# Verifying a set theory statement within a proof

Let $$\dots E_N \subset E_2 \subset E_1$$, $$E = \cap_{i = 1}^{\infty} E_i$$ and $$E_i$$ decreases to $$E$$. $$E$$ is a measurable set, and $$m(E_k) < \infty$$ for some $$k$$.

Show $$E_1 = E \cup \cup_{j = 1}^{\infty} G_j$$

where $$G_j = E_j - E_{j + 1}$$ and $$G_j$$ are disjoint.

We were asked to verify this statement within a proof in my measure theory class, but I am unfortunately quite weak when it comes to set theory. I never took a course in it, and I heard my Discrete Math professor didn't really provide me with sufficient tools. So I'm trying to fill in what I can as problems come up. Can anyone help me with this problem? Thanks!

We have that if $$x\in E\cup \bigcup_{i=1}^{\infty} G_i$$ then since $$E\subseteq E_1$$ and $$G_i=E_i\setminus E_{i+1}\subseteq E_i\subseteq E_1$$ we have $$x\in E_1$$. So $E\cup \bigcup_{i=1}^{\infty} G_i\subseteq E_1$\$
if instead $$x\in E_1$$ then there are two cases. If $$x\in E$$ then $$x\in E\cup \bigcup_{i=1}^{\infty} G_i$$. If $$x\notin E$$ then $$x\notin \bigcap_{i=1}^\infty E_i$$ but $$x\in E_1$$. Since the $$E_i$$ are a downwards chain under inclusion or rather $$E_1\supseteq E_2\supseteq\cdots$$ there must be a smallest $$n$$ such that $$x\notin E_n$$ if there wasn't a smallest $$n$$ then $$x$$ would be in all of the $$E_i$$ and therefore would be in the intersection. Being the smallest $$x\in E_{n-1}$$ so $$x\in E_{n-1}\setminus E_n=G_{n-1}$$ therefore in either case if $$x\in E_1$$ then $$x\in E\cup \bigcup_{i=1}^{\infty} G_i$$. So $$E_1 \subseteq E\cup \bigcup_{i=1}^{\infty} G_i$$.
Every $$x \in E \cup \cup_{j=1}^{\infty} G_j$$ is clearly in $$E_1$$. Now suppose that $$x \in E_1$$. If $$x$$ is in every $$E_i$$ then $$x$$ is in $$E$$ and thus in $$E \cup \cup_{j=1}^{\infty} G_j$$. If $$x$$ is not in every $$E_i$$ then let $$N$$ be the largest $$N$$ such that $$x \in E_N$$. Note then that $$x$$ is not in $$E_{N+1}$$. Such an $$N$$ exists since $$x$$ is not in all $$E_{i}$$ and the sequence is decreasing with respect to set containment. Hence $$x$$ is in $$G_{N}$$ and thus in $$E \cup \cup_{j=1}^{\infty} G_j$$.