Show if $f_n$ are non negative measurable functions: $$\int (\sum_{n=1}^\infty f_n) d\mu = \sum_{n=1}^\infty \int f_n d\mu$$
Does this not just follow from the theorem for two additivity?
Say $\int (\sum_{n=1}^N f_n) d\mu = \int (\sum_{n=1}^{N-1} f_n) + f_N \mbox{ }d\mu = \int (\sum_{n=1}^{N-1} f_n) d\mu + \int f_N \mbox{ }d\mu$ and continue this onwards and take the limit as $N\rightarrow \infty$?
On
Since $f_{n} \geq 0$ by assumption, if we let $g_{m} = \sum \limits_{n = 1}^{m} f_{n}$, we have that $0 \leq g_{m} \leq g_{m + 1}$ for all $m$ (i.e., {$g_{m}$} is a monotonically increasing sequence of measurable functions), so by the monotone convergence theorem, we have that $$ \int \lim \limits_{m \to \infty} g_{m} = \lim \limits_{m \to \infty} \int g_{m}$$
But $\lim \limits_{m \to \infty} g_{m} = \sum \limits_{n = 1}^{\infty} f_{n}$, so the equality above is equivalent to $$\int \sum \limits_{n =1 }^{\infty} f_{n} = \lim \limits_{m \to \infty} \int \sum \limits_{n = 1}^{m} f_{n} $$
Since integration distributes over finite sums, the right hand side equals $ \lim \limits_{m \to \infty} \sum \limits_{n = 1}^{m} \int f_{n} = \sum \limits_{n = 1}^{\infty} \int f_{n}$.
Hint: Let $F_N(s)=\sum_{n=1}^N f_n$ and then use Levi's Theorem.