I need to show that sigma-finiteness implies semifinite. Does the following proof work?
Let $(X,m,\mu)$ be a $\sigma-$finite measure. Let, $E\in m$, and $\mu(E)=\infty$. By $\sigma-$finiteness, $\exists E_{i},\ i\in\mathbb{N},\ \mu(E_{i})\ \text{finite},\ni X=\cup_{i}E_{i}.$ Now, $\infty=\mu(E)=\mu(E\cap(\cup_{i}E_{i}))=\mu(\cup_{i}(E\cap E_{i}))\leq\sum_{i}\mu(E\cap E_{i})$. Now, $\exists i\ \ni\mu(E\cap E_{i})\ne0,$and $\mu(E\cap E_{i})<\mu(E)<\infty$. Take $F=E\cap E_{i*}\subset E $ to complete proof.
Any help is appreciated.
The problem is that we do not know the meaning of $\bigcup_i\mu(E\cap E_i)$. Writing this using limit is probably what was meant in the opening post.
We can write it in the following way: we know that defining $F_n:=E\cap \bigcup_{j=1}^nE_j$, we have $\mu(F_n)\lt\infty$ and $\lim_{n\to\infty}\mu(F_n)=\mu(E)=+\infty$. Therefore, there is some $n_0$ such that $\mu(F_{n_0})\gt 1$ and we have $F_{n_0}\subset E$, and $\mu(F_{n_0})\in (0,\infty)$.