Cauchy definite integral vs Riemann once again

Question

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.

Answer 1

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.

Answer 2

For any $\epsilon > 0$, choose $\delta > 0$ such that for any partition $P$ with norm $\|P\| < \delta$, the inequality $$\left| \sum_{i=1}^{n} f(x_{i-1}) \Delta x_i - I \right| < \frac{\epsilon}{10}$$ holds. Then take a positive integer $N > \max \left\{ \frac{5(b-a) \sup |f|}{\epsilon}, \frac{1}{\delta} \right\}$, and consider the partition $$P_0: x_k^0 = a + (b-a) \frac{k}{2N}, \quad k = 0,1,2,\dots,2N.$$ For any $t_i \in [x_{i-1}^0, x_i^0]$, $i=1,2,\dots,2N$, consider the subpartitions: $$ \begin{aligned} P_1: & \quad x_0^0, t_1, t_2, x_2^0, t_3, t_4, x_4^0, \dots, x_{2N}^0; \\ P_2: & \quad x_0^0, t_1, \quad x_2^0, t_3, \quad x_4^0, \dots, x_{2N}^0. \end{aligned} $$ We have $\|P_1\| < \delta$ and $\|P_2\| < \delta$, and by considering the difference of the Riemann sums corresponding to $P_1$ and $P_2$, we obtain: $$\left| \sum_{i=1}^{N} \bigl[ f(t_{2i-1}) - f(t_{2i}) \bigr] (x_{2i}^0 - t_{2i}) \right| < \frac{2\epsilon}{10}.$$ Similarly, consider the partitions: $$ \begin{aligned} P_3: & \quad x_0^0, x_1^0, t_2, t_3, x_3^0, t_4, t_5, x_5^0, \dots, x_{2N-1}^0, x_{2N}^0; \\ P_4: & \quad x_0^0, x_1^0, t_2, \quad x_3^0, t_4, \quad x_5^0, \dots, x_{2N-1}^0, x_{2N}^0. \end{aligned} $$ We have $\|P_3\| < \delta$ and $\|P_4\| < \delta$, and by considering the difference of the sums, we get: $$\left| \sum_{i=1}^{N-1} \bigl[ f(t_{2i}) - f(t_{2i+1}) \bigr] (x_{2i+1}^0 - t_{2i+1}) \right| < \frac{2\epsilon}{10}.$$ Adding the two inequalities, we obtain: $$\left| \sum_{i=2}^{2N} \bigl[ f(t_{i-1}) - f(t_i) \bigr] (x_i^0 - t_i) \right| < \frac{4\epsilon}{10}.$$ Consider the partition $P_5: x_0^0, t_1, t_2, \dots, t_{2N}, x_{2N}^0$. Since $\|P_5\| < \delta$, we have: $$\left| f(x_0^0) (t_1 - x_0^0) + \sum_{i=1}^{2N-1} f(t_i) (t_{i+1} - t_i) + f(t_{2N}) (x_{2N}^0 - t_{2N}) - I \right| < \frac{\epsilon}{10}.$$ Rewrite the sum from (3) as: $$\left| \sum_{i=1}^{2N-1} f(t_i) (x_{i+1}^0 - t_{i+1}) - \sum_{i=2}^{2N} f(t_i) (x_i^0 - t_i) \right| < \frac{4\epsilon}{10}$$ and add it to (4). After algebraic manipulation, we get: $$\left| \sum_{i=2}^{2N-1} f(t_i) (x_{i+1}^0 - x_i^0) + f(t_1) (x_2^0 - t_1) + f(x_0^0) (t_1 - x_0^0) - I \right| < \frac{5\epsilon}{10}.$$ Then, $$\left| \sum_{i=2}^{2N-1} f(t_i) (x_{i+1}^0 - x_i^0) - I \right| < \frac{5\epsilon}{10} + \left| f(t_1) (x_2^0 - t_1) + f(x_0^0) (t_1 - x_0^0) \right| \leq \frac{5\epsilon}{10} + \frac{3(b-a) \sup |f|}{2N} \leq \frac{5\epsilon}{10} + \frac{3\epsilon}{10} = \frac{8\epsilon}{10}.$$ Since $x_{i+1}^0 - x_i^0 = x_i^0 - x_{i-1}^0$ (uniform partition), we have: $$\left| \sum_{i=2}^{2N-1} f(t_i) (x_i^0 - x_{i-1}^0) - I \right| < \frac{8\epsilon}{10}.$$ Adding the first and last terms: $$\left| \sum_{i=1}^{2N} f(t_i) (x_i^0 - x_{i-1}^0) - I \right| < \frac{8\epsilon}{10} + \left| f(t_1) (x_1^0 - x_0^0) \right| + \left| f(t_{2N}) (x_{2N}^0 - x_{2N-1}^0) \right| \leq \frac{8\epsilon}{10} + \frac{(b-a) \sup |f|}{2N} + \frac{(b-a) \sup |f|}{2N} \leq \frac{8\epsilon}{10} + \frac{2\epsilon}{10} = \epsilon.$$ By the arbitrariness of $t_i$, the oscillation sum satisfies: $$\left| \sum_{i=1}^{2N} w_i(f) (x_i^0 - x_{i-1}^0) \right| < 2\epsilon,$$ where $w_i(f)$ is the oscillation of $f$ on $[x_{i-1}^0, x_i^0]$. This implies $f$ is Riemann integrable, and clearly $I = \int_a^b f(x) dx$.