Proof of Additivity of the Riemann Integral over Adjacent Intervals, Subintervals and Oriented Integrals, and Integrals of Restrictions

Notation. For $u,v\in[a,b]$ with $u<v$ we have $[u,v]\subseteq[a,b]$ by Basic Facts about Intervals of the Real Line (v) and Basic Facts about Intervals of the Real Line (ii); let $f_{uv}=f|_{[u,v]}:[u,v]\to\mathbb{R}$ be the restriction, with $f_{uv}(t)=f(t)$ for $t\in[u,v]$ (Domains and Restrictions of Maps (iii)), so $|f_{uv}(t)|\le B$ and $f_{uv}$ is bounded. Let $\underline{I}(u,v)$ and $\overline{I}(u,v)$ be the lower and the upper integral of $f_{uv}$ over $[u,v]$ (The Riemann Integral on a Closed Interval via Lower and Upper Integrals (iii)); $\underline{I}(u,v)\le\overline{I}(u,v)$ 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 (iv). As $f_{ab}=f$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iii)), by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (iv), The Riemann Integral on a Closed Interval via Lower and Upper Integrals (v) and The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vi): $f$ is integrable on $[u,v]$ if and only if $\underline{I}(u,v)=\overline{I}(u,v)$, and then $\int_{u}^{v}f=\underline{I}(u,v)$. For a bounded map $h$ on a closed interval and a partition $R$ of it, the sums $U(h,R)$ and $L(h,R)$ are those of Upper and Lower Darboux Sums of a Bounded Function on a Partition (ii) and Upper and Lower Darboux Sums of a Bounded Function on a Partition (iii), formed with the increasing listing of $R$, which is unique by The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (i).

Proof of (iv). Let $u,v\in K$ with $u<v$ and $[u,v]\subseteq K$, so $[u,v]\subseteq E$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ix), as $[u,v]\cap K=[u,v]=[u,v]\cap E$, both $(g|_{K})|_{[u,v]}$ and $g|_{[u,v]}$ are maps from $[u,v]$ to $\mathbb{R}$ with value $g(t)$ at every $t\in[u,v]$; so they are equal, by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (i). Hence, by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vi), $g|_{K}$ is integrable on $[u,v]$ if and only if this common map is bounded and integrable on $[u,v]$, that is, if and only if $g$ is integrable on $[u,v]$; and in that case $\int_{u}^{v}g|_{K}$ and $\int_{u}^{v}g$ are both its integral over $[u,v]$, so they are equal, and $\int_{v}^{u}g|_{K}=-\int_{u}^{v}g|_{K}=-\int_{u}^{v}g=\int_{v}^{u}g$ by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vii). For $u\in K$, $u\in E$, and $\int_{u}^{u}g|_{K}=0=\int_{u}^{u}g$ by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vii).

Step 1: joining partitions. Let $c\in\mathbb{R}$ with $a<c<b$, let $P_{1}$ be a partition of $[a,c]$ and $P_{2}$ a partition of $[c,b]$. By The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (vii), $P_{1}\cup P_{2}$ is a partition of $[a,b]$, and 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 (v),

$$U(f,P_{1}\cup P_{2})=U(f_{ac},P_{1})+U(f_{cb},P_{2}),\qquad L(f,P_{1}\cup P_{2})=L(f_{ac},P_{1})+L(f_{cb},P_{2}).$$

Step 2: cutting a partition at $c$. Let $c$ be as in Step 1 and let $P$ be a partition of $[a,b]$. By The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (vi), $P'=P\cup\{c\}$ is a partition of $[a,b]$ and a refinement of $P$; and by The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (viii), $P_{1}=P'\cap[a,c]$ and $P_{2}=P'\cap[c,b]$ are partitions of $[a,c]$ and of $[c,b]$ with $P_{1}\cup P_{2}=P'$. By Step 1 and 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),

$$L(f,P)\le L(f,P')=L(f_{ac},P_{1})+L(f_{cb},P_{2}),\qquad U(f,P)\ge U(f,P')=U(f_{ac},P_{1})+U(f_{cb},P_{2}).$$

Step 3: lower and upper integrals add. Let $c$ be as in Step 1. We show $\underline{I}(a,b)=\underline{I}(a,c)+\underline{I}(c,b)$ and $\overline{I}(a,b)=\overline{I}(a,c)+\overline{I}(c,b)$. By The Riemann Integral on a Closed Interval via Lower and Upper Integrals (iii), $\underline{I}(u,v)$ is the supremum of the set of lower sums of $f_{uv}$ and $\overline{I}(u,v)$ the infimum of its set of upper sums, whose elements are the $L(f_{uv},R)$, respectively the $U(f_{uv},R)$, for the partitions $R$ of $[u,v]$, by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (i) and The Riemann Integral on a Closed Interval via Lower and Upper Integrals (ii). A supremum is an upper bound of the set and lies below each of its upper bounds, and an infimum is a lower bound and lies above each lower bound, by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv), Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iii) and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (i).

