Need help understanding this proof (measure theory related) This proof comes from a probability book I am reading. I understand most of the proof already. The part I'm stuck on might be simply using properties of the symmetric difference, but I can't figure it out. The statement of the lemma is this:
Suppose that $\mathcal{F}_0$ is an algebra that generates the $\sigma$-algebra $\mathcal{F}$, that is, $\mathcal{F} = \sigma\{\mathcal{F}_0\}$. For any set $A \in \mathcal{F}$ and any $\epsilon > 0$, there exists a set $A_\epsilon \in \mathcal{F}_0$ such that
$$
P(A \bigtriangleup A_\epsilon) < \epsilon.
$$
(The notation $A \bigtriangleup B$ is the symmetric difference between $A$ and $B$, that is, $(A \setminus B) \cup (B \setminus A)$, and $P(A)$ is the probability (measure) of event/set $A$.)
The proof begins by fixing $\epsilon>0$ and creating a set $\mathcal{G}$ defined by
$$
\mathcal{G} = \{A \in \mathcal{F} \mid P(A \bigtriangleup A_\epsilon) < \epsilon \text{ for some $A_\epsilon \in \mathcal{F}_0$}\}
$$
and ultimately showing that $\mathcal{G} = \mathcal{F}$. To do this, the author shows that $\mathcal{G}$ is a $\sigma$-algebra by showing that it is closed under taking complements and taking countable unions. The closure under countable union part is where I am having trouble understanding. The part I don't understand is at the very end.
To show closure under countable union, let $A_n$ be a sequence of sets in $\mathcal{G}$. Set $A = \bigcup_{n=1}^{\infty} A_n$ and choose $n_*$ such that
$$
P(A \setminus \bigcup_{n=1}^{n_*}) < \epsilon.
$$
Next, let $\{A_{k,\epsilon} \subset \mathcal{F}_0, 1 \leq k \leq n_*\}$ be such that
$$
P(A_k \bigtriangleup A_{k,\epsilon})<\epsilon \quad \text{for $1 \leq k \leq n_*$}.
$$
Since
$$
\left( \bigcup_{k=1}^{n_*} A_k \right) \bigtriangleup \left( \bigcup_{k=1}^{n_*} A_{k,\epsilon} \right) \subset \bigcup_{k=1}^{n_*} (A_k \bigtriangleup A_{k,\epsilon}),
$$
it follows that
$$
P\left( (\bigcup_{k=1}^{n_*} A_k) \bigtriangleup (\bigcup_{k=1}^{n_*} A_{k,\epsilon}) \right) \leq \sum_{k=1}^{n_*} P(A_k \bigtriangleup A_{k,\epsilon}) < n_* \epsilon,
$$
so that, finally,
$$
P \left( A \bigtriangleup \left( \bigcup_{k=1}^{n_*} A_{k,\epsilon} \right) \right) < (n_* + 1)\epsilon.
$$
I don't understand where this last inequality comes from. I've tried writing $A = (\cup_{k=1}^{n_*} A_k) \cup (A \setminus \cup_{k=1}^{n_*} A_k)$, but that didn't help.
 A: Note that 
$$\begin{align*}
A \bigtriangleup \left( \bigcup_{k=1}^{n_*} A_{k,\epsilon} \right) &= \left(\left(A\setminus\bigcup_{k=1}^{n_*} A_k\right) \bigtriangleup \bigcup_{k=1}^{n_*}A_k\right)\bigtriangleup \left( \bigcup_{k=1}^{n_*} A_{k,\epsilon} \right) \\
&=\left(A\setminus\bigcup_{k=1}^{n_*} A_k\right) \bigtriangleup \left(\bigcup_{k=1}^{n_*}A_k\bigtriangleup  \bigcup_{k=1}^{n_*} A_{k,\epsilon} \right) \\
&\subseteq \left(A\setminus\bigcup_{k=1}^{n_*} A_k\right) \cup \left(\bigcup_{k=1}^{n_*}A_k\bigtriangleup  \bigcup_{k=1}^{n_*} A_{k,\epsilon} \right).
\end{align*}$$
(The first equality uses the fact that if $B$ and $C$ are disjoint then $B\cup C=B\bigtriangleup C$ and the second equality uses the fact that $\bigtriangleup$ is associative.)
Since $$P\left(A\setminus\bigcup_{k=1}^{n_*} A_k\right)<\epsilon$$ and $$P\left(\bigcup_{k=1}^{n_*}A_k\bigtriangleup  \bigcup_{k=1}^{n_*} A_{k,\epsilon} \right)<n_*\epsilon,$$ this gives $$P\left(A \bigtriangleup \left( \bigcup_{k=1}^{n_*} A_{k,\epsilon} \right)\right)<\epsilon+n_*\epsilon=(n_*+1)\epsilon.$$
