# 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?

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

• How are you defining your $x_n$? Jan 16, 2020 at 22:20
• You can prove more. See this answer Jan 19, 2020 at 7:39

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 $$U(f)=L(f)$$.
• You first take $\limsup$ on both sides, and then the infimum on the left. Jan 17, 2020 at 0:49
• I still cannot see why you have $L(f)\leq \lim \sup L(f,P_N)$. Since we define $L(f)=\sup_P L(f,P)$, shouldn't we have $L(f)\geq \lim \sup L(f,P_N)$? (Because the sup is always $\geq$ the limsup.) Jan 17, 2020 at 1:01
• @MartinArgerami, I have 2 remarks, maybe you can correct me if I am wrong: 1.) shouldn't it be $\liminf_{N\to\infty}L(f,P_N)\leq L(f)$?Because $\liminf_{N\to\infty}$ is a sequence of infima which means: all members of this sequence are$\leq L(f)$. Taking the supremum of this sequence means that it attains at most$L(f)$. 2.) Why do you argue with lim inf and lim sup at all?Isn't it possible to do this with the squeeze theorem?If we consider$$L(f,P_N)\leq \sum_{n=1}^{N}f(x_n)\frac{b-a}{N}\leq U(f,P_N),$$where$L(f,P_N)$ and $U(f,P_N)$ are two convergent sequences with the same limit. Jan 19, 2020 at 18:10