# The symmetric difference metric

Jordan Bell
April 12, 2015

Let $(\Omega,\Sigma,\mu)$ be a probability space. For $A,B\in\Sigma$, define

 $d_{\mu}(A,B)=\mu(A\triangle B).$

This is a pseduometric on $\Sigma$:

 $\displaystyle d_{\mu}(A,C)$ $\displaystyle=\mu(A\triangle C)$ $\displaystyle=\mu((A\triangle B)\triangle(B\triangle C))$ $\displaystyle\leq\mu((A\triangle B)\cup(B\triangle C))$ $\displaystyle\leq\mu(A\triangle B)+\mu(B\triangle C)$ $\displaystyle=d_{\mu}(A,B)+d_{\mu}(B,C).$

The relation $A\sim B$ if and only if $d_{\mu}(A,B)=0$ is an equivalence relation on $\Sigma$, and $d_{\mu}([A],[B])=d_{\mu}(A,B)$ is a metric on the collection $\Sigma_{\mu}$ of equivalence classes. We call $d_{\mu}$ the symmetric difference metric.

The following theorem shows that $(\Sigma_{\mu},d_{\mu})$ is a complete metric space.11 1 V. I. Bogachev, Measure Theory, volume I, p. 54, Theorem 1.12.16.

###### Theorem 1.

If $(\Omega,\Sigma,\mu)$ is a probability space, then $(\Sigma_{\mu},d_{\mu})$ is a complete metric space.

###### Proof.

Suppose that $[B_{n}]$ is a Cauchy sequence in $(\Sigma_{\mu},d_{\mu})$. As for any Cauchy sequence in a metric space, there is a subsequence $[A_{n}]$ of $[B_{n}]$ such that $d_{\mu}([A_{k}],[A_{n}])<2^{-n}$ for $k\geq n$. Define

 $E_{n}=\bigcup_{k\geq n}A_{k}.$

We have

 $\displaystyle E_{n}\setminus A_{n}$ $\displaystyle=\bigcup_{k=n+1}^{\infty}(A_{k}\setminus A_{n})$ $\displaystyle=\bigcup_{k=n+1}\left(A_{k}\setminus\bigcup_{j=n}^{k-1}A_{j}\right)$ $\displaystyle\subset\bigcup_{k=n+1}(A_{k}\setminus A_{k-1})$ $\displaystyle=\bigcup_{k=n}^{\infty}(A_{k+1}\setminus A_{k}),$

hence

 $\mu(E_{n}\triangle A_{n})=\mu(E_{n}\setminus A_{n})\leq\sum_{k=n}^{\infty}\mu(% A_{k+1}\setminus A_{k})<\sum_{k=n}^{\infty}2^{-k}=2^{-n+1}.$ (1)

Now, define

 $A=\limsup_{n\to\infty}A_{n}=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_{k}=% \bigcap_{n=1}^{\infty}E_{n},$

for which

 $\displaystyle\mu(A_{n}\triangle A)$ $\displaystyle=\mu(A_{n}\setminus A)$ $\displaystyle=\mu\left(A_{n}\cap\left(\bigcap_{k=1}^{\infty}E_{k}\right)^{c}\right)$ $\displaystyle=\mu\left(A_{n}\cap\bigcup_{k=1}^{\infty}E_{k}^{c}\right)$ $\displaystyle=\mu\left(\bigcup_{k=1}^{\infty}(A_{n}\cap E_{k}^{c})\right)$ $\displaystyle=\lim_{k\to\infty}\mu(A_{n}\cap E_{k}^{c})$ $\displaystyle=\lim_{k\to\infty}\mu\left(\bigcap_{j\geq k}(A_{n}\setminus A_{j}% )\right)$ $\displaystyle\leq\lim_{k\to\infty}\mu(A_{n}\setminus A_{k})$ $\displaystyle<2^{-n}.$

Using (1),

 $d_{\mu}(A_{n},A)\leq\mu(E_{n}\triangle A_{n})+\mu(A_{n}\triangle A)<2^{-n+1}+2% ^{-n}=3\cdot 2^{-n},$

showing that $[A_{n}]$ converges to $[A]$ as $n\to\infty$, and because $[A_{n}]$ is a subsequence of the Cauchy sequence $[B_{n}]$, it follows that $[B_{n}]$ converges to $[A]$ and therefore that $(\Sigma_{\mu},d_{\mu})$ is a complete metric space. ∎

###### Lemma 2.

For $A,B\in\Sigma$,

 $|\mu(A)-\mu(B)|\leq\mu(A\triangle B).$
###### Proof.
 $\displaystyle|\mu(A)-\mu(B)|$ $\displaystyle=|(\mu(A\setminus B)+\mu(A\cap B))-(\mu(B\setminus A)+\mu(B\cap B% ))|$ $\displaystyle=|\mu(A\setminus B)-\mu(B\setminus A)|$ $\displaystyle\leq\mu(A\setminus B)+\mu(B\setminus A)$ $\displaystyle=\mu((A\setminus B)\cup(B\setminus A))$ $\displaystyle\leq\mu(A\triangle B).$

The following theorem connects the metric space $(\Sigma_{\mu},d_{\mu})$ with the Banach space $L^{1}(\mu)$.22 2 John B. Conway, A Course in Abstract Analysis, p. 90, Proposition 2.7.13.

###### Theorem 3.

If $(\Sigma_{\mu},d_{\mu})$ is separable then $L^{1}(\mu)$ is separable.