Lebesgue outer measure and complete metric space Let $A$ be an elementary set. For $E,F\subseteq A$, let $E\sim F$ iff $E\triangle F$ is null. Define $d([E]_{\sim},[E']_{\sim})=m^*(E\triangle E')$. I've shown that $((\mathcal PA)/\sim,d)$ is a metric space, but how do I show that it's complete?
An attempt I had was to take a $d$-Cauchy sequence $([E_n]_{\sim})_{n=1}^{\infty}$ and show that it $d$-converges to $\left[\bigcup_{j=1}^{\infty}\bigcap_{k=j}^{\infty}E_k\right]_{\sim}$ or $\left[\bigcap_{j=1}^{\infty}\bigcup_{k=j}^{\infty}E_k\right]_{\sim}$ using the monotone convergence theorem but it didn't work.
This is from Tao's "An Introduction to Measure Theory" p. 35.
 A: Let $(E_n)_{n=1}^{\infty}$ be a $d-$Cauchy sequence. Using recursively the definition of $(E_n)_{n=1}^{\infty}$ being Cauchy we may find a strictly increasing sequence of integers
$$n_1<n_2<...<n_k<n_{k+1}<...$$
such that
$$\tag{1}d(E_n,E_m)=m^*(E_n\triangle E_m)<\frac{1}{2^k}$$
for every $n,m\geq n_k$. Let $E=\liminf_{k}E_{n_k}$. We claim that $E_n \overset{d}{\to}E$. To see this, first observe that
$$\tag{2} E\triangle E_{n_k}\subseteq \bigcup_{m=k}^{\infty}E_{n_m}\triangle E_{n_{m+1}}$$
Using $(1),(2)$ we obtain
\begin{align}
m^*(E\triangle E_{n_k})&\leq \sum_{m=k}^{\infty}m^*(E_{n_m}\triangle E_{n_{m+1}})\\
&\leq \sum_{m=k}^{\infty}\frac{1}{2^m}=\frac{1}{2^{k-1}}
\end{align}
Hence, for every $n\geq n_k$
\begin{align}
d(E_n,E)&=m^*(E_n\triangle E)=m^*\bigl((E_n\triangle E_{n_k})\triangle (E_{n_k}\triangle E)\bigr)\\
&\leq m^*(E_n\triangle E_{n_k})+m^*(E_{n_k}\triangle E)\\
&\leq \frac{1}{2^k}+\frac{1}{2^{k-1}}
\end{align}
The latter shows that $d(E_n,E)=m^*(E_n\triangle E)\to 0$ as $n\to \infty$.
