Let $ \lambda: \mathcal{M} \longrightarrow [0,+\infty]$ be defined by
$$\lambda(E)= \int_{E} fd\mu \text{ for each } E \in \mathcal{M}. $$
Apply the Monotone Convergence Theorem to show that $\lambda$ is countably additive. That is, prove that if $\{E_{n}\}_{n=1}^{\infty}$ is a disjoint sequence of measurable sets, then
$$\lambda\bigg( \bigcup_{n=1}^{\infty} E_{n} \bigg) = \sum_{n=1}^{\infty} \lambda (E_{n}).$$
I have attempted this problem for a couple hours and have made insights in the theory but cannot solve this problem.
I know from $\mathbf{Lemma\ 2: }$ If $f \in L^{+}(X,\mathcal{M})$, then there exists a sequence $f_{n}$ of simple functions such that
$0 \leq f_{n}(x)\leq f_{n+1}(x)$ for $x \in X, n \in \mathbb{N}$ and $f_{n}$ converges pointwise to $f$.
Somehow I have to get $$\lim_{n\rightarrow \infty}\bigg(\sum_{i=1}^{n}\int_{E_{n}} f_{n}\ d\mu\bigg) =\lim_{n\rightarrow \infty}\bigg(\sum_{i=1}^{\infty}\int f_{n}\chi_{E_{n}}\bigg)=?\sum_{n=1}^{\infty} \int f\ d\mu=\sum_{n=1}^{\infty} \lambda (E_{n})$$
However I cannot the logic for =? and don't know how to get $\lim_{n\rightarrow \infty}\bigg(\sum_{i=1}^{n}\int_{E_{n}} f_{n}\ d\mu\bigg).$
Best Answer
This easily follows from (the union is disjoint):
$$\int_{\bigcup_n E_n} fd \mu = \sum_{n}\int_{E_n} f d \mu$$
which is an easy corollary from the monotone convergence theorem:
Indeed, $$\int_{\bigcup_n E_n} fd \mu = \int_X f I_{\bigcup_n E_n} d \mu = \int_X \sum_n fI_{E_n} d \mu = \sum_n \int _X f I_{E_n} d \mu = \sum_n \int_{E_n} fd\mu$$
where the monotone convergence theorem was used to exchange integral and sum (apply monotone convergence to the partial sums).
Note that the proposition that I made at the beginning of my answer is also true when $f$ also attains negative values.