Proof of Uniform Partitions and Order Bounds for the Riemann Integral

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$, and elementary manipulations of sums, products, reciprocals and the order in the ordered field $\mathbb{R}$ are used as recorded in Elementary Order Arithmetic in an Ordered Field. For a partition $P$, the symbol $|P|$ denotes its mesh, as in Partition of a Closed Interval, never an absolute value.

Claim 1. Fix $n\in\mathbb{N}$, set $x_0=p$, and set $x_i=p+i\,(q-p)/n$ for $i=1,\dots,n$. Then $x_n=p+(q-p)=q$. For each $i$ with $1\le i\le n$,

which is positive: by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field we have $0<n$ in $\mathbb{R}$, and $n^{-1}$ exists with $0<n^{-1}$, so $(q-p)/n$ is a product of positive factors and is positive. Hence $p=x_0<x_1<\cdots<x_n=q$, so $P_n=(x_0,x_1,\dots,x_n)$ is a partition of $[p,q]$, and its mesh $|P_n|$, the maximum of the increments $x_i-x_{i-1}$, which are all equal, is $(q-p)/n$.

Now let $\delta>0$ be real. By claim 2 of The Archimedean Property of the Real Numbers, applied with $x=q-p$ and $\varepsilon=\delta$, there exists $n\in\mathbb{N}$ with $q-p<n\,\delta$; dividing by $n>0$ gives $(q-p)/n<\delta$.

Claim 2. Write $I=\int_p^q f(u)\,du$ for the Riemann integral of $f$ over $[p,q]$, and let $\varepsilon>0$ be real. By the definition of Riemann integrability there exists a real $\delta>0$ with the following property: whenever $P$ is a partition of $[p,q]$ with mesh $|P|<\delta$ and one chooses a tagged partition of $[p,q]$ relative to $P$, the corresponding Riemann sum $S$ of $f$ satisfies $|S-I|<\varepsilon$.

By claim 1 there is a partition $P_n=(x_0,x_1,\dots,x_n)$ of $[p,q]$ with mesh $|P_n|<\delta$. Choose the tags $t_i=x_{i-1}\in[x_{i-1},x_i]$ for $1\le i\le n$; these form a tagged partition of $[p,q]$ relative to $P_n$, and the corresponding Riemann sum of $f$ is

Since $x_i-x_{i-1}>0$ and $m\le f(t_i)\le M$ for each $i$, comparing term by term gives

where $\sum_{i=1}^n(x_i-x_{i-1})=x_n-x_0=q-p$ by cancellation of the telescoping sum (finite induction on $n$).

From $|S-I|<\varepsilon$ we get $S-\varepsilon<I<S+\varepsilon$, and combining with the bounds on $S$,

Since $\varepsilon>0$ was arbitrary, we conclude that $m\,(q-p)\le I\le M\,(q-p)$: if $I<m\,(q-p)$ held, the choice $\varepsilon=m\,(q-p)-I>0$ would give $I<I$, and if $M\,(q-p)<I$ held, the choice $\varepsilon=I-M\,(q-p)>0$ would give $I<I$; both are impossible. $\blacksquare$