TITLE: Why is Riemann integration used in complex analysis and not Lebesgue integration?
QUESTION [10 upvotes]: In the development of complex analysis you use Riemann integration and not Lebesgue integration to define line integrals. My questions are:
Are the theories developed the same? (i.e. does it not matter which integral you use in the development? Since all the functions usually involved are analytic or meromorphic can you use things such as analytic is equivalent to having a power series representation and uniform convergence within the radius of convergence to somehow show that the choice doesn't matter. I feel as if the function involved is analytic and has a finite radius of convergence this should be the case but I'm not so sure about what would happen if the function was meromorphic and/or has an infinite radius of convergence (Am I on the right track?))
If the theories developed are the same, does it become significantly easier to develop the theory with Riemann integration rather than Lebesgue integration.
If they are not the same, what are examples to show that show they are different?

REPLY [5 votes]: Mostly it doesn't matter. Typically, the paths of integration are piecewise continuously differentiable, and the integrated functions are continuous, so for these cases both theories are equivalent. The Riemann integral is simpler, and since not everybody hearing complex analysis may be familiar with the Lebesgue integral, using the Riemann integral may have pedagogical advantages.
There are cases where the two theories differ, though. For example when considering integrals like
$$\int_{-\infty}^\infty \frac{\sin x}{x}\,dx,$$
in the Riemann theory this exists as an ordinary improper integral, since
$$\lim_{\substack{R\to\infty\\S\to -\infty}} \int_S^R \frac{\sin x}{x}\,dx$$
exists when the bounds approach their limits independently. In the Lebesgue theory, the integral does not exist, one has to interpret it as a principal value integral or explicitly as a limit of integrals over bounded intervals.
That is not a big problem, however.
On the other side, if one considers Cauchy integrals of boundary values of holomorphic functions (Hardy spaces for example), one needs the Lebesgue integral (or some other integration theory that makes them well-defined), since the boundary values in general are not regular enough to be Riemann-integrable, but they are Lebesgue-integrable for interesting classes of functions.