Proof of Darboux Sums of Sums, Scalar Multiples and Dominated Functions, and the Oscillation Bound

Notation. For a bounded map $h:[a,b]\to\mathbb{R}$ and $i\in[n]$ let $S_{i}(h)=\{h(t):t\in[x_{i},x_{i+1}]\}$, $M_{i}(h)=\sup S_{i}(h)$ and $m_{i}(h)=\inf S_{i}(h)$ be as in Upper and Lower Darboux Sums of a Bounded Function on a Partition (i), which makes $i\mapsto M_{i}(h)$ and $i\mapsto m_{i}(h)$ maps from $[n]$ to $\mathbb{R}$; and let $\Delta_{i}=x_{i+1}-x_{i}$, which is positive by The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (iii). By 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),

$$U(h,P)=\sum_{i=1}^{n}M_{i}(h)\,\Delta_{i},\qquad L(h,P)=\sum_{i=1}^{n}m_{i}(h)\,\Delta_{i}.$$

Fact A. Let $h:[a,b]\to\mathbb{R}$ be bounded, $i\in[n]$ and $r\in\mathbb{R}$. The elements of $S_{i}(h)$ are the $h(t)$ with $t\in[x_{i},x_{i+1}]$ (Images of Typed Terms: the Set of Values $\{t:u\in D\}$ Is a Set, Finite when $D$ Is Finite, and a Typed Term (i)). (A1) $m_{i}(h)\le h(t)\le M_{i}(h)$ for every $t\in[x_{i},x_{i+1}]$, as $M_{i}(h)$ is an upper bound and $m_{i}(h)$ a lower bound of $S_{i}(h)$, 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). (A2) If $h(t)\le r$ for every $t\in[x_{i},x_{i+1}]$, then $r$ is an upper bound of $S_{i}(h)$, so $M_{i}(h)\le r$, the supremum being the least element of the upper bounds (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv) and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iii)); likewise, if $r\le h(t)$ for every such $t$, then $r\le m_{i}(h)$.

Fact S. Let $\alpha,\beta:[n]\to\mathbb{R}$ and $r\in\mathbb{R}$. (S1) $\sum_{i=1}^{n}(\alpha_{i}+\beta_{i})\Delta_{i}=\sum_{i=1}^{n}\alpha_{i}\Delta_{i}+\sum_{i=1}^{n}\beta_{i}\Delta_{i}$, by Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (v), addition on $\mathbb{R}$ being associative and commutative; and $\sum_{i=1}^{n}(\alpha_{i}-\beta_{i})\Delta_{i}=\sum_{i=1}^{n}\alpha_{i}\Delta_{i}-\sum_{i=1}^{n}\beta_{i}\Delta_{i}$, by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (ii). (S2) $\sum_{i=1}^{n}r\,\alpha_{i}\Delta_{i}=r\sum_{i=1}^{n}\alpha_{i}\Delta_{i}$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (i); in particular $\sum_{i=1}^{n}r\,\Delta_{i}=r\,(b-a)$, by The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (iii). (S3) If $\alpha_{i}\le\beta_{i}$ for every $i\in[n]$, then $\alpha_{i}\Delta_{i}\le\beta_{i}\Delta_{i}$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (v), and so $\sum_{i=1}^{n}\alpha_{i}\Delta_{i}\le\sum_{i=1}^{n}\beta_{i}\Delta_{i}$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (v).

Let $B_{f},B_{g}\in\mathbb{R}$ satisfy $|f(t)|\le B_{f}$ and $|g(t)|\le B_{g}$ for every $t\in[a,b]$.

