Let $A\subset \mathbb{R}^{d}$ be some bounded set and $E_{n}\subseteq A\;\forall n\in\mathbb{N}$. Assume $\{E_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence with respect to metric $\rho$, defined as $\rho(X,Y) = m^{\ast}(X\Delta Y)$, where $m^{\ast}$ is Lebesgue outer measure.
Let me give a proof that there is a subsequence $\{E_{n_{k}}\}_{k=1}^{\infty}$, which converges pointwise almost everywhere: are there any mistakes? I tried to make the reasoning rigorous, so it may be a little bit difficult to read. Is there a way to simplify it?
We start with assigning $\mathbb{1}_{E_{n}}(x)$ to every $x\in A$. Suppose that $F_{\{x_{n}\}}$ is a set of all those points, where $\mathbb{1}_{E_{n}(x)}$ does not converge. If $m^{\ast}(F) = 0$, then there is nothing to prove. Suppose that $m^{\ast}(F) > 0$. We now confine our attention to only this set. For every $n\in\mathbb{N}$ define $G_{1,n}$ to be a set of those points, for which $n$ is minimal number such that $\mathbb{1}_{E_{n}}(x)\neq \mathbb{1}_{E_{n-1}(x)}$. In other words, $n$ is the first time, when indicator changes its value on $x$. Note that $\mathcal{P}_{1} = \{G_{1,n}\}_{n=1}^{\infty}$ is a countable partition of $F$. Define $G_{2,n}$ in the similar fashion: it is the set of points $x$, for which $n$ is the second time when indicator changes its value on $x$. So, we get a sequence of countable partitions $\{\mathcal{P}_{n}\}_{n=1}^{\infty}$ of $F$. For every partition $\mathcal{P}_{n}$ assign a number $m$ to corresponding element $G_{n,m}$. We now perform the following procedure: for every $k\in\mathbb{N}$ define $n_{k}$ to be the smallest integer assigned to any of the elements of $\mathcal{P}_{k}$. The claim is that $E_{n_{k}}$ is the desired subsequence. Indeed, the corresponding sequence of indicators $\mathbb{1}_{E_{n_{k}}}$ does not stabilize a point $x$ if and only if its corresponding sequence $\{i_{k}\}_{k=1}^{\infty}$ (note that for every $k$ a point $x$ is in some $G_{k,l}$ to which we assigned a number) intersect with $\{n_{k}\}_{k=1}^{\infty}$ inifinitely often. Note that the minimum among assigned numbers strictly increases, so if $n_{j} = i_{k}$ then it must be the case that $j=k$. But since on each iteration values assigned to different sets $G_{n,m}$ are different, a point $x$ is a member of $G_{j,n_{j}} = G_{k,i_{k}}$. So it follows that $x$ is a member of $$\bigcap\limits_{k=1}^{\infty}G_{k,i_{k}}$$ Now notice that $$m^{\ast}\left(\bigcap\limits_{k=1}^{\infty}G_{k,i_{k}}\right) = 0$$ Indeed, otherwise $\{E_{n_{k}}\}_{k=1}^{\infty}$ is not Cauchy, since any two consecutive terms differ at least on the set $\bigcap\limits_{k=1}^{\infty}G_{k,i_{k}}$ of measure $m^{\ast}\left(\bigcap\limits_{k=1}^{\infty}G_{k,i_{k}}\right)>0$. But then since there are countable many such intersections and any $x\in F_{x_{n_{k}}}$ is a member of at least one of them, we use countable additivity (since null sets are measurable) to conclude that $F$ is a subset of null set, which implies that $m(F_{x_{n_{k}}}) = 0$ as desired.