Lebesgue integral and sums I'm not particularly well read on the Lebesgue integral, but I have heard that it permits a much wider class of functions, and in particular we can interchange integrals and limits more easily. However, I was wondering, are there any similar benefits with the Lebesgue integral when it comes to sums and integrals. For example, does it make it "easier" to interchange an infinite sum and integral sign?
 A: Yes. For example, if you have a sequence of nonnegative measurable functions $f_1, f_2, \ldots$, then it is always true that
$$\sum_{j=1}^\infty \int_{\Omega} f_j\, d\mu=\int_{\Omega} \sum_{j=1}^\infty f_j\, d\mu.$$
This is one of the consequences of the monotone convergence theorem and it is easy to prove: the sequence of the partial sums $\sum_1^n f_j(x)$ is monotone with respect to $n$ at almost all $x\in \Omega$.
You do not have an analogous result in Riemann theory. 
EDIT As anonymous points out in comments, you do have an analogous result for the Riemann integral, that is 
If a sequence $f_1, f_2 \ldots$ of nonnegative Riemann integrable functions is such that the series
$$\tag{!}\sum_{j=1}^\infty f_n(x)\ \text{is Riemann integrable},$$
then you can interchange sum and integral. 
The main problem with this theorem is the assumption marked with (!): in general, that series (even if convergent) needs not be Riemann integrable. 
This kind of phenomenon can be compared with the non-completeness of the rational system. Namely, a series of nonnegative rational numbers needs not converge in $\mathbb{Q}\cup\{+\infty\}$: take for example $\sum n^{-2}$. 
A: Yes, it does.
Case 1: If $\{f_n\}$ are nonnegative measurable functions, then:
$$
\int_X \sum_{n=1}^\infty f_n \, d\mu = \sum_{n=1}^\infty \int_X f_n \, d\mu
$$
In other words, you can always interchange an infinite sum and the integral sign when dealing with nonnegative functions. This is an immediate result of the monotone convergence theorem.
Case 2: If $\{f_n\}$ are complex measurable functions and:
$$
\sum_{n=1}^\infty \int |f_n| \, d\mu < \infty
$$
Then the series $\sum_{n=1}^\infty f_n(x)$ converges for almost all $x$, and we have:
$$
\int_X \sum_{n=1}^\infty f_n \, d\mu = \sum_{n=1}^\infty \int_X f_n \, d\mu
$$
This is a result of the dominated convergence theorem.
