Prove $s_{n+q} > s_n$ when $ s_n = \frac1n\sum_{k=0}^{n-1} f(\frac{k}{n})$ and $f(x)$ is strictly increasing

Question

While I was doing some exercises about monotonic functions I thought about a proposition, but I don't know really well how to proceed to prove it, or maybe it could be even false :

Take a strictly increasing, and bounded function $f(x)$ on the interval $[0,1]$, and define $\displaystyle s_n = \frac1n\sum_{k=0}^{n-1} f(\frac{k}{n})$, then for each $n \in \mathbb{N}^+$ there exists a $q \in \mathbb{N}$ such that $s_{n+q} > s_n$.

It seems really linked to integrability of monotonic functions but It doesn't seem to me that it is sufficient to prove this one. Any help is appreciated

Answer 1

As you have clarified in the comment, I will assume strict monotonicity of the function. If you are seeking for such $q$ for each $n$, simply put $q=n$ :

$$s_{2n} = \sum_{k=0}^{2n} \frac{1}{2n} f\left(\frac{k}{2n}\right) = \sum_{i=0}^n \frac{1}{2n}\left[ f\left(\frac{2i}{2n}\right) + f\left(\frac{2i+1}{2n}\right)\right] $$ $$> \sum_{i=0}^{n} \frac{1}{2n} \left[ f\left( \frac{2i}{2n}\right)+f\left( \frac{2i}{2n}\right)\right] = s_n. $$

This argument works equally for discontinuous functions.

Answer 2

Another way could be to use integrability of increasing functions. Defining $t_n = \frac1n \sum_{k=1}^n f(\frac kn)$, we have :

$$s_n <\int_0^1 f(x) \ dx < t_n \implies 0 < \int_0^1 f(x) \ dx-s_n < \frac{f(1)-f(0)}{n}$$

Then $\{s_n\}$ converges to $\int_0^1 f(x)$ when $n \to + \infty$ for squeeze theorem, so since $s_{n_1} < \int_0^1 f(x)$ for all $n_1$ there must exist a $n_2 > n_1$ so that $s_{n_2} > s_{n_1}$.