In this context "Cauchy integral" has the meaning you know.
It is a fact that if a function is bounded and Cauchy integrable over $[a,b]$, then it is also Riemann integrable over that interval.
It seems that there is no elementary proof of this theorem.
The proof in Kristensen, Poulsen, Reich A characterization of Riemann-Integrability, The American Mathematical Monthly, vol.69, No.6, pp. 498-505, (theorem 1), could be considered elementary because plays only with Riemann sums but is an indigestible game.
Note that there exist unbounded functions Cauchy integrable.
Also the use of regular partitions is enough to define Riemann integral.
See Jingcheng Tong Partitions of the interval in the definition of Riemann integral, Int. Journal of Math. Educ. in Sc. and Tech. 32 (2001), 788-793 (theorem 2).
I repeat that the use of only left (or right) Riemann sums with only regular partitions doesn't work.
There are several issues with the proposed definition.
When we partition the domain $D$ we usually require that the member of the partition are either not closed or allow them intersects by their boundaries of by subsets of zero measure. The reason for this is that a connected set cannot be a union of finitely many its pair-wise disjoint closed non-empty subsets. Moreover, by the Sierpiński theorem (see Appendix below), a continuum (that is, a connected compact space) is partitioned into countably many pair-wise disjoint closed subsets, then at most one member of the partition is non-empty.
The mesh $\|P\|$ of the partition defined as the measure of is largest member is bad, because then the Riemann sums fail to converge when $\|P\|$ tends to zero even for a continuous non-constant function defined on a square, because we can partition the square into thin strips with big oscillation of the function.
Thus, I think a usual measure of the mesh $\|P\|$ is the diameter of its largest member. For instance, such definitions of Riemann sums was proposed in the book [Fich], which I inherited from my mother. In it the integration domain $D$ was partitioned into finitely many parts by a family of curves in two-dimensional case (see Chapter 16, §1, 586) and of surfaces in three-dimensional case (see Chapter 18, §1, 643).
It is not very natural to define a Riemann integral based on Lebesgue measure. But if the members of partitions are so nice that they are Jordan measurable then they can be approximated (with respect to the measure) by bricks. In this case for a continuous function the limit of the Riemann sums exists and equals to the integral defined via the coverings by bricks (sub-hyperrectangles).
When the domain $D$ is not Jordan measurable then the Riemann integral fail to exists on it even for a non-zero constant function. This can happen even when $D$ is compact and connected. For instance, when $D$ is a cone over the fat Cantor set is not. Its inner Jordan measure vanishes, since its complement is dense; however, its outer Jordan measure does not vanish, since it cannot be less than its Lebesgue measure.
Appendix (from [Eng])
6.1.27. The Sierpiński theorem. If a continuum $X$ has a countable cover $\{X_i\}_{i=1}^\infty$ by pair-wise disjoint closed subsets, then at most one of the sets $X_i$ is non-empty.
Proof Let $X =\bigcup_{i=1}^\infty X_i$, where the sets $X_i$ are closed and $X_i\cap X_j =\varnothing$ whenever $i\ne j$;
assume that at least two of the sets $X_i$ are non-empty. From Lemma 6.1.26 it follows that there exists a decreasing sequence $C_1\supset C_2\supset\dots$ of continua contained in $X$ such that
$$C_i\cap X_i =\varnothing\mbox{ and }C_i\ne\varnothing\mbox{ for } i = 1, 2, \dots\tag{3}$$
The first part of (3) implies that $\left(\bigcap_{i=1}^\infty C_i\right)\cap\left(\bigcup_{i=1}^\infty X_i\right)=\varnothing$, i.e., that $\bigcap_{i=1}^\infty C_i=\varnothing$, and yet from the second part of (3) and compactness of $X$ it follows that
$\bigcap_{i=1}^\infty C_i\ne\varnothing$. $\square$
6.1.26. Lemma. If a continuum $X$ is covered by pair-wise disjoint closed sets $X_1, X_2,\dots$ of which at least two are non-empty, then for every $i$ there exists a continuum $C\subset X$ such that $C\cap X_i=\varnothing$ and at least two sets in the sequence $C\cap X_1, C\cap X_2,\dots$ are non-empty.
