Avner Friendman's book on Analysis has the following theorem
Theorem 1.2.1: Let $\mu$ be a measure with domain $\mathcal{A}$. Then if $\{E_n\}$ is a monotone-increasing sequence of sets of $\mathcal{A}$, then \begin{equation} \lim_n \mu(E_n)=\mu \left(\lim_n E_n\right) \end{equation} Proof: Note that $\lim_n E_n$ belongs to $\mathcal{A}$. Near, warring $E_0=\varnothing$ we have \begin{align*}\mu(\lim_n E_n)&=\mu \left[\bigcup_n (E_n-E_{n-1})\right]=\sum_{n=1}^\infty \mu(E_n-E_{n-1})=\lim_m \sum_{n=1}^m \mu(E_n-E_{n-1})\\ &=\lim_m\mu\left[\bigcup_{n=1}^m(E_n-E_{n-1})\right]=\lim_m\mu(E_m) \end{align*}
My question is: I know that the limit of an increasing sequence of sets is \begin{equation}\lim_n E_n=\bigcup_n E_n \end{equation}
And I don't see any problem with replacing $E_n-E_{n-1}$ throughout the entire proof with just $E_n$. So why do they go through out the entire proof using $E_n-E_{n-1}$? Does something go wrong if we don't do this? I'm just not seeing it.
I'm also wondering why we set $E_0=\varnothing$. Is this just to ensure that we don't start out with an infinitely large set as our first set?
I know these questions seem a bit trivial, but I'm really trying to understand the intricacies of all of the proofs in this book and these little details go a long way. Thanks!