Let $f$ be Riemann integrable on $[a,b]$ then $$ \int_a^b f(x) \, \mathrm{d}x = \lim\limits_{n \to ∞} \dfrac{b-a}{n} \sum\limits_{r=1}^n f \left(a+\dfrac{r(b-a)}{n} \right) $$
I want to prove this, $$ \lim\limits_{n \to ∞} \dfrac{1}{n} \sum\limits_{t=an}^{bn} f \left(\dfrac{t}{n}\right) = \int_a^b f(x) \, \mathrm{d}x$$
Here's my attempt: Let $a,b \in \mathbb{Z}$ and $K=b-a$ then $$ I= \int_a^b f(x) \, \mathrm{d}x = \lim\limits_{n \to ∞} \dfrac{K}{n} \sum\limits_{r=1}^n f \left(a+\dfrac{rK}{n} \right) = \lim\limits_{n \to ∞} \dfrac{K}{n} \sum\limits_{r=1}^n f \left(\dfrac{an+rK}{n}\right) $$
Since $1≤r≤n$ so if $K>0$ then $an + K ≤ an+rK=t ≤ bn$
$$1) \quad I= \lim\limits_{n \to ∞} \dfrac{K}{n} \sum\limits_{t=an+K}^{bn} f \left(\dfrac{t}{n}\right) =\lim\limits_{n \to ∞} \dfrac{K}{n}\left(\sum\limits_{t=an}^{bn}f\left(\dfrac{t}{n}\right)- \sum\limits_{t=an}^{an+K-1} f\left(\dfrac{t}{n}\right) \right)= \lim\limits_{n \to ∞}\dfrac{K}{n}\sum\limits_{t=an}^{bn}f\left(\dfrac{t}{n}\right) $$
Similarly, $1≤r≤n$ so if $K<0$ then $an + K ≥ an+rK=t ≥ bn$ so,
$$2) \quad I = \lim\limits_{n \to ∞}\dfrac{K}{n}\sum\limits_{t=bn}^{an}f\left(\dfrac{t}{n}\right)$$
Almost there, but I don't know how to get rid of $K$ outside the sum in $1$ and $2$. If $K$ is gone, my result is proved. Any help?