Let $(\Omega, \mathcal{A}, \mu)$ be a $\sigma$-finite measure space, i.e there exists a sequence of sets $\Omega^{(n)} \in \mathcal{A}$ such that $$\Omega = \bigcup_n\Omega^{(n)} \quad \text{and} \quad \mu\big(\Omega^{(n)}\big) < \infty $$
Prove that there exists finite measures $\mu_1, \mu_2, \mu_3,…$ on $(\Omega, \mathcal{A})$ such that $\sum_{n=1}^{\infty} \mu_n = \mu$ and that the converse of this statement is not true.
How would one go about proving this? I thought of somehow "decomposing" the measure, but couldn't explicit work it out.
Best Answer
First of all, put
$$B_n = \Omega^{(n)} \smallsetminus \bigcup_{k =1}^{n-1} \Omega^{(k)}.$$
Then, the $B_n$ are disjoint, $\mu(B_n) < \infty$ and $\Omega = \bigcup_n B_n$. If you define
$$\mu_n(A) = \mu(A \cap B_n) \quad \text{for all measurable sets $A \in \mathcal{A}$,}$$
it is easy to show that $$\mu = \sum_{n=1}^\infty \mu_n.$$