This is the classical definition of an integral. Here only given for functions f:R -> R.
Let be (a,b) a closed interval of R. Let d be a finite partition of (a,b) into closed intervals. Let |d| be the the length of the largest interval of d.
A step function on d is a function which is constant on the intervals of d.
The lower Riemann sums L(d) of d is the sum of (the values of a step function s in an interval of d multiplied with the length of the interval), where s is a step function on d with s < f and there is no step function t on d, t < f, with an x in (a,b), s(x) < t(x) < f(x).
The upper Riemann sums U(d) of d is defined analogous with s > f.
The Riemann integral of f exists iff
```  lim   U(d) =   lim  S(d)
|d|-> 0       |d|-> 0
```
and the Riemann integral is lim U(d).

Note: The Riemann integral is not defined on unbounded intervals. If you encounter such "integrals", then it's not Riemann or it's a limit of Riemann integrals. You might even have such effects on open sets, but I'm not sure if this always holds.
The Riemann integral is classical, but it has some severe weaknesses:
You can't take the limit inside unless you have uniform convergence.
You can only integrate "relatively" continuous functions. Finite sets of non-continuous points are allowed, you run in trouble with infinite sets. (However, sometimes the Riemann integral exists anyway.)
You can only integrate in Rn style spaces. (Might work on Banach spaces, too)

Specifically, the Riemann integral of a function f: [a,b] → R exists iff f is bounded and the set of points x at which f is discontinuous has measure 0.

The chief advantages of the Riemann integral are its extensions to the various improper integrals. The limit-based formulation is a particularly nice fit, as it uses the order-based nature of the Riemann integral. The Lebesgue integral, to give the most famous and commonly-used alternative, depends on a measure -- and loses many of the benefits of the complete total order we have on R.

Doing integrals in (infinite-dimensional) Banach spaces is a lot more difficult. Given a "box" of side h, it is not even possible to define a "volume" of the box (the volume of such a box of dimension d is hd for finite dimensions, so we'd get either or infinity for infinite dimension -- and neither will do).

You really need a measure to integrate, and then you might as well do a Lebesgue integral. Even the "box volume" problem above is really a problem defining a measure consistent with the norm.

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