TITLE: If a function $f$ is Riemann integrable on $[a,b]$, then how do I know $\lim_{N\to\infty}\sum_{n=1}^{N} f(x_n)\frac{b-a}{n}$ gives the right answer?
QUESTION [5 upvotes]: If a function $f$ is Riemann integrable on $[a,b]$, then how do I know $$\lim_{N\to\infty}\sum_{n=1}^{N} f(x_n)\frac{b-a}{N}$$ will actually converge to the value $$\int_{a}^{b} f(x) dx?$$ 
I know that If I take the supremum over the lower sums and the infimum of the upper sums, that they exist and are equal:
$$\inf_{P} U(f,P)=U(f)=L(f)=\sup_{P} L(f,P)$$
But how do I know that the for the given partition $P_N$, where each subinterval is equally spaced and of length $\frac{b-a}{N}$, that the sequence of partial sums will actually converge to the value of the integral?
This is confusing me because I cannot simply take a refinement, since I want the subintervals to be of the same size.
The reason I ask is because, in doing an exercise for my class, I want to claim that $$\sum_{n=1}^{N} f(x_n)\frac{b-a}{N}$$ actually converges to $$\int_{a}^{b} f(x)dx$$ when I let $N\to \infty$. It should not matter where I choose $x_i$ to be in each subinterval if I understand correctly.
Thanks

REPLY [3 votes]: It is standard to show that, since $P_{N+1}$ is finer than $P_N$, 
$$
L(f,P_N)\leq L (f,P_{N+1})\leq U(f,P_{N+1})\leq U(f,P_N). 
$$
So the sequence $\{L(f,P_N)\}$ is increasing and bounded; thus convergent. Similarly for $\{U(f,P_N)\}$. 
With a little work, you can show that given an arbitrary partition $P$, you can almost refine it via a $P_N$ (there will be some "problem" points where the partitions don't mesh, but you can take $N$ big enough so that the contribution of the problem intervals is insignificant). As you can find $P$ such that $U(f,P)$ is arbitrary close to $U(f)$, we conclude that $U(f,P_N)\to U(f)$. Similarly, $L(f,P_N)\to L(f)$. 
As
$$
L(f,P_N)\leq \sum_{n=1}^{N} f(x_n)\frac{b-a}{N}\leq U(f,P_N),
$$
we get
 by squeezing 
that the limit exists and equals $U(f)=L(f)$.