Lebesgue integrals hav the countable additivity with respect to the domains of integration. This property is also true with respect to the integrands assuming some additional conditions.
I don't know if the Rieman integrals also have countable additivity with respect to
1) Domain of integration ($I$ = union $I_n$, assume $I, I_n$ are non degenerated disjoint closed intervals): $$(Riemann) \int_If=\sum_{n=1}^\infty\int_{I_n} f$$ 2) Integrands: $$(Riemann) \int_I (\sum_{n=1}^\infty f_n) = \sum_{n=1}^\infty \int_I f_n$$
What happens when I assume that the series on the both sides converges, then are they equal?
For integrands there is no countable additivity:
Say that $\mathbb{Q} = \{q_1,q_2,q_3...\}$. Define $f_n(x) = 1$ when $x = q_n$ and $f_n(x) = 0$ otherwise.
$\int_0^1 f_n(x) dx = 0$ for every $n$.
$\sum_{n=1}^\infty f_n = I_\mathbb{Q}$ (indicator function) which is not Rieamn integrable at all.
For domain of integration the same problem can occur. If the first domain is $\{q_1\}$, the second $\{q_2\}$ and so forth, you will get $\mathbb{Q}$ as your domain and you can't have a Rieman integral over $\mathbb{Q}$