Continuous functions have an oriented Riemann integral

For a partition $P$, the maxima and minima in its sums exist by [L2]. If $Q$ refines $P$, each old subinterval is divided into finitely many pieces; the minimum on each piece is at least the old minimum and the maximum is at most the old maximum. Since the piece lengths sum to the old length, $L(g,P)\le L(g,Q)\le U(g,Q)\le U(g,P)$.

Fix $\epsilon >0$. For each $x\in [a,b]$, continuity [F2] gives ${\delta }_x>0$ such that $\mathrm{\mid }g(t)-g(x)\mathrm{\mid }<\epsilon \mathrm{/}2$ when $t\in [a,b]$ and $\mathrm{\mid }t-x\mathrm{\mid }<{\delta }_x$. The relative neighborhoods $(x-{\delta }_x\mathrm{/}2,x+{\delta }_x\mathrm{/}2)\cap [a,b]$ cover $[a,b]$. Choose a finite subcover with centers $x_1,\dots ,x_r$ by [L1], and put $\delta ={\min }_i{\delta }_{x_i}\mathrm{/}2>0$. If $u,v\in [a,b]$, $\mathrm{\mid }u-v\mathrm{\mid }<\delta$, and $u$ belongs to the neighborhood centered at $x_i$, then both $u$ and $v$ are within ${\delta }_{x_i}$ of $x_i$. By [L4], $\mathrm{\mid }g(u)-g(v)\mathrm{\mid }\le \mathrm{\mid }g(u)-g(x_i)\mathrm{\mid }+\mathrm{\mid }g(v)-g(x_i)\mathrm{\mid }<\epsilon$. Thus $g$ is uniformly continuous.

The range of $g$ has a minimum $m$ and maximum $M$ by [L2]. Thus all lower and upper sums lie between $m(b-a)$ and $M(b-a)$, and the partition set is nonempty. Define $I_{-}:={\sup }_{P}L(g,P)$ and $I_{+}:={\inf }_{P}U(g,P)$; both exist by [F1], applying it to negatives for the infimum. For any partitions $P,Q$, their union is a common refinement, so step 1.1 gives $L(g,P)\le L(g,P\cup Q)\le U(g,P\cup Q)\le U(g,Q)$. Taking the supremum and infimum yields $I_{-}\le I_{+}$.

Apply step 1.2 with tolerance $\epsilon \mathrm{/}(2(b-a))$ and choose an equal partition with mesh smaller than its uniform-continuity radius, using [L3]. On each piece the oscillation $\max g-\min g$ is less than that tolerance, because both extrema are attained by [L2]. Hence $U(g,P)-L(g,P)<\epsilon \mathrm{/}2$. Since $I_{-}\ge L(g,P)$ and $I_{+}\le U(g,P)$, we have $0\le I_{+}-I_{-}<\epsilon$. This holds for every $\epsilon >0$, so $I_{-}=I_{+}$. Their common value defines the integral in the statement.

Fix $c$ with $a<c<b$. Refining any partition by inserting $c$ cannot increase its upper sum by step 1.1, so the infimum of upper sums over all partitions equals the infimum over partitions containing $c$. Each such upper sum is the sum of an upper sum on $[a,c]$ and one on $[c,b]$. Every pair of partitions of these two intervals combines to a partition containing $c$; therefore the infimum of the combined sums is the sum of the two separate infima (choose each sum within $\epsilon \mathrm{/}2$ of its infimum for the reverse inequality). The common upper/lower values from step 2.2 give ${\int }_a^{b}g={\int }_a^{c}g+{\int }_c^{b}g$. Endpoint cases follow from ${\int }_a^{a}g=0$. Reversal is the definition, so the integral is oriented and additive.

The constant function with value $d$ has every upper and lower sum equal to $d(b-a)$, so its integral is $d(b-a)$. More generally, if constants $m,M$ satisfy $m\le g(t)\le M$ on $[a,b]$, every lower sum is at least $m(b-a)$ and every upper sum is at most $M(b-a)$; hence $m(b-a)\le {\int }_a^{b}g\le M(b-a)$. If $m_0$ and $M_0$ are the attained minimum and maximum of $g$, then $C:=\max (\mathrm{\mid }{m}_0\mathrm{\mid },\mathrm{\mid }{M}_0\mathrm{\mid })={\max }_{t\in [a,b]}\mathrm{\mid }g(t)\mathrm{\mid }$, since every value lies between $m_0$ and $M_0$. Applying the comparison with $m=-C,M=C$ gives $\mathrm{\mid }{\int }_a^{b}g\mathrm{\mid }\le C(b-a)$; reversal gives the bound with $\mathrm{\mid }b-a\mathrm{\mid }$ for either orientation.

If continuous $g\le h$ on $[a,b]$, then on each partition interval the minimum of $g$ is at most the minimum of $h$. Thus $L(g,P)\le L(h,P)$ for every partition $P$; taking suprema of the lower sums gives ${\int }_a^{b}g\le {\int }_a^{b}h$.

Let $g$ be continuous on an open interval $J$, fix $x_0\in J$, and put $G(x)={\int }_{x_0}^{x}g(t)dt$. The compact-segment results above define each integral. By additivity from step 3.1 and the constant-integral calculation in step 3.2, for $h\mathrm{\ne }0$ sufficiently small, $\frac{G(x+h)-G(x)}{h}=\frac{1}{h}{\int }_x^{x+h}g(t)dt\mathrm{.}$ For any $\eta >0$, continuity at $x$ gives $\delta >0$ such that $g(x)-\eta <g(t)<g(x)+\eta$ whenever $\mathrm{\mid }t-x\mathrm{\mid }<\delta$. If $0<h<\delta$, step 3.1 and the comparison in step 3.2 place this quotient between $g(x)-\eta$ and $g(x)+\eta$. If $-\delta <h<0$, reverse the integral by step 3.1, apply the same comparison on $[x+h,x]$, and divide by the negative $h$; the inequalities again place the quotient between those same constants. The extrema of $g$ on each such segment exist by [L2], and the largest deviation from $g(x)$ is attained at one of them. Hence its limit is $g(x)$, so $G^{\mathrm{\prime }}(x)=g(x)$.