If $f$ is bounded and monotone in $[0,1]$ show that: $$\int_0^1 f(x)dx - \frac{1}{n}\sum_{k=1}^n f\bigg(\frac{k}{n}\bigg)=\mathcal{O}\bigg(\frac{1}{n}\bigg)$$
Well, by definition $$\int_0^1 f(x)dx=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^n f\bigg(\frac{k}{n}\bigg)$$ Then the identity that is to be proven gets the form $$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^n f\bigg(\frac{k}{n}\bigg)= \frac{1}{n}\sum_{k=1}^n f\bigg(\frac{k}{n}\bigg)+\mathcal{O}\bigg(\frac{1}{n}\bigg)$$ I'm wondering what is next, though.
It is easy to show big 0, but not little 0. Assume monotone increasing (for decreasing reverse order). $\frac{1}{n}\sum_{k=1}^n f(\frac{k}{n})\ge \int_0^1 f(x)dx\ge \frac{1}{n}\sum_{k=0}^{n-1} f(\frac{k}{n})$ The difference between the two bounds $=\frac{1}{n}(f(1)-f(0))$ which is big $0$ of $\frac{1}{n}$, squeezing the integral.