Why doesn't exist a Cousin's lemma for left-tagged partitions?

Question

I am thinking of the possible validity of a statement like this:

given any positive mapping $\delta$ on $[a,b]$, there exists a partition $\, a=a_0<a_1<\cdots<a_n=b \,$ of $ \,[a,b] \,$ so that $$a_r - a_{r-1} < \delta (a_{r-1})\qquad(r=1,\ldots,n)$$ Jean Mawhin says it is not true at the end of his contribution in Fulvia Skof (Ed.) Giuseppe Peano between Mathematics and Logic (2011).

I cannot understand if the reason is banal or not, so I need your help to know why it is not true. Thank you in advance.

(the question is tied to the need of the so-called straddle lemma in the introduction of the Henstock-Kurzweil integral)

Answer 1

A counterexample should illustrate. Consider the interval $[0,2]$ define your function $\delta:[0,2]\to R^+$ by requiring that $\delta(x)< 1-x$ for all $x\in [0,1]$ and any other values you wish elsewhere. If an interval $[a_{r-1},a_r]$ in the first half of the interval satisfies $a_r-a_{r-1}<\delta(a_{r-1})$ then $a_r< a_{r-1} + \delta(a_{r-1})<1$. So you never reach the point $1$ in your partition.

But that does not mean that there is no left-tagged partitions possible if you make suitable adjustments. John Hagood came up with a very nice idea for this and we used it to integrate Dini derivatives of continuous functions. If you care to pursue see

MR2202919 (2006i:26010) Hagood, John W. ; Thomson, Brian S.
Recovering a function from a Dini derivative. Amer. Math. Monthly 113 (2006), no. 1, 34--46. Download link here.

Answer 2

The answer to Tony's question is in the negative, but with a slight adjustment we can do this more positively. Here is Tony's motivation, quoting from his previous post on the subject:

Peano's Exercise: Let $f$ be defined on $[a,b]$ and there differentiable. Show that for every $ϵ>0 $ there exists a partition $ a=a_0<a_1<...<a_n=b $ of $[a,b] $ so that$$ \left|\frac{f(a_{i+1})-f(a_i)}{a_{i+1}-a_i}-f'(a_i) \right|<\epsilon\qquad(i=0,...,n-1)$$

This proof was left as an exercise to the Belgian mathematician P. Gilbert by G. Peano in a quarrel (1884) about a mistake made by C. Jordan in his Cours d'analyse vol.1 (1882). According to Peano, Jordan's proof of the mean value inequality theorem presented a fallacious argument: Gilbert did not agree.

Naturally, Tony who is aware and perhaps even a fan of the Cousin Covering lemma thought that there must be a proof that uses that lemma or something similar. Certainly it looks so. But we need a modification.

Piccolo-Cousin Covering Lemma: Let $\cal C$ be a collection of closed subintervals of $[a,b]$ with the following two properties:

(a) For every $a\leq x < b$ there is a $\delta(x)>0$ so that $[x,x+t] \in {\cal C}$ for all $0<t<\delta(x)$.

(b) For every $a<x\leq b$ there is at least one interval $[c,x]\in {\cal C}$.

Then ${\cal C}$ contains a partition of $[a,b]$.

We call it the "Piccolo" lemma in honor of Tony or, perhaps, because we are thinking of this a "little" version of the Cousin lemma, little (piccolo) because it assumes so little about what is happening on the left at each point. Of course, if you assume less you get less: here we have a partition of $[a,b]$ but not of every subinterval of $[a,b]$.

Solution of Peano's Exercise using Piccolo coverings: Define the collection $${\cal C}=\left\{[u,v]: \left|\frac{f(v)-f(u)}{v-u} - f'(u)\right| <\epsilon \right\}$$ Just verify that $\cal C$ satisfies the two conditions of the lemma. The condition (a) is quite evident. The condition (b) follows from the mean-value theorem but is elementary (Tony's other post shows how). By the lemma there is a partition that satisfies Peano's requirements.

As Fermat once said (roughly), I believe I have valid proofs of these statements but StackExchange allows too few characters to add them here.