Metric assuming the value infinity If we instead define a metric as $d:X\times X \rightarrow [0,\infty]$, do we lose any nice properties of metric spaces?
The reason I ask is that I saw this theorem: Given a finite measure space $(X,\mathcal{M},\mu )$, define equivalence relation $A \sim B$ if $\mu(A \mathbin{\Delta} B)=0$ for $A,B\in \mathcal{M}$. Then $d: \mathcal{M} /{\sim} \rightarrow [0,\infty)  $ defined by $d(A,B)=\mu(A \mathbin{\Delta}  B)$ is a metric. 
But the same argument would go through if $X$ wasn't a finite measure space, the only place where it seems to fail is because metrics are defined not to take the value $\infty$. 

Instead, think about this. Many measure spaces are $\sigma$-finite; this just means that they are countable unions of subsets of finite measure.
Suppose $(\Omega, \mathcal{S}, \mu)$ is a $\sigma$-finite measure space.  Then you can partition it with $\{E_n\}_{n=1}^\infty$ so $\mu(E_n)<\infty$ for all n.
Then for all $n$, $A\mapsto \mu(A\cap E_n)$ is a pseudometric on $\mathcal{S}$.
Now put

$$d(A, B) = \sum_{n=1}^\infty {\mu(A \mathbin{\Delta} B)\cap E_n)\over 2^n(1+ \mu(A\ \mathbin{\Delta} B)\cap E_n))}.\qquad A, B \in \mathcal{S}.$$

This is a pseudometric with many of the qualities you are seeking.  In fact, it can be seen to be a metric on $\mathcal{M}/\sim$.
