If the Darboux sums converge to the integral, does the parameter of the partition tend to zero?

Question

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...

Answer 1

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 $P_n$.

Answer 2

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.