TITLE: Cauchy definite integral vs Riemann once again
QUESTION [9 upvotes]: The problem of showing the equivalence between the Cauchy (definite) integral defined below and the Riemann integral is an exercise with hint in C.R.Rosentrater, Varieties of Integration, 2015.
I am interested in one direction, precisely the one contained in exercise 32 of chapter 3. Its converse, easy to prove, is contained in exercise 38 chapter 2.
In what follows $\mathcal P_L$ indicates the partition $\mathcal P$ tagged by the left endpoints of its subintervals.
The symbol $S_R$ indicates a Riemann sum.
A Cauchy sum is a Riemann sum referred to a partition tagged by the left endpoints.

It is clear that $f$ is Cauchy integrable over $[a,b]\,$ only if,$\,$ for any $\varepsilon>0$, we can find a $\delta>0$ so that, whenever $\mathcal P$ and $\mathcal Q$ are partitions of $[a,b]$ with $\|\mathcal P\|<\delta$ and $\|\mathcal Q\|<\delta$, one has $$|S_R(f,\mathcal P_L)-S_R(f,\mathcal Q_L)|<\varepsilon$$ Author suggests a proof by contraposition using a characterization of Riemann integrability, called there "height-width bounds theorem".
The statement to be proved is
If $f$ is bounded and Cauchy integrable over $[a,b]$, then $f$ is Riemann integrable over $[a,b]$
I restated what is written in exercise 32 because Cauchy integrability doesn't imply boundedness.
This problem appears here and there in the literature but, as far as I know, proofs are not at all elementary. Maybe Rosentrater's hint refers to an elementary one.
I don't know how to use the hint. I am waiting for a further hint at least.
Many thanks in advance.

REPLY [7 votes]: This one is really interesting +1. Based on your earlier answers / comments I did read some proofs of equivalence of Cauchy and Riemann integrals but all of them are so crafty and complicated that I can't explain them to someone else (which implies that I don't understand them myself).
Hint a) appears simple, but I am not sure how to use it with b). The idea is that we can choose $s_k, t_k$ in interval $[x_{k-1},x_{k}]$ so that their values are close to the inf and sup. Ideally we need to involve some arbitrary number $\epsilon>0$ and we can choose $s_{k}, t_{k} $ such that $$f(s_{k}) - \inf_{[x_{k-1},x_{k}]}f<\epsilon >\sup_{[x_{k-1},x_{k}]}f-f(t_k)$$ In fact for a suitable $\epsilon$ it is possible choose the points such that $|f(s_k) - f(t_k) |>h$. The fact that $s_{k+1}=t_{k+1}=x_{k+1}$ automatically ensures that $$\Delta s_{k} =s_{k+1}-s_{k}\geq x_{k+1}-x_{k}=\Delta x_{k} $$ Essentially based on two adjacent subintervals of $\mathcal{P} $ we create a single subinterval of partitions $\mathcal{P}_{l},\mathcal {P}_{u} $. Thus their norms can not exceed $\delta$. Also this exercise can be carried out over at least half of the subintervals where the oscillation of $f$ is greater than $h$. The subintervals of partition $\mathcal{P} $ where the oscillation of $f$ is less than $h$ and if they are not adjacent to those selected half of the subintervals (with oscillation greater than $h$) are retained as it in both partitions $\mathcal{P}_{u}, \mathcal{P}_{l} $. Thus their contribution to the Cauchy sum get canceled when we subtract Cauchy sums over these two partitions. The difference $$|C(\mathcal{P}_{u}) - C(\mathcal{P}_{l}) |$$ depends on those points $s_k, t_k$. The difference $$|f(x_{k-1})(s_k-x_{k-1})-f(x_{k-1})(t_k-x_{k-1})+f(s_k) (s_{k+1}-s_k)-f(t_k)(t_{k+1}-t_k)|$$ needs to be analyzed properly to complete the proof. The above difference should come out to be not less than $h\Delta x_{k} $ and then the difference between Cauchy sums is at least $hl/2$. This is the contradiction needed.