# Prove that: Every $\sigma$-finite measure is semifinite.

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, for $E_i \in \mathcal{M}$, $X = \bigcup_{i=1}^{\infty}E_i$ where $\mu(E_i) < \infty$.

Definition of semifiniteness: 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.

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I think you meant $F\subseteq E$ in the definition of semifinite. Also there are something off in the definition of $\sigma$-finite (e.g. "There are $E_i$ such that...). – Asaf Karagila Nov 11 '12 at 14:30
Yes, Asaf, thanks. – oeda Nov 11 '12 at 14:31
The $E_i$ are not included in $\cal M$, but are elements of $\cal M$. – Davide Giraudo Nov 11 '12 at 15:11
Yes, okay, this was another typo, corrected it, thanks. – oeda Nov 11 '12 at 15:37

We can find $n$ such that $\mu\left(A\cap\bigcup_{j=1}^NE_j\right)>0$, and we have $\mu\left(A\cap\bigcup_{j=1}^NE_j\right) < \mu(A) < \infty$. Furthermore, $A\cap\bigcup_{j=1}^NE_j\subset A$, so $\mu$ is semi-finite.
The converse is not true: counting measure on the subsets of $[0,1]$ is semi-finite but not $\sigma$-finite.