Approximation by an algebra In the book An Introduction To Measure Theory by Terence Tao, there is an exercise:

Exercise 1.4.28(Approximation by an algebra). Let $\mathcal A$ be a Boolean algebra on $X$, and let $\mu$ be a measure on $\langle\mathcal A\rangle$.
(1) If $\mu (X)<\infty$, show that for every $E\in \langle \mathcal A\rangle$ and $\varepsilon >0$ there exists $F\in \mathcal A$ such that $\mu (E\Delta F)<\varepsilon$.
(2)More generally, if $X=\bigcup_{n=1}^{\infty}A_n$ for some $A_1,A_2,\cdots\in \mathcal A$ with $\mu (A_n)<\infty$ for all $n$, $E\in \langle \mathcal A\rangle$ has finite measure, and $\varepsilon >0$, show that there exists $F\in \mathcal A$ such that $\mu (E\Delta F)<\varepsilon$.

Here $\langle\mathcal A\rangle$ means the $\sigma$-algebra generated by $\mathcal A$, and $E\Delta F=(E\setminus F)\cup (F\setminus E)$.
This execise is in the chapter of Abstract measure spaces, and right after the part about completion. The problem is not difficult if we consider the specific measures such as the completion of Jordan measure, but I can't figure out how to deal with abstract measure, at least not for (1). Probably I can solve (2) if I've found how to solve (1). Could anybody give a hint or an answer? Any help will be appriciated.
 A: Since you asked for a hint: consider $$\mathcal{B} = \{ E \subseteq X: \forall \epsilon > 0~ \exists F \in \mathcal{A}: \mu(E\Delta F) < \epsilon   \}$$
Clearly, $\mathcal{A} \subseteq\mathcal{B}$, therefore it suffices to show that $\mathcal{B}$ is a $\sigma$-algebra.
A: To see why 2) follows from 1) note first that for each $n$, $\mathcal{A}\cap A_n$ is an algebra on $A_n$. We claim that $\langle \mathcal{A} \cap A_n \rangle \supseteq\langle \mathcal{A}\rangle  \cap A_n $. Indeed, consider $\mathcal{B}:=\{A\in \langle \mathcal{A}\rangle : A\cap A_n\in \langle \mathcal{A} \cap A_n \rangle\}$. It is then clear that $\mathcal{B}$ contains $\mathcal{A}$ and we claim that $\mathcal{B}$ is a $\sigma$-algebra: if $A\in\mathcal{B}$, then $A^C\cap A_n=(A^C\cup A_n^C)\cap (X\cap A_n)=(A\cap A_n)^C\cap (X\cap A_n)\in\langle \mathcal{A}\cap A_n \rangle$, because $(A\cap A_n)^C\in\langle \mathcal{A}\cap A_n \rangle$  by the assumption that $A\in\mathcal{B}$. Hence $A^C\in\mathcal{B}$. You can also check that $\mathcal{B}$ is closed under countable intersections. It then follows that $\mathcal{B}=\langle \mathcal{A} \rangle$, i.e.  $\langle \mathcal{A} \cap A_n \rangle \supseteq\langle \mathcal{A}\rangle  \cap A_n $, as claimed. 
If we are now given an $\epsilon>0$ $E\in \langle \mathcal{A} \rangle$ with $\mu(E)<\infty$ we see that for each $n$, $E\cap A_n\in\langle\mathcal{A}\rangle  \cap A_n \subset \langle \mathcal{A} \cap A_n \rangle$. Applying 1) to the finite measure space $(A_n, \langle \mathcal{A} \cap A_n \rangle, \mu_{|\langle \mathcal{A} \cap A_n \rangle})$ and the algebra $\mathcal{A} \cap A_n$ we obtain an $F_n\in \mathcal{A} \cap A_n$ such that $\mu(E\cap A_n\triangle F_n)<\epsilon/2^n$, then $F:=\bigcup\limits_{n=1}^{\infty}F_n\in \mathcal{A}$ and will satisfy $\mu(E\triangle F)<\epsilon$.
A: proof:
(1): Let $P(E)$ be the property of sets $E \subset X$ that $\forall \varepsilon > 0$, $\exists F \in \mathcal A$ such that $\mu(E\Delta F)<\varepsilon$. Suppose that $E_1, E_2,... \subset X$ are such that $P(E_n)$ is true for all $n$. ie, $\forall \varepsilon > 0$, $\exists F_n \in \mathcal A$ such that $\mu(E_n \Delta F_n)<\varepsilon$ for all $n$. Since $\mu(X)<\infty$, and the sets $V_j = E_j \setminus \bigcup_{n \neq j}E_n \subset X$ are all disjoint, it must be from countable additivity that $V_j = \emptyset$ for all $j \geq k$ for some number $k$. In particular, $\bigcup_{n}E_n = \bigcup\limits_{n=1}^{k}E_n$. Then we have $(\bigcup_{n} E_n) \Delta \bigcup\limits_{m=1}^{k}F_m = (\bigcup\limits_{n=1}^{k}E_n) \Delta \bigcup\limits_{m=1}^{k}F_m$. Note that: $(\bigcup\limits_{n=1}^{k}E_n) \setminus \bigcup\limits_{m=1}^{k}F_m = \bigcup\limits_{n=1}^{k}(E_n \setminus \bigcup\limits_{m=1}^{k}F_m) \subset \bigcup\limits_{n=1}^{k}(E_n \setminus F_n) \subset \bigcup\limits_{n=1}^{k}E_n \Delta F_n$. Similarly we have $(\bigcup\limits_{m=1}^{k}F_m) \setminus \bigcup\limits_{n=1}^{k}E_n \subset \bigcup\limits_{n=1}^{k}E_n \Delta F_n$. Hence $(\bigcup_{n} E_n) \Delta \bigcup\limits_{m=1}^{k}F_m \subset \bigcup\limits_{n=1}^{k}E_n \Delta F_n$, which can be of arbitrarily small measure. Clearly $\bigcup\limits_{m=1}^{k}F_m \in \mathcal A$ and thus $P(\bigcup_{n}E_n)$ is true.
If $P(E)$ is true for some $E \subset X$ then $P(E^\complement)$ is true since $E^\complement \Delta F^\complement = F \Delta E$ and $F^\complement \in \mathcal A$.
Clearly $P(\emptyset)$ is true and $P(E)$ is true for all $E \in \mathcal A$. By remark $1.4.15$, $P(E)$ is thus true for all $E\in \langle \mathcal A\rangle$.
(2): Solution to part 2 is given above in Jonathan's answer. Except the fact that we only need finitely many $A_n$ in the approximation of $E$ since $E$ has finite measure. More specifically, $\mu(E \cap A_{n+1} \setminus \bigcup\limits_{i=1}^{n}E \cap A_i) = 0$, $\forall n \geq N$ for sufficiently large $N$. That is, $\bigcup\limits_{n=1}^{N}E \cap A_n$ approximates $E$ to within a null set. Then for $1 \leq n \leq N$, we find $F_n$ in $A_n \cap \mathcal A$ that differs from $E \cap A_n$ in measure by at most $\varepsilon/N$ and take $F = \bigcup\limits_{n=1}^{N}F_n$.
