Why every localizable measure space is semifinite measure space?

Question

Let $(X,\mathcal{E},\mu)$ be a measure space.

We recall the following two definitions:

• $(X,\mu)$ is called semifinite measure space if for each $E \in \mathcal{E}$ with $\mu(E) = \infty$ , there exists $F \subset E$ and $F \in \mathcal{E}$ and $0 < \mu(F) < \infty$.

• $(X,\mu)$ is called localizable measure space if it can be partitioned into a (possibly uncountable) family of measurable subsets $X_{\lambda}$ such that

(i) $\mu(X_{\lambda})< \infty$ for all $\lambda$.

(ii) a subset $A\subseteq X$ is measurable if and only if $A\cap X_{\lambda}$ is measurable for all $\lambda$.

(iii) $\mu(A) =\sum_{\lambda}\mu(A\cap X_{\lambda})$ for every measurable $A\subseteq X$.

For more details about localizable measure space one can see here.

Why every localizable measure space is semifinite measure space?

Answer 1

I think it's pretty clear: let $E \in \mathcal{E}$ such that $\mu(E) = \infty$. Let $X_\lambda, \lambda \in \Lambda$ be the partition of $X$ that witnesses localisability. If $\mu(E \cap X_\lambda) = 0$ for all $\lambda \in \Lambda$ then by (iii) we would have that $\mu(E) = 0$, which is not the case; so for some $\mu \in \Lambda$ we have $\mu(E \cap X_\mu) > 0$. But then $F:= E \cap X_\mu \subset E$ is as required: it's measurable (as $X_\mu$ is) and $0 < \mu(F) \le \mu(X_\mu) < \infty$.

Answer 2

Measures called here localizable are called frequently strictly localizable. The name "localizable" is used for those measures such that any collection $\{A\}$ of measurable sets admits a least upper bound, i.e. there exists a measurable set $F$ such that $A\setminus F$ has measure zero and if $G$ is another set with this property, then $F \setminus G$ has measure zero. The two classes of measures are different by a result due to Fremlin, if I am not wrong.