Countable additivity of Rieman integrals with respect to domains and integrands

Question

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?

Answer 1

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 $\mathbb{Q}$