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.
Best Answer
We can find $N$ such that $\mu\left(A\cap E_N\right)>0$ (otherwise, we would have for each $n$ that $\mu\left(A\cap\bigcup_{j=1}^nE_j\right)=0$ and $\mu\left(A\right)=\lim_{n\to +\infty}\mu\left(A\cap\bigcup_{j=1}^nE_j\right)$), and we have $\mu\left(A\cap E_N\right)\leqslant \mu\left( E_N\right)<+\infty$. Furthermore, $A \cap E_N\subset A$, hence the choice $F:=A\cap E_N$ does the job. This proves that $\mu$ is semi-finite.
The converse is not true: counting measure on the subsets of $[0,1]$ is semi-finite but not $\sigma$-finite.