Proof of (i). For $t\in[a,b]$, $(f+g)(t)=f(t)+g(t)$ by Pointwise Operations on Real-Valued Maps (i), so $|(f+g)(t)|\le|f(t)|+|g(t)|\le B_{f}+B_{g}$ by Rules of Arithmetic and Order in an Ordered Field (xi) and Rules of Arithmetic and Order in an Ordered Field (iv); thus $f+g$ is bounded. Let $i\in[n]$. For $t\in[x_{i},x_{i+1}]$, (A1) and Rules of Arithmetic and Order in an Ordered Field (iv) give $m_{i}(f)+m_{i}(g)\le(f+g)(t)\le M_{i}(f)+M_{i}(g)$; so, by (A2), $M_{i}(f+g)\le M_{i}(f)+M_{i}(g)$ and $m_{i}(f)+m_{i}(g)\le m_{i}(f+g)$. By (S3) and (S1),

$$U(f+g,P)\le\sum_{i=1}^{n}\bigl(M_{i}(f)+M_{i}(g)\bigr)\Delta_{i}=U(f,P)+U(g,P),\qquad L(f,P)+L(g,P)=\sum_{i=1}^{n}\bigl(m_{i}(f)+m_{i}(g)\bigr)\Delta_{i}\le L(f+g,P).$$

Proof of (ii). For every bounded $h:[a,b]\to\mathbb{R}$ with $|h(t)|\le B_{h}$ for $t\in[a,b]$, and every $r\in\mathbb{R}$, $(rh)(t)=r\,h(t)$ by Pointwise Operations on Real-Valued Maps (i), so $|(rh)(t)|=|r|\,|h(t)|\le|r|\,B_{h}$ by Rules of Arithmetic and Order in an Ordered Field (x) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (v); thus $rh$ is bounded. In particular $cf$ is bounded.

(N) Nonnegative factors. Let $h$ be bounded, $r\ge0$ and $i\in[n]$; we show $M_{i}(rh)=r\,M_{i}(h)$ and $m_{i}(rh)=r\,m_{i}(h)$. For $t\in[x_{i},x_{i+1}]$, (A1) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (v) give $r\,m_{i}(h)\le r\,h(t)\le r\,M_{i}(h)$, so $M_{i}(rh)\le r\,M_{i}(h)$ and $r\,m_{i}(h)\le m_{i}(rh)$ by (A2). If $r=0$, then, as $x_{i}\in[x_{i},x_{i+1}]$, (A1) and Rules of Arithmetic and Order in an Ordered Field (i) give $m_{i}(rh)\le(rh)(x_{i})=0=r\,m_{i}(h)$ and $r\,M_{i}(h)=0=(rh)(x_{i})\le M_{i}(rh)$. If $r>0$, then $r^{-1}>0$ by Rules of Arithmetic and Order in an Ordered Field (viii), and for $t\in[x_{i},x_{i+1}]$, $h(t)=r^{-1}(rh)(t)$ lies between $r^{-1}m_{i}(rh)$ and $r^{-1}M_{i}(rh)$ by (A1) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (v); by (A2), $M_{i}(h)\le r^{-1}M_{i}(rh)$ and $r^{-1}m_{i}(rh)\le m_{i}(h)$, and multiplying by $r$ (Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (v)) gives $r\,M_{i}(h)\le M_{i}(rh)$ and $m_{i}(rh)\le r\,m_{i}(h)$. In both cases the two inequalities in each direction give the equalities, the order of $\mathbb{R}$ being antisymmetric.

(R) The factor $-1$. Let $h$ be bounded, $k=(-1)h$ and $i\in[n]$; we show $M_{i}(k)=-m_{i}(h)$ and $m_{i}(k)=-M_{i}(h)$. For $t\in[a,b]$, $k(t)=(-1)h(t)=-h(t)$ by Pointwise Operations on Real-Valued Maps (i) and Rules of Arithmetic and Order in an Ordered Field (ii), and $h(t)=-k(t)$. For $t\in[x_{i},x_{i+1}]$, (A1) and Rules of Arithmetic and Order in an Ordered Field (v) give $-M_{i}(h)\le k(t)\le-m_{i}(h)$ and $-M_{i}(k)\le h(t)\le-m_{i}(k)$. By (A2), $M_{i}(k)\le-m_{i}(h)$, $-M_{i}(h)\le m_{i}(k)$, $M_{i}(h)\le-m_{i}(k)$ and $-M_{i}(k)\le m_{i}(h)$; the last two give $m_{i}(k)\le-M_{i}(h)$ and $-m_{i}(h)\le M_{i}(k)$ by Rules of Arithmetic and Order in an Ordered Field (v) and Rules of Arithmetic and Order in an Ordered Field (ii), whence the two equalities.

