Space $M(\lambda)$ of measurable subsets of $[0,1]$ up to null sets is Polish I have encountered the following example of a Polish space:

Let $M(\lambda)$ be the collection of all measurable subsets of $[0,1]$ considered up to null sets, so sets agreeing almost everywhere are identified. The metric here is
  $$d(A,B)=\lambda (A\triangle B),$$
  where $A\triangle B$ is the symmetric difference between $A,B$ and $\lambda$ Lebesgue measure.

I was able to prove that $d$ is a metric.
Questions: 


*

*How to prove that $M(\lambda)$ is complete wrt the metric $d$ defined above? 2. How to prove that $M(\lambda)$ is separable?


Here is my (very poor) attempt about 1.: if $\{A_n\}$ be a Cauchy sequence of measurable subsets, wlog, we may assume $\lambda(A_n\triangle A_{n+1})<2^{-n}$.
Does that work?
 A: 1st Solution using elementary techniques.
We first establish the completeness. Let $(E_n)_{n\geq 1}$ be a Cauchy sequence in $M(\lambda)$. Our aim is to show that this sequence converges in $M(\lambda)$. To this end, it suffices to find a convergent subsequence. We construct such a subsequence by the following algorithm:


*

*Set $n_0 = 0$ for convenience.

*Assume that $k \geq 1$ and $n_{k-1}$ is defined. Then by invoking the Cauchyness, we pick $n_k > n_{k-1}$ such that $d(E_m, E_n) < 2^{-k}$ whenever $ m, n \geq n_k$.


We claim that $(E_{n_k})_{k\geq 1}$ converges. Indeed, define
\begin{align*}
E^{+}
&= \cup_{m\geq 1} \cap_{k\geq m}E_{n_k}
= \{ x \in [0, 1] : \text{$x \in E_{n_k}$ for all but finitely many $k$} \} \\
E^{-}
&= \cap_{m\geq 1} \cup_{k\geq m}E_{n_k}
= \{ x \in [0, 1] : \text{$x \in E_{n_k}$ for infinitely many $k$} \}.
\end{align*}
Then it is clear that both are measurable sets such that $E^{-} \subseteq E^{+}$. Moreover,
\begin{align*}
E^{+}\setminus E^{-}
&= \{ x \in [0, 1] : \text{$x \in E_{n_k}$ for inf. many $k$ and $x \notin E_{n_k}$ for inf. many $k$} \} \\
&= \{ x \in [0, 1] : \text{$x \in E_{n_k} \triangle E_{n_{k+1}}$ for inf. many $k$} \} \\
&= \cap_{m\geq 1}\cup_{k\geq m} E_{n_k} \triangle E_{n_{k+1}}.
\end{align*}
This, together with $\sum_{k=1}^{\infty} d(E_{n_k}, E_{n_{k+1}}) < \infty$, shows that
$$ \lambda(E^{+}\setminus E^{-})
= \lim_{m\to\infty} \lambda(\cup_{k\geq m} E_{n_k} \triangle E_{n_{k+1}})
\leq \lim_{m\to\infty} \sum_{k=m}^{\infty} d(E_{n_k}, E_{n_{k+1}})
= 0.$$
So both $E^{+}$ and $E^{-}$ represent the same element in $M(\lambda)$. Let $E$ denote either of $E^{\pm}$. Then
\begin{align*}
d(E_{n_k}, E)
&= \lambda(E_{n_k}\setminus E^{-}) + \lambda(E^{+} \setminus E_{n_k}) \\
&= \lim_{m\to\infty} \left(\lambda(E_{n_k} \setminus \cup_{j\geq m}E_{n_j}) + \lambda(\cap_{j\geq m}E_{n_j} \setminus E_{n_k} \right) \\
&\leq \varlimsup_{m\to\infty} \left(\lambda(E_{n_k} \setminus E_{n_m}) + \lambda(E_{n_m} \setminus E_{n_k} \right) \\
&\leq 2^{-k},
\end{align*}
and so, letting $k\to\infty$ proves the desired claim.
Finally, the separability of $M(\lambda)$ follows from the standard result that any $E \in M(\lambda)$ can be approximated, under the metric $d$, by a finitely many union of open intervals with rational endpoints.

2nd Solution using $L^p$-theory.
Notice that the metric space $M(\lambda)$ equipped with $d$ is isometric to
$$\widetilde{M}(\lambda) = \{ \mathbf{1}_E : E \in M(\lambda)\} \subset L^1([0,1],\lambda)$$ 
consisting of indicator functions. Indeed, the bijection $\Phi : M(\lambda) \to \widetilde{M}(\lambda)$ defined by $\Phi(E) = \mathbf{1}_E$ gives rise to an isometry, thanks to the identity
$$\lambda(E\triangle E') = \| \mathbf{1}_{E} - \mathbf{1}_{E'} \|_{1}.$$
So it suffices to prove that $\widetilde{M}(\lambda)$ is complete and separable.
Recall that $L^{1}([0, 1], \lambda)$ is complete and separable. Being a subspace of a separable metric space, $\widetilde{M}(\lambda)$ is separable as well. So it suffices to show that $\widetilde{M}(\lambda)$ is complete.
Let $(\mathbf{1}_{E_n})_{n\geq 1}$ be any Cauchy sequence in $\widetilde{M}(\lambda)$. Then this sequence converges in $L^1$. Also, it is well-known that there exists a subsequence $(\mathbf{1}_{E_{n_k}})_{k\geq 1}$ which converges a.e. Since the limiting function takes values in $\{0, 1\}$ a.e., there exists $E \in M(\lambda)$ such that $\mathbf{1}_{E_{n_k}} \to \mathbf{1}_E$ both a.e. and in $L^1$. Since $(\mathbf{1}_{E_n})_{n\geq1}$ is Cauchy, it follows that $\mathbf{1}_{E_n} \to \mathbf{1}_E$ in $L^1$. This proves that $\widetilde{M}(\lambda)$ is complete.
