Let be a measure space (ie. is a set, a -algebra and a measure of sets of ), and let be a complex-valued function. We write , where and are functions (ie. real-valued functions on ). We assume that both and are integrable. Then we define (for a measurable ):
Now for integrals of functions with values in , we have theorems like Fubini's theorem or the dominated convergence theorem. We shall be able to carry them over to the complex situation in a trivial manner.
Let be a complex function. Then
exists if and only if the (real) integral
Assume first that
exists. Then, since and for real, both integrals
exist, and hence, by definition, the integral
Assume now that
exists. This means that
exist, and so does the integral
just by summing the two previous integrals. (Note that we may name the variable of integration arbitrarily.) But clearly, by the triangle inequality, we have , and thus the integral
Theorem 4.3 (Fubini):
Let be measurable (ie. ) and let be another measure space. If additionally, is measurable, then we have for that
provided that one of the three integrals
has a finite value.
Theorem 4.4 (Dominated convergence):
Assume that is a sequence of complex-valued functions converging pointwise to a function , then if there exists a function such that and also , then we will have
as . (It is perhaps more suitable to use this measure-theoretic notation.)
An exception to the general principle that theorems from measure theory regarding functions to functions is the monotone convergence theorem, since it depends on the order structure of the real numbers. The only way to do this which comes to my mind is the case where the real part and the imaginary part of a given sequence of functions converge in a monotonous way. This will be left for the exercises.
Let be a bounded and closed interval (ie. there exist two real numbers , , such that , , or ).
A curve in is simply a continuous function . A path is a curve, where the interval is the unit interval . It is called path since as moves along the interval , the corresponding point will move along the complex plane and travel a "path" from the point to the point .
A contour is a curve (but not necessarily a path), which has the property that it is piecewise smooth. That is, a map is a contour if and only if we have real numbers such that and the restriction (where ) is smooth.
The curves which we will use in this book will all be contours, although the smoothness assumption is a bit of an overkill; in fact, differentiable would have been sufficient. But most curves actually appearing in practice are smooth, and by now the definition of countours is standard, and can not be changed by one book doing it differently, forcing me to use the common one (since if I didn't do that, I'd either have to redefine it, which would create inconsistencies with other texts, or I'd have to invent a new word, but then a reader needing only one theorem will not be able to know what it is, and a person who learns complex analysis from this book will not be familiar with the usual notion of contours).
We now wish to find a way of computing what is indicated in the following motion picture:
That is, we have a curve and a continuous function, evaluate the function along the curve, form the surface enclosed by the curve and the function at the given point of the curve, iron the surface together and we then want to get the area.
Our strategy will be this: We will approximate the curve (let's call it , even though in the picture it's called ) by a piecewise linear curve, and make the pieces of the piecewise linear curve smaller and smaller, so that in the limit we get the original curve. However, in the special case of piecewise linear curves, we will be able to compute the area of the surface as described above. Then we will take the limit of this process and define this to be the area we want. NOTE that this process will require the curve to be differentiable and the function to be continuous.
Indeed, we shall assume that the domain of definition of is ; in the case of other interval boundaries, the construction will be completely analogous.