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.


  • $\begingroup$ How are you defining your $x_n$? $\endgroup$
    – Math1000
    Jan 16, 2020 at 22:20
  • $\begingroup$ You can prove more. See this answer $\endgroup$
    – Paramanand Singh
    Jan 19, 2020 at 7:39

1 Answer 1


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 $U(f)=L(f)$.

  • $\begingroup$ Could you explain more about why the second inequality is true? Its hard for me to see since you are taking a sup and a limsup simultaneously. $\endgroup$ Jan 17, 2020 at 0:44
  • $\begingroup$ You first take $\limsup$ on both sides, and then the infimum on the left. $\endgroup$ Jan 17, 2020 at 0:49
  • $\begingroup$ 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.) $\endgroup$ Jan 17, 2020 at 1:01
  • $\begingroup$ Please see the edit. $\endgroup$ Jan 17, 2020 at 2:01
  • $\begingroup$ @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. $\endgroup$
    – Philipp
    Jan 19, 2020 at 18:10

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