Show that $(\mathcal{M},d)$ is complete metric space

Question

Let $(\Bbb{R},\mathcal{M},\mu)$ be the Lebesgue measure space modulo the equivalence relation $A\sim B$ if $\mu(A\bigtriangleup B)=0$. Let $d(A,B)=\mu(A\bigtriangleup B)$. Show that $(\mathcal{M},d)$ is complete metric space.

I could show that $d$ is a metric on the equivalence classes but how can I show that this metric is complete?

Answer 1

You need to appeal to the definition of completeness. Suppose you are given a Cauchy sequence of sets $A_n$. That is, for any $\epsilon > 0$ one can find a sufficiently large $N$ for which $m,n \geq N$ implies $\mu(A_{n} \bigtriangleup A_{m}) < \epsilon$. We would like to show that a limit set $A_{\infty}$ exists, and is measurable.

Here is a hint: Given a sequence of measurable sets $A_n$, there are a number of different sets one can associate to it, all of which are measurable. One example is the union $\bigcup A_n$, but there are others. Consider these.

Answer 2

Note that $d(A,B) = \| 1_A-1_B\|_1$, so we can use properties of $L^1(\mathbb{R})$ to show completeness.

As Michael noted in the comments above, one needs to be careful with infinities. Suppose we have a $d$-Cauchy sequence of sets $A_n$. Without loss of generality we can presume that $d(A_n,A_1) < 1$ for all $n$.

Let $f_n = 1_{A_n} - 1_{A_1}$. We have $f_n \in L^1(\mathbb{R})$ for all $n$ and $f_n$ is Cauchy. Since $L^1(\mathbb{R})$ is complete, we know that there is some $f \in L^1(\mathbb{R})$ such that $f_n \to f$ (in the $\|\cdot\|_1$ norm, of course).

The only issue remaining is to 'extract' a set $A$ such that $f(x) = 1_A(x)- 1_{A_1}$ ae. $[\mu]$.

A standard result (see https://math.stackexchange.com/a/716328/27978, for example) is that if $f_n \to f$ in $L^1(\mathbb{R})$, then there is a subsequence such that $f_{n_k}(x) \to f(x)$ ae. $[\mu]$. We note that $f_n(x) \in \{-1,0,+1\}$ for all $n$, hence we have $f(x) \in \{-1,0,+1\}$ ae. $[\mu]$. Let $N= f^{-1}(\{-1\})$, $Z= f^{-1}(\{0\})$ and $P= f^{-1}(\{+1\})$ and let $A = (Z \cap A_1) \cup P$, then it is easy to check that $f(x) = 1_A(x) - 1_{A_1}(x)$ ae. $[\mu]$ and $d(A,A_n) \to 0$. (Measurability follows from the fact that $f \in L^1(\mathbb{R})$.)