I was trying to show that if $f$ is Riemann integrable on $[0,1]$ then $$\int_{0}^{1} f dx = \lim \left( \frac{1}{n} \sum_{k=1}^{n} f(\frac{k}{n}) \right)$$
I know that since $f$ is Riemann integrable, $\forall \varepsilon > 0$ $\exists P_\varepsilon$ partition of $[0,1]$ such that if $P \le P_\varepsilon$ then $$\left| S(P;f,x)-\int_{0}^{1} f dx\right| < \varepsilon$$
Then I defined $P \le P_\varepsilon$ as $P=\left\{ x_0=0<x_1<...<x_n=1 \right\}$ and $c_k=\frac{k}{n}$ and I got that $$S(P;f,x)=\sum_{k=1}^{n} f(c_k)[x_k - x_{k-1}] = \sum_{k=1}^{n} f\left(\frac{k}{n}\right)[x_k - x_{k-1}]$$
I think that if the intervals of the partition are equidistant I would get the desired result, but is there anything that restricts me from creating a partition $P$ as such? Or did I made any mistakes in the reasoning explained before?
Thanks in advance!
Following Michael Spivak - Calculus, page 282, 2008 - if $f$ is integrable on $[a,b]$, then for every $\varepsilon > 0$, there is some $\delta>0$ such that, if $P=\{t_0,\cdots,t_n\}$ is any partition of $[a,b]$, with all lengths $t_i-t_{i-1}<\delta$, then $$\left|\sum\limits_{i=1}^{n}f(x_i)(t_i-t_{i-1}) -\int\limits_{a}^{b}f(x)dx \right |<\varepsilon$$ for any Riemman sum formed by choosing $x_i$ in $[t_i,t_{i-1}]$.
As you see nothing prevents you from considering equidistant partition and choosing one border of it as value for function. So, your getting desired result is correct.