I have a question regarding the proof of the following theorem:
If $\mu$ is a measure on a ring $\mathbb{R},$ if $E \in \mathbb{R},$ and if $\left\{E_{i}\right\}$ is a finite or infinite sequence of sets in $\mathbf{R}$ such that $ E \subset \bigcup_{i} E_{i}, \text { then } \quad \mu(E) \leq \sum_{i} \mu\left(E_{i}\right) $
Proof Steps:
If $\left\{F_{i}\right\}$ is any sequence of sets in a ring $\mathbb{R}$, then there exists a disjoint sequence $\left\{G_{i}\right\}$ of sets in $R$ such that $ G_{i} \subset F_{i} \text { and } \bigcup_{i} G_{i}=\bigcup_{i} F_{i}. $ Applying this result to the sequence $\left\{E \cap E_{i}\right\}$, the desired result follows from the countable additivity and monotoneness of $\mu$.
When I tried to apply the result to $\left\{E \cap E_{i}\right\}$, I am stucked. I said there is disjoint $\left\{E \cap F_{i}\right\}$ such that $\mu(E) = \bigcup_{i} \mu(F_{i}).$ And I don't see how to apply monotoness to complete the proof.
You're almost there. By the Lemma you mentioned, extract disjoint $G_i$ such that $G_i \subset E_i$ and $\bigcup_i G_i = \bigcup_i E_i$. Note that $E = \bigcup_i (E \cap G_i )$, and $(E \cap G_i) \subset G_i$, then $$\mu(E) = \mu \left( \bigcup_i (E \cap G_i ) \right) = \sum_i \mu (E \cap G_i) \leq \sum_i \mu(G_i) \leq \sum\mu(E_i),$$
where the second equality uses the countable additivity and the third one uses the monotoneness.