Proposition 5

Let $f$ be a regulated function on an interval I, let $g$ be a primitive of $f$ on I, and $[x_0,\ x]$ a compact interval contained in I. For every $\varepsilon > 0$ there exists a number $\rho > 0$ such that for every subdivision of $[x_0,\ x]$ into intervals of length $\leq \rho$ one has

(1) for every Riemann sum relative to this subdivision.

Indeed, let $f_1$ be a step function such that $\|f(y) - f_1(y)\| \leq \varepsilon$ for every $y \in [x_0,\ x]$; one has, denoting a primitive of $f_1$ on I by $g_1$,

by the mean value theorem, and on the other hand

It thus suffices to prove the proposition when $f$ is a step function. Let $(y_k)_{1 \leq k \leq m}$ be the strictly increasing finite sequence of points of discontinuity of $f$ in $[x_0,\ x]$. For every subdivision of $[x_0,\ x]$ into intervals of length $\leq \rho$ each of the points $y_k$ belongs to at most two of these intervals; there can therefore be no more than $2m$ intervals on which $f$ is not constant; but, on such an interval $[x_i,\ x_{i+1}]$ one has

on denoting by $M$ the least upper bound of $\|f\|$ on $[x_0,\ x]$; on the other hand, when $f$ is constant on $[x_i,\ x_{i+1}]$ one has One thus sees that the difference cannot exceed $4Mm\rho$; it therefore suffices to take $\rho \leq \varepsilon / 4Mm$ to obtain (1).