Let $P_{1}$ and $P_{2}$ be partitions of $[a,c]$ and $[c,b]$. By Step 1, $L(f_{ac},P_{1})+L(f_{cb},P_{2})=L(f,P_{1}\cup P_{2})\le\underline{I}(a,b)$, so $L(f_{ac},P_{1})\le\underline{I}(a,b)-L(f_{cb},P_{2})$ by Rules of Arithmetic and Order in an Ordered Field (iv). Fixing $P_{2}$, $\underline{I}(a,b)-L(f_{cb},P_{2})$ is thus an upper bound of the lower sums of $f_{ac}$, so $\underline{I}(a,c)\le\underline{I}(a,b)-L(f_{cb},P_{2})$, that is, $L(f_{cb},P_{2})\le\underline{I}(a,b)-\underline{I}(a,c)$ (Rules of Arithmetic and Order in an Ordered Field (iv)). As $P_{2}$ was arbitrary, $\underline{I}(a,b)-\underline{I}(a,c)$ is an upper bound of the lower sums of $f_{cb}$, so $\underline{I}(c,b)\le\underline{I}(a,b)-\underline{I}(a,c)$, that is, $\underline{I}(a,c)+\underline{I}(c,b)\le\underline{I}(a,b)$. Conversely, let $P$ be a partition of $[a,b]$, and let $P_{1}$ and $P_{2}$ be the partitions of Step 2; then $L(f,P)\le L(f_{ac},P_{1})+L(f_{cb},P_{2})\le\underline{I}(a,c)+\underline{I}(c,b)$ by Step 2 and Rules of Arithmetic and Order in an Ordered Field (iv). So $\underline{I}(a,c)+\underline{I}(c,b)$ is an upper bound of the lower sums of $f_{ab}=f$, and $\underline{I}(a,b)\le\underline{I}(a,c)+\underline{I}(c,b)$. The two inequalities give the equality for the lower integrals. The argument for the upper integrals is the same, with all inequalities reversed, upper sums in place of lower sums, lower bounds in place of upper bounds and infima in place of suprema.

Proof of (i). For $u<v$ in $[a,b]$ let $D(u,v)=\overline{I}(u,v)-\underline{I}(u,v)\ge0$. By Step 3, $D(a,b)=D(a,c)+D(c,b)$. If $f$ is integrable on $[a,c]$ and on $[c,b]$, then $D(a,c)=D(c,b)=0$, so $D(a,b)=0$ and $f$ is integrable on $[a,b]$. If $f$ is integrable on $[a,b]$, then $0\le D(a,c)=-D(c,b)\le0$, so $D(a,c)=D(c,b)=0$ and $f$ is integrable on $[a,c]$ and on $[c,b]$. In that case, by the notation paragraph and Step 3, $\int_{a}^{b}f=\underline{I}(a,b)=\underline{I}(a,c)+\underline{I}(c,b)=\int_{a}^{c}f+\int_{c}^{b}f$.

Proof of (ii). Let $f$ be integrable on $[a,b]$ and $u,v\in[a,b]$ with $u<v$. First, $f$ is integrable on $[u,b]$: if $u=a$ this is the hypothesis, as $f_{ab}=f$; if $a<u$, then $a<u<b$ and (i) with $c=u$ applies. Let $g=f_{ub}$, bounded and integrable on $[u,b]$. If $v=b$ we are done. If $v<b$, (i), applied to $g$ on $[u,b]$ with $c=v$, shows that $g$ is integrable on $[u,v]$. As $g=f|_{[u,b]}$ and $[u,v]\subseteq[u,b]\subseteq[a,b]$, (iv), with $[a,b]$ for $E$, $[u,b]$ for $K$ and $f$ for $g$, shows that $f$ is integrable on $[u,v]$.

Proof of (iii). Let $f$ be integrable on $[a,b]$. For $x,y\in[a,b]$ let $I(x,y)=\int_{x}^{y}f$: for $x<y$ this is $\int_{x}^{y}f_{xy}$, defined by (ii) and The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vi); $I(x,x)=0$; and for $y<x$, $I(x,y)=-I(y,x)$, as in The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vii). Hence $I(y,x)=-I(x,y)$ for all $x,y\in[a,b]$, checking the cases $x<y$, $x=y$ and $y<x$.

Let $x,y,z\in[a,b]$ with $x<y<z$. By (ii), $h=f_{xz}$ is bounded and integrable on $[x,z]$, and by (i), applied to $h$ on $[x,z]$ with $c=y$, $\int_{x}^{z}h=\int_{x}^{y}h+\int_{y}^{z}h$. Here $\int_{x}^{z}h=I(x,z)$; and, as $h=f|_{[x,z]}$ with $[x,y]\subseteq[x,z]$ and $[y,z]\subseteq[x,z]$, (iv), with $[a,b]$ for $E$, $[x,z]$ for $K$ and $f$ for $g$, gives $\int_{x}^{y}h=\int_{x}^{y}f=I(x,y)$ and $\int_{y}^{z}h=\int_{y}^{z}f=I(y,z)$. Thus $I(x,z)=I(x,y)+I(y,z)$.

For $u,v,w\in[a,b]$ let $E(u,v,w)=I(u,v)+I(v,w)+I(w,u)$. Since $I(w,u)=-I(u,w)$, the assertion is $E(u,v,w)=0$. If two of $u,v,w$ coincide, $E(u,v,w)=0$ by $I(s,s)=0$ and $I(s,t)=-I(t,s)$. Otherwise let $x<y<z$ be $u,v,w$ in increasing order. Then $E(x,y,z)=I(x,y)+I(y,z)-I(x,z)=0$ by the preceding paragraph. $E$ is unchanged when its arguments are rotated, $E(u,v,w)=E(v,w,u)$, and changes sign when the first two are exchanged: $E(v,u,w)=I(v,u)+I(u,w)+I(w,v)=-E(u,v,w)$. Every arrangement of $x,y,z$ is a rotation of $(x,y,z)$ or of $(y,x,z)$, so $E(u,v,w)=\pm E(x,y,z)=0$.