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 $X,Y$.

$$
\begin{eqnarray*}
0{}={}\mathbb{E}\left[\ \big| X-Y\big|\ \right]&{}={}&\mathbb{E}\left[\ \big| X-Y\big|\ \bigg| \ X\ne Y\right]P\left(X\ne Y\right){}+{}\mathbb{E}\left[\ \big| X-Y\big|\ \bigg| \ X= Y\right]P\left(X= Y\right)\newline
&&\newline
&{}={}&\mathbb{E}\left[\ \big| X-Y\big|\ \bigg| \ X\ne Y\right]P\left(X\ne Y\right)\,.
\end{eqnarray*}
$$
Therefore,

$$
\mathbb{E}\left[\ \big| X-Y\big|\ \bigg| \ X\ne Y\right]\ne0\implies P\left(X\ne Y\right)=0\implies P\left(X= Y\right)=1\,.
$$

