TITLE: Prove that: Every $\sigma$-finite measure is semifinite.
QUESTION [18 upvotes]: I am trying to prove every $\sigma$-finite measure is semifinite. This is what I have tried:

Definition of $\sigma$-finiteness: Let $(X,\mathcal{M},\mu)$ is a measure space. Then, $ \mu$ is  $\sigma$-finite if $X = \bigcup_{i=1}^{\infty}E_i$ where $E_i \in \mathcal{M}$ and $\mu(E_i) < \infty$ for all $ j \in N$. (Real Analysis: Modern Techniques and Their Applications 2nd Edition by Foland).
Definition of semifiniteness: $\mu $ is simifinite if for each $E \in \mathcal{M}$ with $\mu(E) = \infty$ $\exists$ $F \subset E$ and $F \in \mathcal{M}$ and $0 < \mu(F) < \infty$.

So, take $A$ s.t. $\mu(A) = \infty$. We know $X \cap A = A$. Then, $A = A \cap \bigcup E_j$ hence $A = \bigcup E_j \cap A$. By subadditivity,
$$\infty = \mu(A) = \mu\left(\bigcup E_j \cap A\right) \leq \sum_1^{\infty} \mu(E_j \cap A) $$
OK, I am here. But I do not understand how to continue, or even this is a right approach. Thanks.

REPLY [2 votes]: Is it possible to think in this direction:
From the definition of $\sigma-$ finite measure the $E_i$'s are finite measure since $\nu(E_i)<\infty\:\forall\,i$. Now we know from the definition that $X=\bigcup_{I=1}^{\infty}E_i$ hence for arbitrary E u pick from the sequence of $E_i, E\subset X$. We only need to show that $\nu(X)=\infty$.
$\nu(X)=\nu\left(\bigcup_{I=1}^{\infty}E_i\right)$
$=\sum_{i=1}^{\infty}\nu(E_i)=\infty$.