Let $\mathbf E$ be a Banach space, let $a<b$ be real numbers, let
$f : [a,b] \to \mathbf E$ be a function.
A partition $\pi$ of $[a,b]$ is a finite subset, $\{a,b\} \subseteq \pi \subset [a,b]$, usually written
in order: $a = x_0 < x_1 < \dots < x_n = b$. Tags for a partition
$\pi$ as above are points $t_i$ such that $x_{i-1} \le t_i \le x_i$
for $1 \le i \le n$.
Partition $\pi_1$ refines partition $\pi_2$
iff $\pi_1 \supseteq \pi_2$, remembering that a partition is
a finite set.
Definition. Let $f$ be as above, and let $\mathbf u \in \mathbf E$.
We say that $f$ is Riemann integrable on $[a,b]$ and $\mathbf{u}$
is its integral iff: for every $\epsilon > 0$, there is a
partition $\pi_0$ of $[a,b]$ such that for all refinements
$\pi=(x_i)_{i=0}^n$ of $\pi_0$ and all tags $(t_i)_{i=1}^n$ for $\pi$,
$$
\left\|\mathbf u - \sum_{i=1}^n f(t_i)\;(x_{i}-x_{i-1})\right\| < \epsilon.
$$
Lemma $f$ is integrable iff: for every $\epsilon > 0$
there is a partition $\pi = (x_i)_{i=0}^n$ such that
for any two choices $(t_i)_{i=1}^n, (s_i)_{i=1}^n$ of tags
for $\pi$, we have
$$
\left\|\sum_{i=1}^n \big(f(t_i)-f(s_i)\big)\;(x_{i}-x_{i-1})\right\| < \epsilon.
$$
Proof. Cauchy criterion.
Theorem. Let $f : [a,b] \to \mathbf E$ be bounded and
continuous except on a set $N\subseteq [a,b]$
of measure zero. Then $f$ is integrable.
Proof. Add $\{a,b\}$ to the null set $N$ to avoid special
cases for endpoints.
Let $\epsilon>0$. Say $f$ is bounded by $M$,
$\|f(x)\| \le M$. Let $\alpha > 0$ be so small that
$2M\alpha + \alpha(b-a) < \epsilon$. For an open interval
$(u,v)$ we say $f$ has oscillation at most $\alpha$ on $(u,v)$
if for all $x,y \in (u,v)$, $\|f(x)-f(y)\| \le \alpha$.
If $f$ is continuous at a point $s$, then there is an
inverval $(u,v)$ with rational endpoints, $s \in (u,v)$,
so that $f$ has oscillation at most $\alpha$ on $(u,v)$.
So there is a countable union of such intervals $(u,v)$
that contains $[a,b] \setminus N$, and thus has full measure.
So there is a finite list $(u_j,v_j)$ of intervals where
$f$ has oscillation at most $\alpha$, and their union has
measure greater than $b-a-\alpha$. Then there is a partition
$\pi = (x_i)_{i=0}^n$ of $[a,b]$ such that each subinterval
$[x_{i-1},x_i]$ from the partition either is contained in
an interval where $f$ has oscillation at most $\alpha$,
or is an "exceptional" interval. The total length of all
the exceptional intervals is ${}< \alpha$. Now
let $(t_i)$ and $(s_i)$ be two choices of tags for the
partition $\pi$. Now we must consider
$$
\left\|\sum_{i=1}^n\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\|
\le
\sum_{i=1}^n\left\|\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\| .
$$
Consider the term
$(f(t_i)-f(s_i))(x_i-x_{i-1})$. If the subinterval $[x_{i-1},x_i]$
is not exceptional, then
$$
\left\|\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\|
\le \alpha (x_{i}-x_{i-1}) ,
$$
so the total of all terms for non-exceptional intervals
is at most $\alpha (b-a)$. If the subinterval $[x_{i-1},x_i]$
is exceptional, then
$$
\left\|\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\|
\le 2M (x_{i}-x_{i-1}) ,
$$
so the total of all terms for exceptional intervals
is at most $2M\alpha$. Thus
$$
\left\|\sum_{i=1}^n\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\|
\le \alpha(b-a)+2M\alpha < \epsilon .
$$
The upper and lower Darboux integrals do not depend on the partitions, they are the infimum and supremum of all upper and lower Darboux sums for all possible partitions of your block. That is, $U(f)=\inf_{P}U(f,P)$ and $L(f)=\sup_{P}L(f,P)$. We know that for any pair of partitions $L(f,P)\leqslant U(f,P')$. If for some partition, $L(f,P)=U(f,P)$, then it follows that $f$ is integrable, for it will follow $L(f)=\sup_P L(f,P)=U(f,P)\geqslant \inf_P U(f,P)=U(f)$. Since we always have $L(f)\leqslant U(f)$, we have $L(f)=U(f)$ and $f$ is Riemann integrable.
Best Answer
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
$$\underline{\int_Q} \hat{f} = \overline{\int_Q} \hat{f}$$