Proving completeness of Nikodym Metric

I'm trying to prove completeness directly of the metric given by $d(A, B) = \mu (A \triangle B)$ on a finite measure space $(X, M, \mu)$.

Edit: I should make clear that I'm referring to completeness of the actual Nikodym Metric $M/\sim$ under the relation $A \sim B$ if and only if $\mu(A \triangle B) = 0$. I leave a number of minor details left implicit with respect to this relation (like when I'm talking about a class vs a representative), but they're all rather mundane.

Let $M_1, M_2, ...$ be a Cauchy sequence in our metric space.

Take a subsequence such that for all $m, p > n$, $d(A_m, A_p) < 1/2^n$, where $A_1, A_2, ...$ denotes our subsequence.

Let $A = \limsup M_i$.

Then $A$ is also the $\limsup$ of the $A_i$'s, since $x$ belongs to infinitely many sets in one sequence if and only if it belongs to infinitely many in the other.

Edit 2: The above bolded argument is erroneous. The $\limsup$ of the $A$'s should differ from the $\limsup$ of the $M$'s only on a set of measure zero, but we may very well lose elements when we pass to a $\limsup$ of the subsequence.

Let $n \in N$.

We use measure continuity from above on $A \setminus A_n$ and from below on $A_n \setminus A$.

\begin{align*} \mu(A \setminus A_n) &= \mu(( \bigcap_{k \geq 1}^\infty \bigcup_{j=k}^\infty A_j \setminus A_n)) \\ &= \mu( \bigcap_{k \geq 1}^\infty \bigcup_{j=k}^\infty (A_j \setminus A_n)) \\ &= \lim_{k \rightarrow \infty} \mu(\bigcup_{j=k}^\infty (A_j \setminus A_n)) \end{align*} At this point I'm stuck - using subadditivity of $\mu$ does not help, as $A_j$ might grow further from $A_n$ as $j \rightarrow \infty$, so that the measure of their difference will grow, making the infinite sum infinite for every $k$.

Bounding the other difference is quite simple, however. Using DeMorgan's laws: \begin{align*} \mu(A_n \setminus A) &= \mu(A_n \setminus ( \bigcap_{k \geq 1}^\infty \bigcup_{j=k}^\infty A_j)) \\ &= \mu(A_n \cap ( \bigcap_{k \geq 1}^\infty \bigcup_{j=k}^\infty A_j)^c) \\ &= \mu( \bigcup_{k \geq 1}^\infty \bigcap_{j=k}^\infty (A_n \setminus A_j)) \\ &= \lim_{k \rightarrow \infty} \mu(\bigcap_{j=k}^\infty (A_n \setminus A_j)) \\ &< 1/2^n \end{align*}

If we had the same bound on the first difference as well, then we'd have

$\mu(A_n \triangle A) = \mu(A_n \setminus A) + \mu(A \setminus A_n) < 1/2^{n-1} \rightarrow 0$ as $n \rightarrow \infty$

Which would prove convergence of the subsequence to $A$.

But a Cauchy sequence with a convergent subsequence converges to the same limit, which would prove the result.

I was working with a similar problem recently. I will replace notation $M$ by $\mathcal{M}$ to distinguish a set and a $\sigma$-algebra. Here's my opinion:

1. $d(A,B):=\mu(A\Delta B)$ is just a pseudometric since it doesn't guarantee that \begin{align} d(A,B)=0\iff A=B. \end{align} To make $d$ a metric, we should consider an equivalence relation $\sim$ defined by \begin{align} A\sim B\quad\mbox{if}\quad \mu(A\Delta B)=0 \end{align} and define the metric $d$ on $\mathcal{M}/\sim$ by \begin{align} d(\tilde{A},\tilde{B})=\mu(A\Delta B) \end{align} where $\tilde{A}$, $\tilde{B}$ are equivalence classes in $\mathcal{M}/\sim$ and $A$, $B$ are their representatives, respectively. Therefore, I think you need to change your argument accordingly.
2. I would set \begin{align} B_n=\bigcup_{k=n}^\infty{A_k}. \end{align} Two things need to be proved: \begin{align} \mu(B_n\Delta A)\rightarrow0\\ \mu(A_n\Delta B_n)\rightarrow0 \end{align} The first one follows from that measure is continuous from above. As for the second one, note that \begin{align} \bigcup_{k=n}^\infty{A_k}\setminus A_n&=(A_{n+1}\setminus A_n)\cup(A_{n+2}\setminus A_n)\cup\dots\\ &=(A_{n+1}\setminus A_n)\cup(A_{n+2}\setminus(A_n\cup A_{n+1}))\cup\dots\\ &\subset(A_{n+1}\setminus A_n)\cup(A_{n+2}\setminus A_{n+1})\cup\dots. \end{align} Next, we have \begin{align} \mu(A_n\Delta A)\leq\mu(A_n\Delta B_n)+\mu(B_n\Delta A) \end{align} and the result follows.

Let's denote our $\sigma$-algebra by $\mathcal A$ where we define $\mathcal A/\!\sim$ to be the set of equivalence relations on $\mathcal A$ such that $E\sim F$ iff $\mu(E\,\Delta \,F)=0.$

Note that $E \Delta G \subseteq E \Delta F \cup F \Delta G$ since:

• Let $x \in E \Delta G = (E \cup G) \setminus (E \cap G) = (E \setminus G) + (G \setminus E)$. Suppose, without loss of generality, that $x \in E$ and $x \notin G$. If $x \in F$, then $x \in F \cup G$ and $x \notin F \cap G$, so $x \in (F \cup G) \setminus (F \cap G)$. If $x \notin F$, then $x \in E \cup F$, but $x \notin (E \cap F)$, so $x \in (E \cup F) \setminus (E \cap F)$. Hence, we can conclude that $(E \cup G) \setminus (E \cap G) \subseteq (E \cup F) \setminus (E \cap F) \cup (F \cup G) \setminus (F \cap G)$.

