Proving completeness of Nikodym Metric

Question

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.

Answer 1

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.