For $(X, M, \mu)$, if $\mu$ is $\sigma$-finite, then does it imply that $\exists E_1 \subset E_2 \subset \cdots \text{ with } \cup_i E_i = X$ and $\mu(E_i) < \infty$?
In my opinion, the statement isn't right. It's equivalent to saying it's finite measure? Since with this statement, by continuity from below, $\mu(X) = \mu(\cup_i E_i) = \lim_i \mu(E_i) < \infty$
This is actually part of one solution given by our prof. similar to this for the problem but instead of assuming $E_i$ being partitions.
Best Answer
Consider on $\mathbb{R}$ with Lebesgue measure the sets $E_i=[-i,i]$.
The crux is that each individual set $E_i$ having finite measure does not mean that the limit of the measures is finite. This has little to do with measure theory, it's about real-valued sequences and their limits.