Let $f:[a,b] \to \mathbb{R}$ be a (Riemann) integrable function, which is nowhere locally constant (i.e $f$ is not constant on any open subinterval).

Let $P_n$ be a sequence of partitions of $[a,b]$. Suppose that the upper-Darboux sums $\lim_{n \to \infty}U(f,P_n)=\int_a^b f(x)dx$. Is it true that $\lambda(P_n) \to 0$?

($\lambda(P_n)$ is the parameter of the partition, i.e the maximal length of a subinterval in the partition).

One could probably replace the upper-sums with lower-sums or even arbitrary Riemann sum, to get variants of the question...

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    $\begingroup$ I think that this isn't true for arbitrary Riemann sums. For example take $f:[-1,1] \rightarrow \mathbb{R}$ to be defined by $f(x)=x$. $f$ is Riemann integrable and nowhere locally constant. However, if we take $P_n$ to be $\{-1,0,1\}$ for all $n$. We can choose $x_1 \in [-1,0]$ to be $-1$ and $x_2 \in [0,1]$ to be $1$. Then the Riemann sum is $1f(-1) + 1f(1) = 0 = \int_a^b f(x)dx$ for all $n$. While $\lambda(P_n) = 1$ for all $n$. This counterexample doesn't work when upper or lower sums are used instead. I look forward to seeing a full answer to this question. $\endgroup$ May 14, 2017 at 11:44
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    $\begingroup$ As proved in this answer math.stackexchange.com/a/2047959/72031 we can only conclude that for any sequence $P_{n} $ with $\lambda(P_{n}) \to 0$ we have the fact that $U(f, P_{n}) $ tends to upper Darboux integral for $f$. The convergence can't imply that $\lambda(P_{n}) \to 0$. $\endgroup$
    – Paramanand Singh
    May 14, 2017 at 12:11
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    $\begingroup$ The same linked answer shows that the limit is valid under refinement also and not just when parameter tends to $0$. So there is no a priori reason to suspect that convergence necessarily implies that parameter tends to $0$. $\endgroup$
    – Paramanand Singh
    May 14, 2017 at 12:15
  • $\begingroup$ @ParamanandSingh Thanks. Your other answer is very interesting. $\endgroup$ May 15, 2017 at 7:36

2 Answers 2


I almost wrote down a proof of your statement, but then realize that the negative of the Thomae's function $f$ is a counterexample: It's nowhere locally constant, and

$$\sup_{x\in [c,d]} f(x) = 0$$

for all intervals $[c,d]$. Thus

$$U(f,P_n) = 0 = \int_0^1 f$$

for all partition $P_n$.

  • $\begingroup$ Thanks! Maybe there is a chance your attempted proof would work if we assume in addition that $f$ is continuous? or monotone? (Then our geometric intuition should reflect more reliably the mathematical reality, I hope:) $\endgroup$ May 15, 2017 at 7:54
  • $\begingroup$ Yes, the attempted proof works for both continuous and monotone case. One basically use a contradiction argument and show that there is a FIXED interval $[c,d]$ contains in all the partitions $P_n$. @AsafShachar $\endgroup$
    – user99914
    May 15, 2017 at 8:31
  • $\begingroup$ Hmmm... sounds interesting. I see why you can obtain a (sub)sequence of partitions $P_n$ with parameter $\lambda(P_n) \ge \epsilon > 0$. I can also see why there is a fixed common interval to a (sub)sequence of the $P_n$, but my argument for that is cumbersome. Do you have a slick argument for that? $\endgroup$ May 15, 2017 at 9:27
  • $\begingroup$ probably I do not have a slick argument. I posted it anyway. @AsafShachar $\endgroup$
    – user99914
    May 16, 2017 at 6:01
  • $\begingroup$ Thanks. Your argument is slick enough:) (better than mine, anyway). $\endgroup$ May 16, 2017 at 6:35

As suggested, we show that the statement is true when $f$ is either continuous or monotone. Indeed it is sufficient to assume $f$ satisfies

$$\tag{1} (d-c)\sup_{x\in [c,d]} f(x) > \int_c^d f(x) dx.$$

for all subintervals $[c,d]\subset [a,b], c<d$. Clearly (1) is satisfied when $f$ is continuous or monotone.

Next we argue by contradiction (we write "btsin" as "by taking a subsequence if necessary").

Assume the contrary that $\lambda(P_n)$ does not converge to zero. btsin, assume $\lambda(P_n) \ge \epsilon_0$ for some $\epsilon_0$. Thus there is a subinterval $[a_n,b_n]$ in the partition $P_n$ with $b_n - a_n \ge \epsilon_0$. btsin, assume $a_n \to a_\infty \in [a,b]$. Then btsin, there are $[c,d]$ with $$a_n \le c < d\le b_n$$ Now consider the refinement $\bar{P_n}$ of $P_n$ by adding $c,d$. Then $\lambda (\bar P_n) \ge d-c$ and

$$U(f,P_n) \ge U(f,\bar P_n) \ge \int_a^b f(x)dx \Rightarrow U(f, \bar P_n) \to \int_a^b f(x)dx. $$

However, by (1), we have

$$\begin{split} U(f,\bar P_n) &= U(f, P_n \cap [a,c]) + (d-c)\sup_{x\in [c,d]}f(x) + U(f, P_n\cap [d,b]) \\ &> \int_a^c f(x) dx + \int_c^d f(x) dx + \delta + \int_c^d f(x)dx \\ &= \int_a^b f(x)dx + \delta. \end{split}$$

Here $\delta$ is a positive number that can be added to the RHS of (1) and still preserve the inequality. It does not depend on $n$.

Thus we arrive at a contradiction and the statement is proved.


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