Generalising Riemann integral to functions with values in a Banach space I found the following idea for generalising the Riemann integral to functions with values in Banach spaces, but I get stuck when discontinuous functions come into play:
We fix an interval $ [a,b] $ on which we'll be integrating and a Banach space X. Define $ S[a,b] $ as the set of all step functions on [a,b] with values in X. We have a very obvious definition of Riemann integral for such a function, where we take values of the function on subintervals, multiplied by the length of the corresponding subinterval and sum all of those. The function $ I $ which to a step function $ f $ assings a value $ I(f) $ as described above is linear and continuous on $ S[a,b] $ as a subspace of the Banach space of bounded function from $ [a,b] $ to $ X $, so we can uniquely extend $ I $ to a linear operator on the closure of $ S[a,b] $. So $ I $ gives us a notion of integrability.
Continuous functions are clearly integrable according to the construction above, so all's fine up to this point. But for this to be a proper generalisation, we would need functions continuous almost-everywhere to be integrable(at least when $ X = \mathbb{R} $), so possible to approximate uniformly by step functions, which is not the case.
I thought about using topology of pointwise convergence instead of topology of uniform convergence induced by the typical norm on the space of bounded functions. But from what I've found, we can approximate almost-everywhere continuous functions with step functions almost-everywhere, but not everywhere, so even with this topology these functions might not be integrable.
Is there a simple way to fix this construction to make almost-everywhere continuous real-valued functions integrable? Or is there a different way to define the integral which will work better?
 A: Every regulated function $$f:[a,b] \rightarrow F$$ ( a function wich has a left and right limit in every point) can uniformly be approximated by step functions.
The integration of a regulated functions is defined as:
$$\int_{a}^{b}{f(x)dx} = \lim_{n \rightarrow \infty}\int_{a}^{b}{f_{n}(x)dx}$$
where $(f_{n})_{n \in \mathbb{N}}$ is any sequence of step functions wich converges uniformly to $f$. 
An example of a regulated function:
$$f: \mathbb{R} \rightarrow \mathbb{R}: x \rightarrow \left\{ \begin{array}{ll}
             q  & \text{if }  x=\frac{p}{q} \:,\: gcd(p,q)=1, x \in \mathbb{Q} \\
              0 & \text{if } x \in \mathbb{R}\backslash\mathbb{Q} \\
             \end{array}
   \right.$$
Wich is called the Thomae function.
A: I have have found a definition of Riemann-integrability for functions with  values in an arbitrary Banach space $E$ for which even Lebesgue’s criterion holds:
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 $f:\left[ a,b \right]\to E$ I define
$$\omega\left( f,P \right)=\sum_{i=1}^{n}\left( x_{i}-x_{i-1} \right)\cdot sup\left\{ \left\| f\left(s \right)-f\left( t \right) \right\|:s,t\in \left( x_{i-1},x_{i} \right) \right\}$$
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!
Surprisingly even Lebesgue’s criterion for Riemann-integrability holds: 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 in mind is a very useful extension of the well known one for real-valued functions.
