# A $\sigma$-finite measure as infinite sum of finite measures

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.

$$B_n = \Omega^{(n)} \smallsetminus \bigcup_{k =1}^{n-1} \Omega^{(k)}.$$
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 $$\mu = \sum_{n=1}^\infty \mu_n.$$