Given a measure $\mu$ on some $\sigma$-algebra $A$, prove that $d(x,y)=\mu((x-y)\cup(y-x))$ is a metric.

Question

Given a measure $\mu$ on some $\sigma$-algebra $A$, prove that $d:A\times A\to[0,\infty)$ defined as $d(x,y)=\mu((x-y)\cup(y-x))$ is a metric.

I started by noticing that $d(x,y)=\mu(x-y)+\mu(y-x)$ because of disjoint set property of measures. But I don't know how to follow from there. Any help would be appreciated. Thanks.

Answer 1

This needs extra conditions on $\mu$.

It must be a finite measure and it must satisfy $\mu(x)=0\implies x=\varnothing$.

This to accomplish $d(x,y)=0\implies x=y$.

Further the symmetry $d(x,y)=d(y,x)$ is obvious and the triangle inequality follows from $$x\Delta z\subseteq(x\Delta y)\cup(y\Delta z)$$ combined with the subadditivity of $\mu$.