$$ \mbox{One has}\quad\int_{0}^{1}\operatorname{f}\left(\,{x}\,\right)\,{\rm d}x = \lim_{n \to \infty}\ \frac{2}{n^{2}} \sum_{i\ =\ 1}^{n}\ \sum_{j\ =\ 0}^{i} \operatorname{f}\left(\,{j \over i}\,\right)\quad \mbox{for any continuous}\ \operatorname{f}. $$
- This can be checked by direct computation when $\operatorname{f}$ is a polynomial, and then extended to continuous functions by the Weierstrass approximation theorem.
- Does it hold for every Riemann integrable function $?$.
- Clearly not for all Lebesgue integrable functions, since they can be changed on the set of rational $x$ without changing the integral.
If $f$ ia any Riemann integrable function on $[0,1]$, then
$$\lim_{n \to \infty} A_n:= \lim_{n \to \infty} \frac{1}{n}\sum_{j=0}^n f\left(j/n\right)=\lim_{n \to \infty} \frac{1}{n}\sum_{j=1}^n f\left(j/n\right) = \int_0^1 f(x) \, dx$$
Now,
$$\frac{2}{n^2}\sum_{i=1}^n\sum_{j=0}^if(j/i) = \frac{2}{n^2}\sum_{i=1}^ni \cdot \frac{1}{i}\sum_{j=0}^if(j/i)= 2\frac{\sum_{i=1}^niA_i}{n^2}$$
By the Stolz-Cesaro theorem,
$$\lim_{n \to \infty}2\frac{\sum_{i=1}^niA_i}{n^2}=2\lim_{n \to \infty}\frac{\sum_{i=1}^niA_i-\sum_{i=1}^{n-1}iA_i}{n^2 - (n-1)^2}= 2 \lim_{n\to \infty}\frac{nA_n}{2n-1}\\= \lim_{n\to \infty}\frac{A_n}{1- \frac{1}{2n}}= \int_0^1 f(x) \, dx$$
Although related, a Cesaro average of Riemann sums would be written as
$$\frac{1}{n}\sum_{i=1}^n\frac{1}{i}\sum_{j=1}^if(j/i)$$