Proof of Rules for the Riemann Integral: Sums, Scalar Multiples, Monotonicity, Bounds and Constants

Notation and a fact. For a bounded map $h:[a,b]\to\mathbb{R}$ and a partition $P$ of $[a,b]$, $U(h,P)$ and $L(h,P)$ are the upper sum and the lower sum of $h$ with respect to $P$; on a fixed partition, the sums of $f+g$, of $cf$ and of maps bounded by one another are compared with those of $f$ and $g$ by Darboux Sums of Sums, Scalar Multiples and Dominated Functions, and the Oscillation Bound (i), Darboux Sums of Sums, Scalar Multiples and Dominated Functions, and the Oscillation Bound (ii) and Darboux Sums of Sums, Scalar Multiples and Dominated Functions, and the Oscillation Bound (iii).

Fact B. Let $p,q\in\mathbb{R}$, and suppose that for every real $\varepsilon>0$ there are $\ell,u\in\mathbb{R}$ with $\ell\le p\le u$, $\ell\le q\le u$ and $u-\ell<\varepsilon$. Then $p=q$. Indeed, for every $\varepsilon>0$ we get $p\le u<\ell+\varepsilon\le q+\varepsilon$, so $p\le q$ by Arbitrary Positive Slack, and Approximation of Suprema and Infima, in the Real Numbers (i); likewise $q\le p$.

We also use that $L(h,P)\le U(h,P)$ for every bounded $h$ and every partition $P$, by Bounds for Darboux Sums, Monotonicity under Refinement, the Supremum of the Lower Sums and the Infimum of the Upper Sums, and Concatenation of Partitions (iii).

Proof of (i). By Darboux Sums of Sums, Scalar Multiples and Dominated Functions, and the Oscillation Bound (i), $f+g$ is bounded, and, with Bounds for Darboux Sums, Monotonicity under Refinement, the Supremum of the Lower Sums and the Infimum of the Upper Sums, and Concatenation of Partitions (iii) for $f+g$, every partition $P$ of $[a,b]$ satisfies

$$L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P)\le U(f,P)+U(g,P).\tag{1}$$

Let $\varepsilon>0$. By Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (i), choose partitions $P_{1}$ and $P_{2}$ of $[a,b]$ with $U(f,P_{1})-L(f,P_{1})<\varepsilon/2$ and $U(g,P_{2})-L(g,P_{2})<\varepsilon/2$, and let $P=P_{1}\cup P_{2}$, a partition of $[a,b]$ that refines $P_{1}$ and $P_{2}$ by The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (v). By Bounds for Darboux Sums, Monotonicity under Refinement, the Supremum of the Lower Sums and the Infimum of the Upper Sums, and Concatenation of Partitions (ii), $U(f,P)-L(f,P)\le U(f,P_{1})-L(f,P_{1})<\varepsilon/2$ and likewise for $g$ with $P_{2}$. Put $\ell=L(f,P)+L(g,P)$ and $u=U(f,P)+U(g,P)$; then $u-\ell<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ii) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vii), and by (1), $U(f+g,P)-L(f+g,P)\le u-\ell<\varepsilon$. As $\varepsilon$ was arbitrary, $f+g$ is integrable on $[a,b]$ by Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (i). Moreover, for $\varepsilon$, $P$, $\ell$ and $u$ as above, Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (ii) and (1) give $\ell\le L(f+g,P)\le\int_{a}^{b}(f+g)\le U(f+g,P)\le u$, and Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (ii) for $f$ and for $g$ gives $\ell\le\int_{a}^{b}f+\int_{a}^{b}g\le u$. By Fact B, $\int_{a}^{b}(f+g)=\int_{a}^{b}f+\int_{a}^{b}g$.

Proof of (ii). By Darboux Sums of Sums, Scalar Multiples and Dominated Functions, and the Oscillation Bound (ii), $cf$ is bounded, and for every partition $P$ of $[a,b]$,

$$\text{if }c\ge0:\quad L(cf,P)=cL(f,P),\ U(cf,P)=cU(f,P);\qquad\text{if }c<0:\quad L(cf,P)=cU(f,P),\ U(cf,P)=cL(f,P).\tag{2}$$

