How does one integrate a function with values in a general vector space?

Answer 1:
Pick a basis, and integrate in each coordinate.

Answer 2:
Recall that one way to think of integrating functions is:
push forward the measure to the target space,
and take the relevant sum/expectation (integrate "x" over the vector space,
w/r/t that measure).
So the idea that comes from this is:
divide the -vector space- into pieces, P_i.
Now you have the measure of each of these parts a_i = u(P_i),
and the actual integral is
a_i x_i^*,
where x_i^* is "the average value on that piece"
  (the -correct- choice)
[this is a rather Riemannian way of putting it]
...but anyway you can bound the -possibilities-:
the -possible- integrals, given a partition, are
$\set{\sum a_i x_i | x_i \in P_i}$

In the 1-dimensional case, if you take the pieces to be intervals,
this gives -exactly- lower sums and upper sums
(the possible integrals are exactly the interval defined
 by these lower and upper sums).

Oh cool -- and now my remark about using compactness to prove
that intergrals exist actually -is- relevant:
given any finite number of partitions (2 is enough, by induction),
one can take the common refinement, which has a non-empty set of possible
integrals
(and is closed, if you feel like it)
...and thus the overall integral (intersection over all partitions)
is non-empty; the function has a unique, "well-defined" integral
iff this is a single point.
...and this works on any locally compact topological vector space?
[Yay: 2006-Aug: I finally understand 1st year (grad) analysis.]

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Ok, so Jordan's big idea was to see integration not as partitioning the -domain- but as partitioning the -target-
  (still "into intervals")
...and then measuring how much falls in that part of the target
by measuring the pullback.
This is -exactly- a round about way of saying:
"push forward the measure and take expectation".
[Indeed, "push forward" is more intuitive, albiet technically less direct,
 since it's a shrike kinda map (is it?)]

NB: one can -alternately- say: "You are simply doing Darboux,
but cutting up the -domain- into measurable sets, not just intervals".
What's the connection?
If you cut up the -target- into intervals,
and then -pull back-, the resulting partition of the -domain-
(recall that \cP^* f is mono, so partitions pull back to partitions)
is into -measureable- sets.

[And the asymmetry between measure spaces (domains) and vector spaces
(targets) explains the wacky asymmetric definition of a measurable function.]