I am trying to prove the following:

Suppose $ f:[a,b]\rightarrow\mathbb{R} $ is bounded. Then $ f $ is Riemann integrable if and only if for each sequence of marked partitions $\{P_n\}$ with $\{\mu(P_n)\}\rightarrow0$, the sequence $\{S(P_n,f)\}$ is convergent

,where $\mu(P)$ is the mesh of partition $P$ and $S(P,f)$ is the Riemann sum of $f$ over partition $P$.

My attempt at a solution:

Suppose for each sequence of marked partitions $\{P\}$ with $\{\mu(P_n)\}\rightarrow0,$ $ \{S(P_n,f)\}$ converges.

Let $\epsilon>0$ be given. Then there is an $A\in\mathbb{R}$ and $N\in\mathbb{N}$ such that when $n>N$, there exists $\delta$ such that $\mu(P_n)<\delta\implies |S(P_n,f)-A|<\epsilon$

Then, by the theorem provided by leo below, the existence of $A$ implies that $f$ is Riemann integrable.

Now suppose $f$ is integrable. Then given $\epsilon>0$, there exists $A\in\mathbb{R}$ such that there exists $\delta$ for which $\mu(P)<\delta\implies |S(P,f)-A|<\epsilon, \forall P$.

Then for each sequence each sequence of marked partitions $\{P\}$ with $\{\mu(P_n)\}\rightarrow0,$ $\mu(P_n)<\delta$.

Then, $|S(P_n,f)-A|<\epsilon$ which means that $\{S(P_n,f)\}$ converges to A. Also by the theorem below, $A=\int f dt$

  • 1
    $\begingroup$ What is $\mu(P_n)$? $\endgroup$
    – leo
    Feb 1, 2012 at 2:13
  • $\begingroup$ This is a notation I have seen for the mesh of a partition; this is the length of the longest interval in the partition. $\endgroup$ Feb 1, 2012 at 2:16
  • $\begingroup$ Sorry, clarified $\endgroup$ Feb 1, 2012 at 2:21
  • $\begingroup$ Use the definition that says: $f$ is integrable over $[a,b]$ if there exist a number $I$ such that for every $\epsilon\gt 0$ exist $\delta\gt 0$ s.t. if $P$ is partition of $[a,b]$ with $\mu(P)\lt\delta$, then $$|S(P_n,f)-I|\lt\epsilon.$$ Now I'm tired. I'll post an answer tomorrow if nobody does. $\endgroup$
    – leo
    Feb 1, 2012 at 3:10
  • 1
    $\begingroup$ Converges is enough: if two sequences converge to some different limits, interlace them to get a divergent sequence, which contradicts the hypothesis. $\endgroup$
    – Did
    Feb 25, 2012 at 12:02

1 Answer 1


Your statement should be an easy corollary of the Du bois-Reymond and Darboux integration theorem. The proof of the theorem is rather cumbersome. So here is a reference: the proof can be found in Analysis by Its History by Hairer and Wanner.

And by the way, you need to define what convergence as $\{\mu(P_n)\}\to 0$ and what marked partition mean.


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