Let $\ell_{P}=L(cf,P)$ and $u_{P}=U(cf,P)$ be these values. By Absolute Value in an Ordered Field (i), $u_{P}-\ell_{P}=|c|\,(U(f,P)-L(f,P))$ in both cases, and multiplying $L(f,P)\le\int_{a}^{b}f\le U(f,P)$, which holds by Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (ii), by $c$ gives $\ell_{P}\le c\int_{a}^{b}f\le u_{P}$: for $c\ge0$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (v), and for $c<0$ by the same with the factor $-c>0$ (Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (iii)) followed by negation, Rules of Arithmetic and Order in an Ordered Field (v) and Rules of Arithmetic and Order in an Ordered Field (ii). Let $\varepsilon>0$. As $|c|+1>0$, by Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (i) choose a partition $P$ with $U(f,P)-L(f,P)<\varepsilon/(|c|+1)$. Since $U(f,P)-L(f,P)\ge0$, $u_{P}-\ell_{P}\le(|c|+1)(U(f,P)-L(f,P))<\varepsilon$, that is, $U(cf,P)-L(cf,P)<\varepsilon$. Hence $cf$ is integrable on $[a,b]$ by Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (i). For $\varepsilon$ and $P$ as above, Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (ii) and (2) give $\ell_{P}\le\int_{a}^{b}cf\le u_{P}$, and $\ell_{P}\le c\int_{a}^{b}f\le u_{P}$; so $\int_{a}^{b}cf=c\int_{a}^{b}f$ by Fact B.

Proof of (iii). As $f(t)\le g(t)+0$ for every $t\in[a,b]$, Darboux Sums of Sums, Scalar Multiples and Dominated Functions, and the Oscillation Bound (iii) with $c=0$ gives $U(f,P)\le U(g,P)+0\,(b-a)=U(g,P)$ for every partition $P$ of $[a,b]$. Let $\varepsilon>0$, and by Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (i) choose a partition $P$ with $U(g,P)-L(g,P)<\varepsilon$. By Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (ii) for $f$ and for $g$,

$$\int_{a}^{b}f\le U(f,P)\le U(g,P)<L(g,P)+\varepsilon\le\int_{a}^{b}g+\varepsilon .$$

As $\varepsilon$ was arbitrary, $\int_{a}^{b}f\le\int_{a}^{b}g$ by Arbitrary Positive Slack, and Approximation of Suprema and Infima, in the Real Numbers (i).

Proof of (iv). The set $\{a,b\}$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iii) and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (v), and $a,b\in[a,b]$ by Basic Facts about Intervals of the Real Line (v), so $\{a,b\}$ is a partition of $[a,b]$. By Bounds for Darboux Sums, Monotonicity under Refinement, the Supremum of the Lower Sums and the Infimum of the Upper Sums, and Concatenation of Partitions (i), applied with $\ell=m$ and $C=M$, and Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (ii),

$$m\,(b-a)\le L(f,\{a,b\})\le\int_{a}^{b}f\le U(f,\{a,b\})\le M\,(b-a).$$

Proof of (v). By Rules of Arithmetic and Order in an Ordered Field (x), $-B\le f(t)\le B$ for every $t\in[a,b]$. By (iv) with $m=-B$ and $M=B$, $-(B\,(b-a))\le\int_{a}^{b}f\le B\,(b-a)$, and so $\bigl|\int_{a}^{b}f\bigr|\le B\,(b-a)$ by Rules of Arithmetic and Order in an Ordered Field (x).

Proof of (vi). Let $k:[a,b]\to\mathbb{R}$ be the constant map with value $c$ of Pointwise Operations on Real-Valued Maps (ii). Then $|k(t)|=|c|$ for every $t\in[a,b]$, so $k$ is bounded. For every partition $P$ of $[a,b]$, Bounds for Darboux Sums, Monotonicity under Refinement, the Supremum of the Lower Sums and the Infimum of the Upper Sums, and Concatenation of Partitions (i) with $\ell=C=c$ gives $c\,(b-a)\le L(k,P)\le U(k,P)\le c\,(b-a)$, so $U(k,P)-L(k,P)=0$. As $\{a,b\}$ is a partition of $[a,b]$ (proof of (iv)), for every $\varepsilon>0$ there is a partition $P$ with $U(k,P)-L(k,P)=0<\varepsilon$, and $k$ is integrable on $[a,b]$ by Riemann's Criterion for Integrability, and the Integral Lies between Lower and Upper Sums (i). Applying (iv) to the bounded integrable map $k$ in place of $f$, with $m=M=c$, gives $c\,(b-a)\le\int_{a}^{b}k\le c\,(b-a)$, so $\int_{a}^{b}c\,dt=\int_{a}^{b}k=c\,(b-a)$.