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!
 A: 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 $E \cup \cup_{j=1}^{\infty} G_j$.
A: 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 $E_1 \subseteq  E\cup \bigcup_{i=1}^{\infty} G_i$.
This means that the two sets are subsets of each other and thus must be equal
