Riemann vs. Darboux integrability for multiple integrals It is well-known that for bounded functions of one variable, Darboux integrability (via upper and lower sums) is equivalent to Riemann integrability (via Riemann sums.)
I tried to generalize this to functions of two variables, and I encountered some technical issues; consider, for example, a bounded function in two variables defined on a unit disk; the lower sum is built from the little squares inside the disk and infimum of function values, while the upper sum is built from the little squares intersecting the disk and supremum of function values; (Notice that there are more little squares for the upper sum than for the lower sum.) then one defines the Darboux integrability as the lower sum and upper sum being equal in some limit as the little squares become smaller; to make this sandwiching to work via monotone convergence theorem, one needs to assume that the function is nonnegative. A general bounded function is Darboux integrable if it is the difference between two nonnegative Darboux integrable functions.
As for the Riemann integrability, the Riemann sum is built from the little squares inside the disk and  function values at an arbitrary point in each little square. It is straightforward to show that Darboux integrability implies Riemann integrability. What I'm having trouble with is the converse: Does Riemann integrability imply Darboux integrability? I can prove this only if the function is nonnegative; since upper and lower sums are used for nonnegative functions only, the usual technique cannot be applied for signed functions. Is there a way around it? Or, could it be that Riemann integrability does not imply Darboux integrability for functions of more than one variable?
 A: The Riemann integral of $f$ over an  arbitrary bounded region $D \subset \mathbb{R}^2$ is defined as
$$\int_Df = \int_Q\hat{f},$$
$\hat{f}(x) =  f(x)$ for $x \in D$ and $\hat{f}(x) = 0$ for $x \notin D$, and $Q$ is any rectangle containing $D$. It is straightforward  to show that the choice for $Q$ is arbitrary.
Assume that $\hat{f}$ is Riemann integrable over $Q$. This means there exists a real number $I$ -- the integral-- such that for any $\epsilon > 0$ there exists a partition $P$ with subrectangles $\{R_1, \ldots, R_n\}$ such that
$$\tag{1}- \frac{\epsilon}{2} < S(P,f) - I < \frac{\epsilon}{2}$$
Here $S(P,f) = \sum_{j=1}^n f(t_j) \, vol(R_j)$ is a Riemann sum with any choice of intermediate points $t_j \in R_j$ for $j = 1, \ldots, n$.
Recall that upper and lower Darboux sums are given by
$$U(P,f) = \sum_{j=1}^n  M_j\, vol(R_j), \quad M_j := \sup_{x \in R_j}f(x),\\ L(P,f) = \sum_{j=1}^n  m_j\, vol(R_j), \quad m_j := \inf_{x \in R_j}f(x)$$
By the properties of supremum and infimum there exists $\xi_j, \eta_j \in R_j$ such that
$$M_j - \frac{\epsilon}{2 \,vol(Q)} < f(\xi_j) \leqslant M_j, \quad m_j  \leqslant f(\xi_j) < m_j + \frac{\epsilon}{2 \,vol(Q)}$$
Summing over all subrectangles of the partition we get
$$\tag{2}U(P,f) - \frac{\epsilon}{2} < \sum_{j=1}f(\xi_j) \, vol(R_j), \quad \sum_{j=1}f(\eta_j) \, vol(R_j)< L(P,f) + \frac{\epsilon}{2}$$
The sums in (2) are Riemann sums with intermediate points $\xi_j$ and $\eta_j$, respectively.  Applying the inequalities in (1), it follows that $U(P,f) - \frac{\epsilon}{2} < I + \frac{\epsilon}{2}$ and $I - \frac{\epsilon}{2} < L(P,f) - \frac{\epsilon}{2}$.
Hence,
$$I - \epsilon < L(P,f) \leqslant U(P,f) < I + \epsilon$$
There are several ways to proceed in showing that this proves Darboux integrability.  One is to note that the upper and lower Darboux integrals are sandwiched between any upper and lower Darboux sums, and, thus,
$$I- \epsilon < \underline{\int_Q} \hat{f} \leqslant \overline{\int_Q} \hat{f} < I + \epsilon$$
Since $\epsilon > 0$ can be chosen arbitrarily close to $0$, Darboux integrability holds with
$$\underline{\int_Q} \hat{f} = \overline{\int_Q} \hat{f}$$
A: Let $D\subseteq\mathbb{R}^2$ be a disk, and $f:D\to\mathbb{R}$ a Riemann integrable function. Let $S\subseteq\mathbb{R}^2$ be a square containing $D$, and extend $f$ by zero to $\hat{f}:S\to\mathbb{R}$. Then $\hat{f}$ is Riemann integrable, and so are $\hat{f}^+$ and $\hat{f}^-$. Note that $\hat{f}^\pm=\widehat{f^\pm}$. Hence, $f^\pm$ are Riemann integrable. Since $f^\pm$ are nonnegative, they are Darboux integrable. Since $f=f^+-f^-$, $f$ is Darboux integrable.
Alternatively, let $p$ be a positive real number such that $f+p\geq0$. Then $f+p$ is nonnegative and Riemann integrable, hence Darboux integrable. Since $p$ is Darboux integrable, so is $f=(f+p)-p$.
