Riemann integrable as a convergent series

Question

If $f$ is integrable on $[0,1]$ show that: $$\int_0^1f(x)dx=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n}f(k/n)$$

how can I prove this?

Answer 1

Since $f(x)$ is integrable in $[0,1]$, than for any partition $\{0,t_{1}, t_{2}, ... , t_{n-1}, 1\}$ any for any choose of the element $x_{i}$ from each interval $x_{i} \in [t_{i},t_{i+1}]$, the Riemann sum defined $\sum_{i=0}^{n-1}f(x_{i})(t_{i+1}-t_{i})$ converges to the quantity denoted $\int_{0}^{1}f(x)dx$ which in unique. So you can choose arbitrary the partition and the element of each interval.

Now note that for your case the partition is $\{0,\frac{1}{n},\frac{2}{n}, ... , \frac{n-1}{n}, 1\}$ and choosen elements are exactly the left bounds of that intervals.

So this is in some sense particural case (for the definition).

The only thing i'm stuck with whether the sum should go form 0 to n, or to n-1 (or maybe form 1 to n).