There has already been some discussion on this topic. However my question is about a specific solution to this problem and for the benefit of readers I think it is better to add some context (even though it means repetition of some stuff mentioned in the linked question).
In what follows $f$ is a function of type $f:[a, b]\to\mathbb{R}$ and $f$ is bounded. A partition $P$ of $[a, b]$ is a set of type $$P = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\}$$ where $$a = x_{0} < x_{1} < x_{2} < \dots < x_{n} = b$$ The norm $||P||$ of partition $P$ is defined by $||P|| = \max_{i = 1}^{n}(x_{i} - x_{i - 1})$. We define the following sums for $f$ over $P$ \begin{align} C(f, P) &= \sum_{i = 1}^{n}f(x_{i - 1})(x_{i} - x_{i - 1})\notag\\ S(f, P) &= \sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1})\notag\\ U(f, P) &= \sum_{i = 1}^{n}M_{i}(x_{i} - x_{i - 1})\notag\\ L(f, P) &= \sum_{i = 1}^{n}m_{i}(x_{i} - x_{i - 1})\notag \end{align} where $t_{i}$ are arbitrary points in $[x_{i - 1}, x_{i}]$ and $$M_{i} = \sup\,\{f(x)\mid x\in [x_{i - 1}, x_{i}]\},\,m_{i} = \inf\,\{f(x)\mid x\in [x_{i - 1}, x_{i}]\}$$ The sum $C(f, P)$ is called (left) Cauchy sum for $f$ over $P$. The Riemann sum $S(f, P)$ depends on choice of tags $t_{i}$ but this dependence in not shown in the notation and should be evident from the context. And finally $U(f, P), L(f, P)$ are upper and lower Darboux sums for $f$ over $P$.
Cauchy Integral: The function $f$ is said be said to be Cauchy integrable over $[a, b] $ with Cauchy integral $I$ if for every $\epsilon >0$ there is a number $\delta > 0$ such that $|C(f, P) - I| < \epsilon$ whenever $P$ is a partition of $[a, b]$ with $||P|| < \delta$.
A similar definition is available for Riemann integral if $C(f, P)$ is replaced by $S(f, P)$. Both these notions are equivalent and since every Cauchy sum is also a Riemann sum, the inference from Riemann to Cauchy is trivial. The converse appears to be hard and perhaps not popular enough to be seen in textbooks.
User Tony Piccolo in his answer gives three references for the proof that Cauchy integrability implies Riemann integrability.
It is the second proof from that answer which I want to discuss here (as other two proofs use somewhat complicated ideas and some very non-obvious tricks). This is provided as a hint that
Given any partition $P$ of $[a, b]$ and a number $\epsilon > 0$ there is a partition $Q\supseteq P$ of $[a, b]$ such that $C(f, Q) > U(f, P) - \epsilon$.
Using the counterpart equation $C(f, P) < L(f, P) + \epsilon$ we can easily show that difference $U(f, P) - L(f, P)$ can be made small if sums $C(f, P)$ tend to a finite limit and thus we get Riemann integrability (via Darboux integrability, also this link between Darboux and Riemann integral is popular and available in good textbooks).
Here are my questions:
It is easy to prove that we can choose tags $t_{i}$ such that $S(f, P) > U(f, P) - \epsilon$. We just have to choose tags so that $f(t_{i})$ is sufficiently near $M_{i}$. My hunch is that if we add the tags $t_{i}$ to $P$ we get a partition $Q\supseteq P$ and that is the needed partition which ensures $C(f, Q) > U(f, P) - k\epsilon$ where $k$ is some fixed positive constant. Is this correct? And if so how do we go about proving this?
Another doubt is whether the relation between $C(f, P)$ and $U(f, P)$ is valid in general? Or does it hold only for Cauchy integrable functions? My guess is that it holds only for Cauchy integrable functions. Is this correct?
Let $f(x) = x$ on the interval $[0,1]$ and $P = (0,1)$.
Given any refinement $Q = (x_0,x_1, \ldots, x_{n-1},x_n)$ we have, since $f$ is increasing,
$$U(f,P) - C(f,Q) = 1 - \sum_{k=1}^n x_{k-1}(x_k - x_{k-1}) > 1 - \int_0^1x \,dx = 1/2.$$
Take $\epsilon < 1/2$ and we see that the conjectured result cannot be true.