If we showed that $\mu(F_n)<\infty$ for all $n\in \mathbb{N}$, can we get $\cup_{n \in \mathbb{N}}F_n<\infty$? If we showed that $\mu(F_n)<\infty$ for all $n\in \mathbb{N}$, can we get $\cup_{n \in \mathbb{N}}F_n<\infty$?
The problem is the following:
In the solution of Folland chapter 1 exercise 14,
Suppose $F^*=${$F:F\subset E, 0<\mu(F)<\infty$} ,  $\alpha:=sup_{F\in F^*}${ $ \mu(F)$} $<\infty$. Then for every $n$ there exists $E_n\in F^*$   with $\alpha-1/n\leq \mu(E_n) \leq \alpha <\infty$,. Let $F_n=\cup^n_1 E_j$. Then $\mu(F_n)\geq\alpha-1/n$ for every $n\in \mathbb{N}$. Also $F_n \subset E$ and $\mu(F_n)<\infty$, so $F_n\in F^*$ for every $n\in \mathbb{N}$...
(the whole solution is in Question from Folland Chapter 1 Exercise 14)
Here is the problem, since we can get that $\mu(F_n)<\infty$ for every $n\in \mathbb{N}$ since  $\mu(F_n)=\mu(\cup^n_1 E_j) \leq \sum^n_1\mu(E_j)\leq n\alpha<\infty$. What will happen when $n=\infty$? I can't prove that $\mu(F_n)<\infty$ when $n=\infty$. So how can I prove this question so that the solution will be more precise?? 
 A: Since $F_n \in F^*$ for all $n \in \mathbb{N}$, it follows directly from the definition of $\alpha$ that
$$\mu(F_n) \leq \sup_{F \in F^*} \mu(F) = \alpha<\infty.$$
Since this holds for all $n \in \mathbb{N}$ and $F_n \uparrow F_{\infty} := \bigcup_{j \in \mathbb{N}} E_j$, the continuity of the measure $\mu$ implies
$$\mu \left( \bigcup_{j \in \mathbb{N}} E_j \right) = \lim_{n \to \infty} \mu(F_n) \leq \alpha,$$
i.e.
$$\bigcup_{j \in \mathbb{N}} E_j \in F^*.$$
A: Let $$\mathcal{F} = \{F\subset E: F \ \ \text{is measurable and} \ 0 < \mu(F) < \infty\}$$
Since $\mu$ is semi-finite, $\mathcal{F}$ is not empty. Let 
$$s = \sup\{\mu(F):F\in\mathcal{F}\}$$
It suffices to show that $s = \infty$. Choose a sequence $\{F_n\}_{n\in\mathbb{N}}$ such that $\lim_{n\to \infty}\mu(F_n) = s$. Then
$$F = \bigcup_{1}^{\infty}F_n\subset E$$
and $\mu(F) = s$.
If $s < \infty \Rightarrow \mu(E\setminus F) = \infty$ and hence there exists $F'\subset E\setminus F$ such that $0 < \mu(F') < \infty$. Then, $F\cup F'\subset E$ and $s < \mu(F\cup F') < \infty$,i.e., $F\cup F'\in\mathcal{F}$ which contradicts the definition of $s$. Thus $s = \infty$.
