Lebesgue integral and sums

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.

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 $\sum n^{-2}$.