Lebesgue's criterion for Riemann-integrability of Banach-space-valued functions? Lebesgue's criterion for Riemann-integrability says that a function $f:[a,b]\to\mathbb{R}$ is Riemann-integrable iff it is bounded and the set of points at which it is not continuous has measure zero.
This can be easily extended to functions with values in $\mathbb{R}^n$. However, is there an equivalent criterion for functions taking values in a Banach space or, making some more assumptions, in a (separable) Hilbert space?

To avoid confusion: The Riemann integral for Banach space valued functions is (obviously) not defined via upper and lower sums, but as described in http://en.wikipedia.org/wiki/Riemann_integral#Definition for real-valued functions
 A: 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
$$
 \left\|\sum_{i=1}^n\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\|
 \le \alpha(b-a)+2M\alpha < \epsilon .
$$
A: Whether it is true that Lebesgue’s criterion for Riemann-integrability also holds for functions with  values in a Banach space $E$ depends strongly on the way the definition for real-valued functions is extended to this wider class. I am thinking of the following: As usual, by a partition of the interval $\left[ a,b \right]$ I understand a finite set $P=\left\{ x_{0}, x_{1}\cdot \cdot \cdot \cdot , x_{n-1}, x_{n} \right\}$ such that $a=x_{0}\lt x_{1}\cdot \cdot \cdot \cdot \lt x_{n-1}\lt x_{n}=b$. For such a partition and any bounded function $f:\left[ a,b \right]\to E$ the weighted sum of oscillations of the function on the sub-intervals of the partition is defined by
$$\omega\left( f,P \right)=\sum_{i=1}^{n}\left( x_{i}-x_{i-1} \right)\cdot sup\left\{ \left\| f\left(s \right)-f\left( t \right) \right\|:s,t\in \left( x_{i-1},x_{i} \right) \right\}$$
Now by definition a bounded function $f:\left[ a,b \right]\to E$ is called Riemann-integrable if the following holds:
(D)
For every $\epsilon\gt 0$ there exists a partition $P=\left\{ x_{0}, x_{1}\cdot \cdot \cdot \cdot , x_{n-1}, x_{n} \right\}$ such that $\omega\left( f,P \right)\lt \epsilon$.
Why is this a natural extension of the definition for real-valued functions? On $\mathbb{R}$ regarded as a one-dimensional real Banach space the norm is just the absolute value, and with the simple fact that for a bounded subset $B\subseteq \mathbb{R}$ we have $sup\left\{ \left| s-t \right|:s,t\in B \right\}=sup\: B-inf\: B$, we may write
$$\omega\left( f,P \right)=\sum_{i=1}^{n}\left( x_{i}-x_{i-1} \right)\cdot sup\left\{ f\left( s \right):s\in \left( x_{i-1},x_{i} \right) \right\}-
\sum_{i=1}^{n}\left( x_{i}-x_{i-1} \right)\cdot inf\left\{ f\left( s \right):s\in \left( x_{i-1},x_{i} \right) \right\}$$
Now it is easy to see that definition (D) is actually equivalent to Darboux’s definition of Riemann-integrability in the case the Banach space is $\mathbb{R}$: it amounts to the equality of the upper and lower integral of $f$.
What about the value of the integral if $f:\left[ a,b \right]\to E$ is Riemann-integrable according to definition (D) in the general case? For a given partition $P=\left\{ x_{0}, x_{1}\cdot \cdot \cdot \cdot , x_{n-1}, x_{n} \right\}$ I call every sum of the kind $\sum_{i=1}^{n}\left( x_{i}-x_{i-1} \right)\cdot f\left( s_{i} \right)$ with $s_{i}\in \left( x_{i-1},x_{i} \right)$ a Riemann sum belonging to partition $P$. According to definition (D) we may choose for every $n\in \mathbb{N}$ a partition $P_{n}$ such that $\omega\left( f,P_{n} \right)\lt \frac{1}{n}$. If then for every $n\in \mathbb{N}$ any Riemann sum $r_{n}$ belonging to the partition $P_{n}$ is selected, it is not hard to show using (D) again that $\left( r_{n} \right)_{n\in \mathbb{N}}$ is a Cauchy-sequence in $E$ and thus converges to some $r\in E$. This limit does actually not depend on the special sequence of partitions $P_{n}$ and the Riemann sums $r_{n}$ belonging to them, once again due to (D). Thus the value of the integral is well defined by this limit $r\in E$.
It is not hard to verify that this definition of the Riemann-integral preserves all the good properties we know for real-valued functions: the set of all Riemann-integrable functions in this sense form a vector subspace of the set of all bounded functions into $E$ and the integral is a linear mapping from this space into $E$. Furthermore, if $f:\left[ a,b \right]\to E$ is Riemann-integrable then so is the real-valued function $\left\| f\left( \cdot \right) \right\|$ and the inequality $\left\| \int_{a}^{b}f \right\|\leqslant \int_{a}^{b}\left\| f \right\|$ holds, which is very important. All this works for an arbitrary Banach space $E$, no additional properties whatsoever of this space are required!
And now comes the surprise: Even Lebesgue’s criterion holds for my definition of Riemann-integrability: For a bounded function to be Riemann-integrable it is necessary and sufficient that the function is continuous almost everywhere. I have a nice proof for this which directly shows this equivalence; it is not needed to treat the sufficiency and the necessity parts separately.
So it seems that the definition of Riemann-integrability I have given above is the “right” extension of the well known one for real-valued functions compared to the definition given in the answer above which is implied by my definition but not vice versa, i.e. these two definitions are not equivalent.
