Definition 1:
Say $X = \bigcup_{n=1}^{\infty} E_n $ where $E_n \in \mathcal{M}$ and $\mu( E_n ) < \infty $ for all $n$, we call $\mu$ $\sigma$-finite. More generally, if $E = \bigcup^{\infty} E_n $ where $E_n \in \mathcal{M}$ for all $n$ and $\mu(E_n) < \infty $ for all $n$ then $E$ is said to be $\sigma-$finite for $\mu$.
Definition 2:
If for each $E \in \mathcal{M}$ with $\mu(E) = \infty$, there exists $F \in \mathcal{M}$ with $F \subset E $ and $0 < \mu(F) < \infty$, then we cal $\mu$ semifinite.
Problem:
Every $\sigma-$finite measure is semifinite.
Attempt
Let $\mu$ be a $\sigma-$finite measure on $X$. Take $E \in \mathcal{M}$ arbitrary with $\mu(E) = \infty $. Write
$$ E = \bigcup^{\infty} E_n \; \; \; \; E_n \in \mathcal{M} \; \; forall \; \; n $$
(Here is where I am not sure I am doing the problem correctly. Can I assume that I can write $E$ in such a form? )
Next, there is some $k$ such that $E_k \subset E $. Hence by monotonicity,
$$ 0 \leq \mu(E_k) < \mu(E) = \infty $$
Best Answer
Just prove it by the contradiction. Suppose that $\mu$ is not semifinite, so there is a set $E\in\mathcal M$ such that $ \mu(E)=+\infty $ and for any $F\in\mathcal M$ with $F\subset E$, we have $\mu(F)\in\{0,\infty\}$.
Suppose that $\mu$ is $\sigma$-finite, then there are $E_1, E_2, ..., E_n,... \in \mathcal M$ such that $X=\bigcup_{n=1}^\infty E_n$ and $\mu(E_n)<\infty$ for any $n\in\mathbb N$. Then we have $$ E=\bigcup_{n=1}^\infty (E\cap E_n)\ \Rightarrow\ \mu(E)\leq \sum_{n=1}^\infty \mu(E\cap E_n) $$ and since $\mu(E\cap E_n)\leq\mu(E_n)<\infty$ and $\mu(E\cap E_n)\in\{0,\infty\}$, $\mu(E\cap E_n)=0$ and consequently, $\mu(E)=0$, a contradiction.