The basic way the one-dimensional Riemann integral is extended to multiple integrals over bounded domains $D \subset \mathbb{R}^n$ is as follows. We extend the function $f:D \to \mathbb{R}$ we want to integrate to all of $\mathbb{R}^n$, by defining $f(x)=0$ for all $x \in \mathbb{R}^n \setminus D$. We enclose $D$ in a hyperrectangle $H \supset D$, say $H=[a_1, b_1] \times [a_2, b_2] \cdots \times [a_n \times b_n]$. A partition $P$ of $H$ is an $n$-tuple $(I_1, I_2, …, I_n)$ where each $I_i=\{S_{i,0}, S_{i,1}, \cdots, S_{i,k_i}\}$ $(k_i \geq 1)$ is a usual one-dimensional partition of $[a_i, b_i]$ into disjoint subintervals $S_{i,j}$. Each such partition partitions $H$ into various sub-hyperrectangles, and a Riemann sum can be defined in the usual way by choosing "tags" in each such sub-hyperrectangle. If there is a meaningful limit that all such Riemann sums approach, independent of the choice of "tags", as the maximal subrectangle hyperrarea of the partitions approaches $0$, we say $f$ is Riemann integrable on $D$, and the value of the limit is the value of the integral.
Problem: can this be defined without hyperrectangles?
Somehow, I don't feel that they are truly needed here, and any partitions of $D$ will do. Specifically, I would propose the following. We assume that $D$ is a "nice", closed and bounded region (perhaps homeomorphic to the closed $n$-ball, but the answerer can propose any reasonable definition of "niceness"). A partition of $D$ is a finite collection of disjoint, nonempty, compact (hence Lebesgue measurable), connected sets $\{S_i\}$ whose union is $D$, and the norm $||P||$ of $P$ is defined to be $\max_{i} \mu(S_i)$ where $\mu$ is the $n$-dimensional Lebesgue measure. We define Riemann sums as usual, namely sums of the form $\sum_{i} f(t_i) \mu(S_i)$, $t_i \in S_i$, and if a limit $\ell$ is approached as $||P|| \to 0$, independent of the choice of tags, we say $f$ is Riemann integrable on $D$ with integral $\ell$.
The basic concern I have with this, however, is that now we're working with much more general partitions, and hence there is a possibility of pathological situations. In particular, there may be functions integrable with respect to the original definition, but not integrable with respect to the second definition.
Question: does this work, and is it equivalent?
Best Answer
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).