Real Analysis, Folland Proposition 1.7 elementary family Definition - An elementary family is a collection $\varepsilon$ of subsets of $X$ such that 
i.) $\emptyset\in \varepsilon$
ii.) if $E,F\in \varepsilon$ then $E\cap F\in \varepsilon$
iii.) if $E\in \varepsilon$ then $E^c$ is a finite disjoint union of members of $\varepsilon$

1.7 Proposition - If $\varepsilon$ is an elementary family, the collection $\mathcal{A}$ of finite disjoint unions of members of $\varepsilon$ is an algebra. 

Attempted proof:
i.) If $A,B\in \varepsilon$, set $B^c = \bigcup_{1}^{n} C_j$ where $\{C_j\}_{1}^{n}\subseteq \varepsilon$. Then $$A\setminus B = A\cap B^c = A\cap \left(\bigcup_{1}^{n}C_j\right) = \bigcup_{1}^{n}C_j\cap A$$ which is a finite disjoint union of members of $\varepsilon$. Thus we have closure under complements.
ii.) Now we can write $A\cup B = (A\setminus B)\cup B$ and we can see that $A\cup B\in \mathcal{A}$ (If more details needs to be provided please let me know)
iii.) Suppose for a fixed $n$ we always have $\bigcup_{1}^{n} A_j\in \mathcal{A}$ whenever $\{A_j\}_{1}^{n}\subset \varepsilon$. Thus, $$\bigcup_{1}^{n+1}A_j = A_{n+1}\cup \bigcup_{1}^{n}A_j$$
I am not sure where to go from here.
Any suggestions on this is greatly appreciated. Please let me know if something is not clear in what I have written so far or if this can be more neatly presented.
 A: Your proof is in the right direction.  Here is it in a complete form.

1.7 Proposition - If $\varepsilon$ is an elementary family, the collection $\mathcal{A}$ of finite disjoint unions of members of $\varepsilon$ is an algebra.

Proof:
We begin by proving two auxiliary results:
i.) $\varepsilon \subseteq  \mathcal{A}$
Note that if $A\in \varepsilon$, then consider $\{C_j\}_1^1$ such that $C_1=A$. Then $\{C_j\}_1^1$ is a disjoint family of sets in $\varepsilon$, so $A=\bigcup_{j=1}^1C_j\in \mathcal{A}$. So $\varepsilon \subseteq  \mathcal{A}$.
ii.) Now let us prove that  for any  $\{A_j\}_{1}^{n}\subset \varepsilon$, ($\{A_j\}_{1}^{n}$ don't need to be disjoint), we have
$\bigcup_{1}^{n}A_j \in \mathcal{A}$.
We will divide the proof in two steps:
ii.a.) First we prove that: if $A \in \mathcal{A}$ and $B\in \varepsilon$, then $A\cup B\in \mathcal{A}$.
If $A \in \mathcal{A}$ and $B\in \varepsilon$, then we have $A=\bigcup_{k=1}^m D_k$ where $\{D_k\}_{1}^{m}$ is a  family of disjoint sets in $\varepsilon$ and  $B^c = \bigcup_{j=1}^{n} C_j$ where $\{C_j\}_{1}^{n}\subseteq \varepsilon$ and $\{C_j\}_{1}^{n}$ is a disjoint family of sets. Then $$A\setminus B = A\cap B^c =\left( \bigcup_{k=1}^m D_i\right)\cap \left(\bigcup_{j=1}^{n}C_j\right) = \bigcup_{k=1}^m\bigcup_{j=1}^{n}C_j\cap D_k$$ which is a finite disjoint union of members of $\varepsilon$. So $A\setminus B \in \mathcal{A}$.
Now we can write $A\cup B = (A\setminus B)\cup B$. Since $A\setminus B$ is finite disjoint unions of members of $\varepsilon$, $B\in \varepsilon$ and $(A\setminus B)\cap B =\emptyset$,  we can see that $A\cup B\in \mathcal{A}$.
ii.b.) It follows by induction that, for all $n\geqslant 2$, if $\{A_j\}_{1}^{n}\subset \varepsilon$, , then $\bigcup_{1}^{n}A_j \in \mathcal{A}$.
In fact, for $n=2$ the result is follows immediately from items i.) and ii.a.).
Assume $n>2$ and that we know by induction hypothesis that $\bigcup_{1}^{n-1}A_j \in \mathcal{A}$, then just apply item ii.a.) to $\bigcup_{1}^{n-1}A_j$ and $A_n$ and we get that $\bigcup_{1}^{n}A_j= \left(\bigcup_{1}^{n-1}A_j\right)\cup A_n\in \mathcal{A}$.
So $\mathcal{A}$ is the the collection of finite unions of members of $\varepsilon$.
Now let us prove that $\mathcal{A}$ is an algebra.

*

*$\emptyset \in \varepsilon \subseteq \mathcal{A}$. So $\emptyset \in \mathcal{A}$.


*Since $\mathcal{A}$ is the the collection of finite unions of members of $\varepsilon$, it is immediate that $\mathcal{A}$ is closed under finite unions. (finite union of finite unions are still just a finite union).


*If $A \in \mathcal{A}$, then we have $A=\bigcup_{k=1}^m D_k$ where $\{D_k\}_{1}^{m}$ is a family of disjoint sets in $\varepsilon$, then we have, for each $k\in \{1,\dots,m\}$,
$$D_k^c= \bigcup_{j=1}^{n_k}B_{k,j}$$ where $B_{k,j}\in \varepsilon$. So we have
$$A^c=\bigcap_{k=1}^m D_k^c=\bigcap_{k=1}^m\left( \bigcup_{j=1}^{n_k}B_{k,j} \right)= \bigcup\{B_{1,j_1}\cap \cdots \cap B_{m,j_m} : 1\leqslant j_k \leqslant n_k, 1 \leqslant k \leqslant m \}$$
So $A^c$ is a finite union of of members of $\varepsilon$. Thus $A^c \in \mathcal{A}$.
So we have proved that $\mathcal{A}$ is an algebra.
