Expectation value of absolute value of difference of two random variables I do not really know how to prove the following statement:

If $E(|X-Y|)=0$  then $P(X=Y)=1$.

The main problem is how to handle the absolute value $|X-Y|$. 
If I say that $|X-Y| \geq 0$ it follows that $E(X)=E(Y)$ which is 
also the result for $|X-Y|<0$. But then the expectation values are equal
and you can show that $P(X=Y)=1$ is not true for all $X,Y$.

Simple proof by contradiction:
Assume that $P(X=Y)<1$, then there has to exists $x',y'$ such that $x'\neq y'$ and $P(X=x'\wedge Y=y')>0$.
But then if we look at the random variable $Z=|X-Y|$, we get that
$$E(Z)=\sum_{x,y}P(X=x\wedge Y=y)\cdot|x-y|$$
$$=P(X=x'\wedge Y=y')\cdot|x'-y'|+\sum_{x,y\neq(x',y')}P(X=x\wedge Y=y)\cdot|x-y|$$
Since $$P(X=x'\wedge Y=y')\cdot|x'-y'|$$ is strictly positive and 
$$\sum_{x,y\neq(x',y')}P(X=x\wedge Y=y)\cdot|x-y|$$
Is non-negative, it follows that $E(Z)>0$.
