L1 Metric on Closed Real Interval is Metric

Theorem
Let $S$ be the set of all real functions which are continuous on the closed interval $\closedint a b$.

Let $d: S \times S \to \R$ be the $L^1$ metric on $\closedint a b$:
 * $\ds \forall f, g \in S: \map d {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$

Then $d$ is a metric.

So holds for $d$.

So holds for $d$.

So holds for $d$.

From Zero Definite Integral of Nowhere Negative Function implies Zero Function we have that:
 * $\map d {f, g} = 0 \implies f = g$

on $\closedint a b$.

So holds for $d$.