Proposition: Riemann Sum Converging To the Riemann Integral

Let $a < b$, $[a,b]$ be a closed real interval and let $f:[a,b]\to\mathbb R$ be Riemann-integrable. If the mesh \(\mu\) of the partition \(a=x_0 < x_1 < \ldots < x_{n-1} < x_n=b\), with respect to which the Riemann sum. \[\sum_{k=1}^n f(\xi_k)(x_k-x_{k-1})\] is defined, converges to $0$, then the above Riemann sum of $f$ converges against the Riemann-integral of $f$.

More strictly, for every $\epsilon > 0$ there is a $\delta > 0$ such that for every partition \(a=x_0 < x_1 < \ldots < x_{n-1} < x_n=b\) with the mash $\mu \le \delta$ and every choice of points $\xi_i\in[x_{k-1},x_k]$ we have

$$\left|\int_a^bf(x)dx-\sum_{k=1}^nf(\xi_k)(x_k-x_{k-1})\right|\le \epsilon.$$

Proofs: 1

Proofs: 1 2

Thank you to the contributors under CC BY-SA 4.0!




  1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983