Background
I have some trouble understanding a step of the proof of the following proposition:
Proposition$\quad$Let $X$ be a measure space, and let $p$ satisfy $1\leq p<+\infty$. If $\mu$ is $\sigma$-finite and $\mathscr{A}$ is countably generated, then $L^p(X,\mathscr{A},\mu)$ is separable.
The proof of the proposition uses the following lemma:
Lemma$\quad$Let $(X,\mathscr{A},\mu)$ be a measure space. Suppose that $\mathscr{A}_0$ is an algebra of subsets of $X$ such that
$\sigma(\mathscr{A}_0) = \mathscr{A}$, and
$X$ is the union of a sequence of sets that belong to $\mathscr{A}_0$ and have finite measure under $\mu$.
Then for each positive $\epsilon$ and each set $A$ that belongs to $\mathscr{A}$ and satisfies $\mu(A)<+\infty$ there is a set $A_0$ that belongs to $\mathscr{A}_0$ and satisfies $\mu(A\triangle A_0)<\epsilon$.
Here is the proof of the proposition:
Proof$\quad$ We can choose a countable subfamily $\mathscr{C}$ of $\mathscr{A}$ that generates $\mathscr{A}$ and contains sets $B_n$, $n=1,2,\dots$, that have finite measure under $\mu$ and satisfy $X=\bigcup_nB_n$. Let $\mathscr{C}^+$ consist of the sets in $\mathscr{C}$, together with their complements, and let $\mathscr{A}_0$ be the algebra (not the $\sigma$-algebra) generated by $\mathscr{C}$. Then $\mathscr{A}_0$ is the set of finite unions of sets that have the form\begin{align*} C_1\bigcap C_2\bigcap\dots\bigcap C_N\end{align*}for some choice of $N$ and some choice of sets $C_1,\dots,C_N$ in $\mathscr{C}^+$. Clearly $\mathscr{A}_0$ is countable and satisfies the hypotheses of Lemma 3.64.
Let $\mathscr{S}$ be the collection of all finite sums\begin{align*} \sum_jd_j\chi_{D_j}\end{align*}where each $d_j$ is a rational number (when dealing with the complex $L^p$ spaces let each $d_j$ be a complex number whose real and imaginary parts are rational) and $D_j$ belongs to $\mathscr{A}_0$ and satisfies $\mu(D_j)<+\infty$. The set $\mathscr{S}$ is countable and is included in $\mathscr{L}^p(X,\mathscr{A},\mu)$; we will show that is determines a dense subset of $L^p(X,\mathscr{A},\mu)$.
Let $f$ belong to $\mathscr{L}^p(X,\mathscr{A},\mu)$, and let $\epsilon$ be a positive number. Then there is a simple function $g$ that belongs to $\mathscr{L}^p(X,\mathscr{A},\mu)$ and satisfies $\|f-g\|_p<\epsilon$ (Proposition 3.53). Suppose that the simple function $g$ has the form $\sum_ja_j\chi_{A_j}$, where each $A_j$ belongs to $\mathscr{A}$ and satisfies $\mu(A_j)<+\infty$. We can choose rational numbers $d_j$ such that\begin{align*} \left\|\sum_ja_j\chi_{A_j} - \sum_jd_j\chi_{A_j}\right\|_p \leq\sum_j|a_j-d_j|\|\chi_{A_j}\|_p<\epsilon,\end{align*}and then we can produce sets $D_j$ in $\mathscr{A}_0$ such that $\left\|\sum_jd_j\chi_{A_j} - \sum_jd_j\chi_{D_j}\right\|_p<\epsilon$ (use the above lemma). Since $f$ and $\epsilon$ are arbitrary, and since $\sum_jd_j\chi_{D_j}$ belongs to $\mathscr{S}$ and satisfies\begin{align*} \left\|f-\sum_jd_j\chi_{D_j}\right\|_p &\leq\ \|f-g\|_p + \left\|g-\sum_jd_j\chi_{A_j}\right\|_p\\&\ \ \ \ \ + \left\|\sum_jd_j\chi_{A_j}-\sum_jd_j\chi_{D_j}\right\|_p\\&<\ 3\epsilon,\end{align*}the proof is complete.
My Question
I have difficulties understanding the step where the proof uses the lemma to produce sets $D_j$ in $\mathscr{A}_0$ such that $\left\|\sum_jd_j\chi_{A_j} - \sum_jd_j\chi_{D_j}\right\|_p<\epsilon$. I couldn't see why this is so.
Intuitively, my understanding is that for each $j$ and for each $\epsilon>0$ if $\mu(A_j\triangle D_j)<\epsilon$ then $A_j$ and $D_j$ are essentially overlapped, and so the two functions $\sum_jd_j\chi_{A_j}$ and $\sum_jd_j\chi_{D_j}$ are very close in the $\mathscr{L}^p$ space, and thus the distance between them can be smaller than any positive $\epsilon$. But I am having a hard time to prove it formally.
Could someone please help me with it? Thank you very much in advance!
Notation$\quad$$E\triangle F$ is the symmetric difference of $E$ and $F$.
Definition$\quad$ Let $\mathscr{A}$ be a $\sigma$-algebra on the set $X$. Then $\mathscr{A}$ is countably generated if there is a countable subfamily $\mathscr{C}$ of $\mathscr{A}$ such that $\mathscr{A}=\sigma(\mathscr{C})$.