Now we want to show completeness. Let $(A_n)$ be a Cauchy sequence under the metric, then given $\varepsilon > 0$, there exists $N > 0$ such that if $n, m > N$, then $d(A_n, A_m) < \varepsilon$. Let $j \in \mathbb N$, then there exists $n_j > 0$ such that if $n, m > n_j$, then $d(A_n, A_m) < \frac{1}{2^j}$. Without loss of generality, choose $n_j < n_{j + 1}$. Notice that if $i < j$, then $n_i < n_j$ and then $d(A_{n_i}, A_{n_j}) < \frac{1}{2^i}$. Define $A = \liminf A_{n_k} = \bigcup_{k = 1}^\infty \bigcap_{j = k}^\infty A_{n_j}$ and define $B_k = \bigcap_{j = k}^\infty A_{n_j}$. Let $n > n_k$ and observe that $d(A, A_n) \leq d(A, A_{n_k}) + d(A_{n_k}, A_n)$. Since $n > n_k$, then $d(A_{n_k}, A_n) < \frac{1}{2^k}$. Observe that $A \Delta A_{n_k} \subseteq A \Delta B_k \cup B_k \Delta A_{n_k} \subseteq \bigcup_{j = k}^\infty A_{n_j} \Delta A_{n_{j + 1}}$. The latter inclusion takes some showing:

• ($A \Delta B_k \subseteq \bigcup_{j = k}^\infty A_{n_j} \Delta A_{n_{j + 1}}$): Let $x \in A \Delta B_k$. Observe that if $x \in B_k$, then immediately $x \in A$ since $A = \bigcup_{k = 1}^\infty B_k$, so we can't have any elements such that $x \in B_k$ and $x \notin A$. So consider $x \in A = \bigcup_{i = 1}^\infty \bigcap_{j = i}^\infty A_{n_j}$, but $x \notin B_k = \bigcap_{j = k}^\infty A_{n_j}$. This means there exists $k'$ such that $x \in \bigcap_{j = k'} A_{n_j} = B_{k'}$. We can find the smallest $k'$ such that $x \in B_{k'}$, but $x \notin B_{k' - 1}$ with $k' - 1 \geq k$. This means that $x \in \bigcap_{j = k'}^\infty A_{n_j}$ but $x \notin \bigcap_{j = k' - 1}^\infty A_{n_j}$ or in other words, $x \in A_{n_{k'}}$ but $x \notin A_{n_{k' - 1}}$ so $x \in A_{n_{k' - 1}} \Delta A_{n_{k'}}$. Since $k' - 1 \geq k$, we can conclude that $x \in \bigcup_{j = k}^\infty A_{n_j} \Delta A_{n_{j + 1}}$.
• ($B_k \Delta A_{n_k} \subseteq \bigcup_{j = k}^\infty A_{n_j} \Delta A_{n_{j + 1}}$): Let $x \in B_k \Delta A_{n_k}$. Observe that if $x \in B_k$, then immediately $x \in A_{n_k}$ since $B_k = \bigcap_{j = k}^\infty A_{n_j}$, so we can't have any elements such that $x \in B_k$ and $x \notin A_{n_k}$. So consider $x \in A_{n_k}$, but $x \notin B_k = \bigcap_{j = k}^\infty A_{n_j}$. This means that there exists $k' > k$ such that $x \notin A_{n_{k'}}$. Find the smallest $k'$ such that $x \notin A_{n_{k'}}$ but $x \in A_{n_{k' - 1}}$ with $k' - 1 \geq k$. This means that $x \in A_{n_{k' - 1}} \Delta A_{n_{k'}}$. Since $k' - 1 \geq k$, we can conclude that $x \in \bigcup_{j = k}^\infty A_{n_j} \Delta A_{n_{j + 1}}$.

Now, using monotonicity and subadditivity of measure, observe that \begin{align*} d(A_, A_{n_k}) &= \mu(A \Delta A_{n_k}) \\ &\leq \mu(A \Delta B_k \cup B_k \Delta A_{n_k})\\ &\leq \mu \bigg( \bigcup_{j = k}^\infty A_{n_j} \Delta A_{n_{j + 1}}\bigg) \\ &\leq \sum_{j = k}^\infty \mu(A_{n_j} \Delta A_{n_{j + 1}})\\ &= \sum_{j = k}^\infty d(A_{n_j}, A_{n_{j + 1}})\\ &< \sum_{j = k}^\infty \frac{1}{2^j} = 1 - \sum_{j = 1}^{k - 1} \frac{1}{2^j} = 1 - \frac{2^{k - 1} - 1}{2^{k - 1}} = \frac{2^{k - 1} - 2^{k - 1} + 1}{2^{k - 1}} = \frac{1}{2^{k - 1}} \end{align*}

Hence, if $n > n_k$, then $d(A, A_n) \leq d(A, A_{n_k}) + d(A_{n_k}, A_n) < \frac{1}{2^{k - 1}} + \frac{1}{2^k}$ which converges to zero as $k \to \infty$. Thus by definition $A_n \to A$ under the metric. Conclude that every Cauchy sequence converges to an element of our space, and hence $\mathcal A /\! \sim$ is a complete metric space.