Proof. If $X_i =\varnothing$ we let $C = X$; thus we can assume that $X_i\ne\varnothing$. Take a $j\ne i$ such that $X_j\ne\varnothing$ and any disjoint open sets $U, V\subset X$ satisfying $X_i\subset U$ and $X_j\subset V$. Let $x$ be a point of $X_j$ and $C$ the component of $x$ in the subspace $\overline{V}$. Clearly, $C$ is a continuum, $C\cap X_i =\varnothing$ and $C\cap X_j\ne\varnothing$. Since $C\cap\operatorname{Fr}\overline{V}\ne\varnothing$, by virtue of the previous lemma, and since $X_j\subset \operatorname{Int}\overline{V}$, there exists a $k\ne j$ such that $C\cap X_k\ne\varnothing.$ $\square$
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. III, 4-th edition, Moscow: Nauka, 1966, (in Russian).
Best Answer
The idea is pretty clear, but we just have to write a lot, and use a lot of notation.
Definitions. A partition of $[a,b]$ is a finite sequence $\pi = (a_j)_{j=0}^J$ with $a=a_0 < a_1 < \cdots < a_J = b$.
The norm of $\pi$ is $\|\pi\| = \max_{1 \le j \le J} |a_j-a_{j-1}|$.
A tagged partition of $[a,b]$ is a pair $(\pi,\gamma)$ where $\pi = (a_j)_{j=0}^J$ is a partition of $[a,b]$ and the finite squence $\gamma = (c_j)_{j=1}^J$ satisfies $a_{j-1} \le c_j \le a_j$ for all $j$.
Let $f : [a,b] \to \mathbb R$. Let $\pi = (a_j)_{j=0}^J$ be a partition of $[a,b]$.
The upper sum is $$ U(f,\pi) = \sum_{j=1}^J M_j(f)\;\cdot(a_j-a_{j-1})\qquad\text{where}\qquad M_j(f) = \sup \{f(x) : a_{j-1} \le x \le a_j\} . $$
The lower sum is $$ L(f,\pi) = \sum_{j=1}^J m_j(f)\;\cdot(a_j-a_{j-1})\qquad\text{where}\qquad m_j(f) = \inf \{f(x) : a_{j-1} \le x \le a_j\} . $$
Let $(\pi,\gamma)$ be a tagged partition. The Riemann sum is $$ R(f,\pi,\gamma) = \sum_{j=1}^J f(c_j)\cdot (a_j-a_{j-1}) $$
We say that $f$ is Riemann integrable, and its integral is $V$ iff
for every $\epsilon > 0$ there exists $\delta > 0$ such that for every tagged partition $(\pi,\gamma)$ of $[a,b]$, if $\|\pi\| < \delta$, then $$ \big |R(f,\pi,\gamma) - V \big| < \epsilon . $$
Alternatively (from a Cauchy criterion), $f$ is Riemann integrable if and only if: for every $\epsilon > 0$ there exists $\delta > 0$ such that for every partition $\pi$ of $[a,b]$, if $\|\pi\| < \delta$ then $$ U(f,\pi) - L(f,\pi) < \epsilon . $$ If so, then the integral $V$ is the unique number with $L(f,\pi) \le V \le U(f,\pi)$ for all partitions $\pi$ of $[a,b]$.
Theorem A
Let $f_1,f_1,\dots, f_K$ be functions on $[a,b]$ such that $f_1, f_2,\dots, f_K$ and the product $F = f_1 f_2\cdots f_K$ are all Riemann integrable.
Claim: for any $\epsilon > 0$ there exists $\delta > 0$ such that: if $\pi$ is a partition of $[a,b]$ and $\|\pi\| < \delta$ and $\gamma^{(k)}=(c_j^{(k)})_{j=1}^J$ , $1 \le k \le K$, are $K$ choices of tags so that for all $k$, the pair $(\pi,\gamma^{(k)})$ is a tagged partition of $[a,b]$, then $$ \left|U(F,\pi) - \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_K(c_j^{(K)})\cdot (a_j-a_{j-1})\right| < \epsilon \tag1$$ and $$ \left|\sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_K(c_j^{(K)})\cdot (a_j-a_{j-1})-L(F,\pi)\right| < \epsilon \tag2$$ and $$ U(F,\pi) - L(F,\pi) < \epsilon. \tag3$$
Case 1: All $f_k \ge 0$.
So, $\epsilon > 0$ is given. Since all the functions $f_k$ are Riemann integrable, they are bounded. There is a single bound $A > 0$ with $$ |f_k(x)|\le A \qquad\text{for}\qquad k=1,\dots,K,\qquad a\le x \le b . $$
Let $$\epsilon' = \frac{\epsilon}{K A^{K-1}}.$$ Since $f_1,\dots,f_K$ are all Riemann integrable, there is $\delta > 0$ so that for any partition $\pi = (a_j)_{j=0}^J$ of $[a,b]$, if $\|\pi\| < \delta$, then $$ U(f_k,\pi) - L(f_k,\pi) < \epsilon'\qquad\text{for } k=1,\dots,K . $$ Let $(\pi, \gamma^{(k)})$ be tagged partitions as above. Assume $\|\pi\| < \delta$. Write $\widetilde{M}_j = M_j(f_1)M_j(f_2)\cdots M_j(f_K)$ and $\widetilde{m}_j = m_j(f_1)m_j(f_2)\cdots m_j(f_K)$.
