I am pretty confused. The following comes from Folland's "Real Analysis," in my version on page 24. I'm paraphrasing slightly, and I've changed some of the variable names because of the typographic limitations presented here.
He defines an elementary family to be a collection of sets $\mathcal E$ such that $ \emptyset \in \mathcal E$, $E,F \in \mathcal E \implies E \cap F \in \mathcal E$, and $E \in \mathcal E \implies E^c = \bigcup_{j=1}^n F_j$ for disjoint $F_j \in \mathcal E$.
He then claims that if $\mathcal A$ is the set of finite disjoint unions of elements of $\mathcal E$, then $\mathcal A$ is an algebra. The part I am unfortunately confused on is this:
he's shown that if $C,D \in \mathcal E$ then $C \cup D \in \mathcal A$. Then, he says
It now follows by induction that if $A_1, \dots, A_n \in \mathcal E$, then $\bigcup A_i \in \mathcal A$; indeed, by the inductive hypothesis we may assume that $A_1, \dots, A_{n-1}$ are disjoint.
What exactly is the inductive hypothesis in this case, and why does it allow us to assume that the sets are disjoint?
What we are proving by induction is the statement below:
The base case can be the $n=1$ case: if $A_1 \in \mathcal E$, then $A_1 \in \mathcal A$. Well, $A_1$ is a disjoint union of elements of $\mathcal E$: just the union of itself and nothing else.
To prove this by induction, we assume that $\bigcup_{j=1}^{n-1} A_j \in \mathcal A$, and try to prove that $\bigcup_{j=1}^n A_j \in \mathcal A$. I disagree slightly with the phrasing in the proof: it's not that
but rather that, because $\bigcup_{j=1}^{n-1} A_j \in \mathcal A$, this union has some other representation $$ \bigcup_{j=1}^{n-1} A_j = \bigcup_{j=1}^{m} A'_j $$ where $A_1', \dots, A_m'$ are disjoint elements of $\mathcal E$. But still, the argument from here is the same, we want to show that $$ \left(\bigcup_{j=1}^{m} A'_j\right) \cup A_n $$ is the disjoint union of elements of $\mathcal E$. This union is the same set as $$ \left(\bigcup_{j=1}^{m} (A'_j \setminus A_n) \right) \cup A_n $$ where this is now a union of disjoint sets. By a claim earlier in this proof, each $A_j' \setminus A_n$ is in $\mathcal A$, and therefore can be rewritten as a disjoint union of elements of $\mathcal E$.
Doing this for each of them gives us a representation of the whole thing as a disjoint union of elements of $\mathcal E$, completing the inductive step.