Finite-measure sets are approximable in measure by sets from a countable generating algebra

Let $\mathcal{G}$ be a countable generator of $\mathcal{A}$, and choose a [L1, L2, L3, given, choose] sigma-finite exhaustion $(X_m)$ by [L2]. Apply [L1] to the countable family $\mathcal{G}\cup \{X_m:m\ge 1\}$ to obtain a countable algebra ${\mathcal{A}}_0\subseteq \mathcal{A}$. Because $\mathcal{G}\subseteq {\mathcal{A}}_0\subseteq \mathcal{A}$, [L3] gives $\sigma ({\mathcal{A}}_0)=\mathcal{A}$.

Fix $m$ and let ${\mathcal{M}}_m$ be the family of measurable subsets [L4, algebra] $B\subseteq X_m$ for which every $\epsilon >0$ admits $A\in {\mathcal{A}}_0$ with $A\subseteq X_m$ and $\mu (B\mathrm{△}A)<\epsilon$. Because $X_m\in {\mathcal{A}}_0$, the family ${\mathcal{M}}_m$ contains ${\mathcal{A}}_0\cap X_m$. On the finite-measure space $X_m$, the class ${\mathcal{M}}_m$ is a sigma-algebra: complements are handled inside $X_m$, and increasing unions are handled by approximating one large stage and using continuity from below. Therefore ${\mathcal{M}}_m$ is a sigma-algebra containing the trace of ${\mathcal{A}}_0$, so it contains every measurable subset of $X_m$.

Now let $E\in \mathcal{A}$ with $\mu (E)<\mathrm{\infty }$ and let $\epsilon >0$. [L4, choose, algebra] By [L4], choose $m$ so large that $\mu (E\setminus X_m)<\epsilon \mathrm{/}2$. Since $E\cap X_m$ is a measurable subset of $X_m$, the construction above yields $A\in {\mathcal{A}}_0$ with $A\subseteq X_m$ and $\mu ((E\cap X_m)\mathrm{△}A)<\epsilon \mathrm{/}2$. Then $\mu (E\mathrm{△}A)\le \mu (E\setminus X_m)+\mu ((E\cap X_m)\mathrm{△}A)<\epsilon \mathrm{.}$ So finite-measure sets are approximable by members of the countable algebra ${\mathcal{A}}_0$.