Let $\left(X,\mathcal{F},\mu\right)$ be a finite measure space and define a relation on $\mathcal{F}$ by $A\sim B\iff\mu\left(A\triangle B\right)=0$ . It can be shown this is an equivalence relation, that $d\left(A,B\right):=\mu\left(A\triangle B\right)$ is a metric on the quotient $\mathcal{F}^{*}:=\frac{\mathcal{F}}{\sim}$ and that this metric space is complete.
Now let $\mathcal{A}\subseteq\mathcal{F}$ be an algebra s.t $\mathcal{F}=\sigma\left(A\right)$ and let $\mathcal{A}^{*}:=\frac{\mathcal{A}}{\sim}$ , I want to show that $\mathcal{A}^{*}$ is dense in $\left(\mathcal{F}^{*},d\right)$ . To do that it would suffice to show that for any $F\in\mathcal{F}$ and any $\varepsilon>0$ there exists $A\in\mathcal{A}$ s.t $\mu\left(A\triangle F\right)<\varepsilon$ but I haven't been able to do that directly.
This is very reminiscent of something I saw in a proof of Caratheodory's theorem. If you consider $\mu$ to be a measure on $\mathcal{A}$ , disregarding $\mathcal{F}$, define an outer-measure $\mu_{0}$ on $\mathcal{P}\left(X\right)$ by $$\mu_{0}\left(B\right):=\inf\left\{ \sum_{j=1}^{\infty}\mathbb{\mu}\left(A_{j}\right)\ |\ B\subseteq\bigcup_{j=1}^{\infty}A_{j}\ ,A_{j}\in\mathcal{A}\ \forall j\right\} $$ And define a set $E\in\mathcal{P}\left(X\right)$ to be $\mu_{0}$ -measurable if for all $\varepsilon>0$ there exists $A\in\mathcal{A}$ s.t $\mu_{0}\left(A\triangle E\right)<\varepsilon$ . Then it can be shown that the collection $\mathcal{M}$ of $\mu_{0}$ -measurable sets is a $\sigma$ -algebra containing $\mathcal{F}=\sigma\left(\mathcal{A}\right)$ which proves that for any $F\in\mathcal{F}$ and any $\varepsilon>0$ there exists $A\in\mathcal{A}$ s.t $\mu_{0}\left(A\triangle F\right)<\varepsilon$ . I'm guessing one can combine this with the fact that in fact $\mu_{0}\equiv\mu$ on $\mathcal{F}$ (easily proven by $\pi-\lambda$ theorem) and get the required result. But this is a really convoluted way of proving something I feel should be fairly simple....
