If $...E_3\subset E_2 \subseteq E_1$ and $\mu(E_i)<\infty$ then $\mu(\cap E_n)=lim(\mu(E_n))$

Question

Let $...E_3\subset E_2 \subseteq E_1$ be a decreasing sequence and $\mu(E_i)<\infty$ then $\mu(\cap E_n)=lim(\mu(E_n))$

I know that there is a prove based on $\mu(\cup E_n)=lim(\mu(E_n))$ for increasing sequence but I want to understand the following proof (if it has not errors)

$$\mu(E_1\setminus\cap E_n)=\mu(E_1)-\mu(\cap E_n)\iff\mu(\cap E_n)=\mu(E_1)-\mu(E_1\setminus\cap E_n)=_{(1)}$$

$$=_{(1)}\mu(E_1)-\mu(E_1\setminus\cup E_n^C )=\mu(E_1)-\mu(\cup (E_1\cup E_n^C ))=$$

$$=\mu(E_1)-\mu(\cup (E_1\setminus E_n))$$

How did we go from $(1)$ and forward? shouldn't it be

$$\mu(E_1)-\mu(E_1\setminus\cap E_n)=_{(1)}\mu(E_1)-\mu(E_1\setminus(\cup E_n^C)^C )$$

Answer 1

There is a small typo in the first step after (1), which should explain it. We have

$$E_1\setminus \cap E_n = E_1\setminus(\cup E_n^c)^c= E_1\cap (\cup E_n^c) = \cup (E_1\cap E_n^c).$$

Does that help?