In Theorem 1.8 of Folland's Real Analysis, we are given a measure $ \mu $ and a sequence of increasing measurable sets $ \{E_i \} $. We want to show that $ \mu(\bigcup_{i = 1}^{\infty} E_i) = \lim_{i \to \infty} \mu(E_i) $.
My attempt: For every integer $ n $, we have $ \bigcup_{i = 1}^{n} E_i = E_n $ since the sequence is increasing, so $ \mu(\bigcup_{i = 1}^{n} E_i) = \mu(E_n) $. Taking limit both sides yields $ \lim_{n \to \infty} \mu(\bigcup_{i = 1}^{n} E_i) = \lim_{n \to \infty} \mu(E_n) $. (We know that this limit exists in $ \overline{\mathbb{R}}_{\geq 0} $ since $ \{\mu(E_i) \} $ is an increasing sequence in $ \overline{\mathbb{R}}_{\geq 0} $.) Now $ \lim_{n \to \infty} \mu(\bigcup_{i = 1}^{n} E_i) = \mu(\bigcup_{i = 1}^{\infty} E_i) $, so the equality follows. I'm not really certain about my last claim that $ \lim_{n \to \infty} \mu(\bigcup_{i = 1}^{n} E_i) = \mu(\bigcup_{i = 1}^{\infty} E_i) $. Is this just the definition of an infinite union or do I need some form of justification? If it is not the definition of $ \mu(\bigcup_{i = 1}^{\infty} E_i) $, how should I proceed to show that $ \lim_{n \to \infty} \mu(\bigcup_{i = 1}^{n} E_i) = \mu(\bigcup_{i = 1}^{\infty} E_i) $?
A measure is a countable additive set function on a sigma algebra. You have to prove this result using only countable additivity. You cannot assume that $\lim \mu (\cup_{i=1}^{n}E_i)=\mu (\cup_{i=1}^{\infty}E_i)$. For a correct proof construct a disjoint sequence from the $E_i$'s as follows: let $A_1=E_1,A_2=E_2\setminus E_1$ and $A_n=E_n\setminus \cup_{i=1}^{n-1}E_i$ for $i >2$. We get $\mu (\cup_{i=1}^{\infty}A_i)=\sum_{i=1}^{\infty} \mu (A_i))$ by countable additivity. Observe that $\cup_{i=1}^{\infty}A_i=\cup_{i=1}^{\infty}E_i$. Finally, $\sum_{i=1}^{\infty} \mu (A_i)=\lim_{k \to \infty} \sum_{i=1}^{k} \mu (A_i)=\lim_{k \to \infty} \mu(\cup_{i=1}^{k} A_i) =\lim \mu(E_k)$ where we used the fact that $\cup_{i=1}^{k}A_i=E_k$.