts about Intervals of the Real Line (v), and $f|_{K}$ is integrable on $[s,t]$ by Additivity of the Riemann Integral over Adjacent Intervals, Subintervals and Oriented Integrals, and Integrals of Restrictions (ii), so by Additivity of the Riemann Integral over Adjacent Intervals, Subintervals and Oriented Integrals, and Integrals of Restrictions (iv), with $K$, $I$ and $f$ for its $K$, $E$ and $g$, $\int_{s}^{t}f|_{K}=\int_{s}^{t}f$ and $\int_{t}^{s}f|_{K}=\int_{t}^{s}f$; and $\int_{s}^{s}f|_{K}=\int_{s}^{s}f$ for $s\in K$ by the same clause. So each of the three integrals of $f|_{K}$ above equals the corresponding integral of $f$, which is the assertion.

Evaluation ((vi)). Let $I$, $f$, $u$ and $v$ be as there. If $u=v$, then $\int_{u}^{v}f'=0=f(v)-f(u)$. Let $s,t\in I$ with $s<t$, $[s,t]\subseteq I$, and $f'$ integrable on $[s,t]$; we show $\int_{s}^{t}f'=f(t)-f(s)$. Let $J=[s,t]$, an interval containing two distinct points by Basic Facts about Intervals of the Real Line (v), with $(s,t)\subseteq J$ by the same clause. By The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vi), $f'|_{J}$ is bounded and integrable on $[s,t]$ and $\int_{s}^{t}f'=\int_{s}^{t}f'|_{J}$. Let $F=f|_{J}:J\to\mathbb{R}$. For every $x\in J$, $f$ is differentiable at $x$, hence continuous at $x$ relative to $I$ by The Increment Form of the Derivative, and Differentiability Implies Continuity (ii), so $F$ is continuous at $x$ relative to $J$ by Restriction Stability of Continuity and of the Derivative (i); thus $F$ is continuous on $J$. For every $x\in(s,t)$, $F$ is differentiable at $x$ with $F'(x)=f'(x)=f'|_{J}(x)$, by Restriction Stability of Continuity and of the Derivative (ii). So Fundamental Theorem of Calculus, Part II: Evaluating an Integral by an Antiderivative (i), with $s$ and $t$ for $a$ and $b$, $f'|_{J}$ for its $f$ and $F$ for its $F$, gives

$$\int_{s}^{t}f'=\int_{s}^{t}f'|_{J}=F(t)-F(s)=f(t)-f(s).$$

If $u<v$, then $\int_{u}^{v}f'$ being defined means that $[u,v]\subseteq I$ and $f'$ is integrable on $[u,v]$ (The Riemann Integral on a Closed Interval via Lower and Upper Integrals (viii)), and the above with $u$ and $v$ for $s$ and $t$ is the assertion. If $v<u$, then $[v,u]\subseteq I$ and $f'$ is integrable on $[v,u]$, so $\int_{v}^{u}f'=f(u)-f(v)$, and by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (ii)

$$\int_{u}^{v}f'=-\bigl(f(u)-f(v)\bigr)=f(v)-f(u).$$