Now let $i\in[n]$. If $c\ge0$, then $M_{i}(cf)=c\,M_{i}(f)$ and $m_{i}(cf)=c\,m_{i}(f)$ by (N). If $c<0$, then $-c>0$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (iii); with $k=(-1)f$ we have $((-c)k)(t)=(-c)(-f(t))=c\,f(t)=(cf)(t)$ for $t\in[a,b]$, by Pointwise Operations on Real-Valued Maps (i) and Rules of Arithmetic and Order in an Ordered Field (ii), so $(-c)k=cf$ by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv); and by (N), applied to $k$ and $-c$, and (R), $M_{i}(cf)=(-c)M_{i}(k)=(-c)(-m_{i}(f))=c\,m_{i}(f)$ and $m_{i}(cf)=(-c)m_{i}(k)=(-c)(-M_{i}(f))=c\,M_{i}(f)$, by Rules of Arithmetic and Order in an Ordered Field (ii). Summing against the $\Delta_{i}$ by (S2): if $c\ge0$, $U(cf,P)=c\,U(f,P)$ and $L(cf,P)=c\,L(f,P)$; if $c<0$, $U(cf,P)=\sum_{i=1}^{n}c\,m_{i}(f)\Delta_{i}=c\,L(f,P)$ and $L(cf,P)=\sum_{i=1}^{n}c\,M_{i}(f)\Delta_{i}=c\,U(f,P)$.

Proof of (iii). Let $i\in[n]$. For $t\in[x_{i},x_{i+1}]$, $f(t)\le g(t)+c\le M_{i}(g)+c$ by (A1) and Rules of Arithmetic and Order in an Ordered Field (iv), so $M_{i}(f)\le M_{i}(g)+c$ by (A2). Also $m_{i}(f)-c\le f(t)-c\le g(t)$ by (A1) and Rules of Arithmetic and Order in an Ordered Field (iv), so $m_{i}(f)-c\le m_{i}(g)$ by (A2), that is, $m_{i}(f)\le m_{i}(g)+c$. By (S3), (S1) and (S2),

$$U(f,P)\le\sum_{i=1}^{n}\bigl(M_{i}(g)+c\bigr)\Delta_{i}=U(g,P)+c\,(b-a),\qquad L(f,P)\le\sum_{i=1}^{n}\bigl(m_{i}(g)+c\bigr)\Delta_{i}=L(g,P)+c\,(b-a).$$

Proof of (iv). Let $i\in[n]$ and $s,t\in[x_{i},x_{i+1}]$. As $|f(s)-f(t)|\le\eta$, $f(s)-f(t)\le\eta$ by Rules of Arithmetic and Order in an Ordered Field (x), so $f(s)\le f(t)+\eta$ by Rules of Arithmetic and Order in an Ordered Field (iv). Fixing $t$ and letting $s$ vary, (A2) gives $M_{i}(f)\le f(t)+\eta$, that is, $M_{i}(f)-\eta\le f(t)$; as this holds for every $t\in[x_{i},x_{i+1}]$, (A2) gives $M_{i}(f)-\eta\le m_{i}(f)$, that is, $M_{i}(f)-m_{i}(f)\le\eta$. By (S1), (S3) and (S2),

$$U(f,P)-L(f,P)=\sum_{i=1}^{n}\bigl(M_{i}(f)-m_{i}(f)\bigr)\Delta_{i}\le\sum_{i=1}^{n}\eta\,\Delta_{i}=\eta\,(b-a).$$