Write $$ \widetilde{U} = \sum_{j=1}^J \widetilde{M}_j\cdot(a_j-a_{j-1}),\qquad \widetilde{L} = \sum_{j=1}^J \widetilde{m}_j\cdot(a_j-a_{j-1}) $$ Claim 1: $\widetilde{U}-\widetilde{L} < \epsilon$
Claim 2: All three of $$ U(F,\pi), \quad L(F,\pi), \quad \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_J(c_j^{(K)})\cdot (a_j-a_{j-1}) $$ are between $\widetilde{U}$ and $\widetilde{L}$.
Consequently, we obtain $(1), (2), (3)$, as required.
Proof of Claim 1.
The difference $\widetilde{U}-\widetilde{L}$ is written as a sum of $L$ terms. First, fix $j$ between $1$ and $J$. Then \begin{align} \widetilde{M}_j &= M_j(f_1)M_j(f_2)M_j(f_3)\cdots M_j(f_K) \\ & \ge m_j(f_1)M_j(f_2)M_j(f_3)\cdots M_j(f_K) \\ & \ge m_j(f_1)m_j(f_2)M_j(f_3)\cdots M_j(f_K) \\ & \ge \cdots \\ & \ge m_j(f_1)m_j(f_2)\cdots m_n(f_K) = \widetilde{m}_j \end{align} From each row to the next, one $M_j$ changes to an $m_j$. Then \begin{align} \widetilde{M}_j - \widetilde{m}_j &= (M_j(f_1)-m_j(f_1)) M_j(f_2)M_j(f_3)\cdots M_j(f_K) \\ &+ m_j(f_1)(M_n(f_2)-m_j(f_2)))M_j(f_3)\cdots M_j(f_K) \\ &+ \cdots \\&+ m_j(f_1) m_j(f_2)\cdots m_j(f_{K-1})(M_j(f_K)-m_j(f_K)) \\ &\le A^{K-1}\sum_{k=1}^K (M_j(f_k) - m_j(f_k)) \end{align} Now multiply by $a_j-a_{j-1}$ and sum on $j$. The left side is $$ \sum_{j=1}^J (\widetilde{M}_j - \widetilde{m}_j)\cdot(a_j-a_{j-1}) =\widetilde{U} - \widetilde{L}. $$ The right side is \begin{align} A^{K-1}\sum_{k=1}^K &\sum_{j=1}^J (M_j(f_k) - m_j(f_k))\cdot(a_j-a_{j-1}) = A^{K-1}\sum_{k=1}^K \big(U(f_k,\pi) - L(f_k,\pi)\big) \\ & < A^{K-1} K \epsilon' =\epsilon \end{align} This completes the proof of Claim 1.
Proof of Claim 2.
To prove $\widetilde{U} \ge U(F,\pi) \ge \widetilde{L}$ note that $\widetilde{M}_j \ge M_j(F) \ge \widetilde{m}_j$. To prove $\widetilde{U} \ge L(F,\pi) \ge \widetilde{L}$ note that $\widetilde{M}_j \ge m_j(F) \ge \widetilde{m}_j$. To prove $$ \widetilde{U} \ge \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_J(c_j^{(K)})\cdot (a_j-a_{j-1}) \ge \widetilde{L} $$ note \begin{align} \widetilde{M}_j &= M_j(f_1)M_j(f_2)\cdots M_j(f_K) \\ &\ge f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_K(c_j^{(K)}) \\ &\ge m_j(f_1)m_j(f_2)\cdots m_j(f_K) = \widetilde{m}_j \end{align} multiply by $a_j-a_{j-1}$ and sum on $j$.
Case 2. arbitrary signs.
Write $f_k = f_k^+ - f_k^-$ in positive and negative parts. Then $f_1 f_2\cdots f_k$ is a linear combination of $2^K$ such sums with nonnegative functions. Both $$ \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_K(c_j^{(K)})\cdot (a_j-a_{j-1}) \qquad\text{and}\qquad \int_a^b f_1(x)f_2(x)\cdots f_K(x)\;dx $$ are linear combinations of $2^K$ terms of the same form, but involving only nonnegative functions. Theorem A for $f_1f_2\cdots f_k$ follows from those $2^k$ cases of Theorem A for nonnegative functions.