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 $\infty$. 

We can use infinite metrics to complete metric spaces in an easy manner:
Given any set $X$, let $\mathbb{R}^X$ be the space of functions $X\to \mathbb{R}$, equipped with the (infinite-valued) uniform metric $d_\infty(f,g)=\sup_{y\in X}|f(y)-g(y)|$. The usual proof shows that $\mathbb{R}^X$ is complete.
Now let $(M,d)$ be a metric space. Set $D\colon M\to \mathbb{R}^M$ as $D(x)=d_x=d(x,\cdot)\colon y\mapsto d(x,y)$ for all $x\in M$. Then $D$ is an isometry:
$$|d_x(y)-d_{x'}(y)|=|d(x,y)-d(x',y)|\leq d(x,x')$$
and
$$|d_x(x)-d_{x'}(x)|=d(x,x'),$$
which shows that $d_\infty(D(x)-D(x'))=d(x,x')$. The closure $\overline{D(M)}$ is a completion of $M$ (which is unique and satisfies the usual universal property as per general theory).
Note this avoid instances of AC as in the usual construction of a completion (as a quotient of the space of Cauchy sequences on $M$, etc.)
