Let $A$ be some elementary subset of $\mathbb{R}^d$, and define an equivalence relation on $2^A$ by declaring two sets $E,F\subset A$ equivalent if $m^*(A \triangle B) = 0$, where $m^*$ is Lebesgue outer measure. We can make $X := 2^A/\sim$ into a metric space if we let $d([E],[F]) = m^*(E \triangle F)$. I'd like to show this metric space is complete. Here's my work so far.
Let $([E_n])_{n=1}^\infty$ be a Cauchy sequence in $X$. I want to show the sequence converges to $[E]$, where $E := \bigcup_{n=1}^\infty \bigcap_{m=n}^\infty E_n$. Fix $\epsilon >0$ and pick $K$ such that for all $k_1,k_2 \geq K$, $m^*(E_{k_1} \triangle E_{k_2}) \leq \epsilon$. We need a bound on the measure of $E \triangle E_k$, when $k \geq K$. We can write $$ E \triangle E_k = \bigcup_{n=1}^\infty\bigcap_{m=n}^\infty (E_m \setminus E_k) \cup \bigcap_{n=1}^\infty\bigcup_{m=n}^\infty (E_k \setminus E_m), $$ so $$ m^*(E \triangle E_k) \leq \lim_{n \to \infty} m^*\left(\bigcap_{m = n}^\infty(E_m \setminus E_k)\right) + \lim_{n\to\infty} m^*\left(\bigcup_{m=n}^\infty(E_k \setminus E_m)\right). $$
The Cauchy criterion lets us bound the first term easily, but the second term is a lot harder. Of course the measure of each $E_k \setminus E_m$ is less than $\epsilon$ for large enough $m$, but this doesn't give us a bound on the measure of an infinite union of these sets. How do we get around this? Could we use an $\epsilon/2^n$ trick?