TITLE: Existance of such sets that $X=\cup_{j=1}^{\infty}A_j$ on $\sigma$-finite space with premeasure.
QUESTION [0 upvotes]: I am taking a measure course this semester and stumbled into the following problem.
Let $(X, \mathfrak{A}, \mu)$ be a $\sigma$-finite space with premeasure. Prove that there exists such sets that
$$A_j \in \mathfrak{A},\quad \mu(A_j)<\infty,\quad j=1,2,..., \quad A_1 \subset A_2 \subset A_3..., \text{ such that } X=\cup_{j=1}^{\infty}A_j.$$
It is relevantly easy for me to show that if instead of subsets we are dealing with pairwise disjoint sets. Unfortunately, at this moment, I am a bit stuck to construct the suitable sets to show the exact same property here.
My idea is to use the property that subsets can be constructed by pairwise disjoint sets such that $A_j = \cup_{j=1}^{\infty}B_j$. Is this even right and one possibility to continue or is there an easier way?

REPLY [1 votes]: $\sigma$-finiteness means there are countably many sets $X_j \in \mathfrak{A}, j \in \mathbb{N}$ such that
$$ X = \bigcup_{j = 1}^\infty X_j \quad \text{and} \quad \mu(X_j) < \infty, \forall j \in \mathbb{N} $$
Just set
$$ A_j = \bigcup_{i=1}^j X_i $$
and the result follows.
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