p-average compound metric I'm trying to prove that probability space metric defined as
$d(X,Y)=(\mathbb{E}|X-Y|^p)^{1/p}$
is a metric indeed.
Literature states that $d(X,Y)=0$ implies $Pr(X=Y)=1$, but no further explanations about that. Is it really so obvious?
 A: The Minkowsky inequality states that for $1\le p \le \infty $ and $V,W\in L^p(\Omega)$ ($V$, $W$ are measurable functions on $[\Omega,\mathscr A, \mu]$) the following inequality holds
$$\left(\int \mid V+W\mid^p \ d\mu\right)^{\frac{1}{p}}\le\left(\int \mid V\mid^p \ d\mu\right)^{\frac{1}{p}}+\left(\int \mid W\mid^p \ d\mu\right)^{\frac{1}{p}}.$$
In probability language (when $\mu (\Omega)=1)$ the Minkowsky inequality tells that
$$\mathbb E\left[\mid V+W\mid^p\right]^\frac{1}{p}\le \mathbb E\left[\mid V \mid^p\right]^\frac{1}{p}+ \mathbb E\left[\mid W\mid\mid^p\right]^\frac{1}{p}.$$
If we let $V= X-Z$ and $W=Z-Y $ ($Z,Y,X \in L^p(\Omega)$) then we see that 
$$d(X,Y)=\mathbb E\left[\mid X-Z+Z-Y\mid^p\right]^\frac{1}{p}\le  d(X,Z)+d(Z,Y).$$
This is the triangle inequality.
So, $d$ is non negative, $d(X,X)=0$, and the triangle inequality holds. The question remains: why $d(X,Y)=0 \rightarrow X=Y? $ (with probability 1) Suppose that there is a set of positive probability over which $X \not =Y$. It is clear that then the integral of the absolute value of the difference could not be zero.
