Let $f \in \Re[0,1]$. Prove that $\lim_{n \to \infty} \sum_{k=1}^n f(\frac{k}{n}) \frac{1}{n} = \int_0^1f.$
I want to prove this using Riemann-Stieltjes sums. Here is what I thought so far:
I know by a theorem that $\lim_{norm P \to \infty} S(f,P,T) = \int_a^b f d \alpha$
The given equation tells me that $t_i$ is $\frac{k}{n}$ and $\Delta \alpha_i = \frac{1}{n}$. Now I'm trying to figure out, what do I do with $norm P \to \infty$. What exactly does that mean? Is it the same as approaching the upper bound of 1 on $f$?
Hint: You can do this with normal Riemann integration:
Take the following partition $$x_0=0<\frac{1}{n}<\frac{2}{n}<\frac{3}{n}<...<\frac{n-1}{n}<x_n=1$$ or $x_k=\frac{k}{n}$.
And see what happens to left Riemann sum and right Riemann sum.
Verify using the fact that $f$ is Riemann integrable, the limits of both left and right sums exist and are equal to $\int_{0}^{1}f(x)dx$.