Why is $\mu(E_{1}) + \lim\limits_{N \rightarrow \infty} \sum\limits_{n=1}^{N} \mu(E_{n+1} -E_{n}) = \lim\limits_{N \rightarrow \infty} \mu(E_{N})$

Question

I am learning Continuity of Set Functions. There is a proof about a theorem which I don't fully understand. more specifically, I can't see why $\mu(E_{1}) + \lim\limits_{N \rightarrow \infty} \sum\limits_{n=1}^{N} \mu(E_{n+1} -E_{n}) = \lim\limits_{N \rightarrow \infty} \mu(E_{N})$, Maybe it is clear for some of you, so any help would be appreciated, Thank you!

I have also attached the image about the theorem and its partial proof.

Answer 1

This is just the telescope sum trick for sets / measures. By additivity, we have $$ \mu(E_{i+1}-E_i) + \mu(E_i) = E_{i+1}, $$ hence $$ \mu(E_1) + \sum_{i=1}^n \mu(E_{i+1}-E_i) = \mu(E_1) + \sum_{i=1}^n ( \mu(E_{i+1}) -\mu(E_i) ) = \mu(E_n). $$ Since $\mu(E)>-\infty$ for all $E$, no indefinite expressions appear.