how to prove that $\Bbb E(XY) =\Bbb E(X) \Bbb E(Y)$? I have to prove that $\Bbb E(XY) =\Bbb E(X) \Bbb E(Y)$, for any pair $X, Y$ of independent random variables on $(\Omega, \mathcal{F}, P)$ which are in $L^1(\Omega)$.
I would say that the claim is the same as saying (since $X, Y$ independent): $\Bbb E[H(X)K(Y)] = \Bbb E[H(X)] \Bbb E[K(Y)]$ for any functions $ H, K:  \mathbb R \rightarrow \mathbb R$.
But...is this really correct? I wouldn't use any actual stuff we're learning, but some stuff we had some weeks ago. 
 A: Here you have a proof:
$$E\left(X\right)E\left(Y\right)=\left(\sum_{x\in X}x\cdot P\left(X=x\right)\right)\left(\sum_{y\in Y}y\cdot P\left(Y=y\right)\right)=$$
$$=\sum_{x\in X}\sum_{y\in Y}x\cdot y\cdot P\left(X=x\right)P\left(Y=y\right)$$
Now, since the variable are independent, the last expression becomes:
$$E\left(X\right)E\left(Y\right))=\sum_{x\in X}\sum_{y\in Y}x\cdot y\cdot P\left(X=x\text{ and }Y=y\right)=$$
$$=\sum_{z\in X\cdot Y}z\sum_{xy=z}\cdot P\left(X=x\text{ and }Y=y\right)=\sum_{z\in X\cdot Y}zP(X\cdot Y=z)=E\left(X\cdot Y\right)$$
This is a standart proof for this claim. Now if you are in $L^1(\Omega)$ you just have to change the summations by integrals.
A: I would say: first prove it when $X$ and $Y$ are indicator functions.  Then prove it when $X$ and $Y$ are finite linear combinations of indicator functions.  Then use definition of $\mathbb E$.  But of course that depends on what your definition of $\mathbb E$ was.  And your definition of $X,Y$ independent (as noted by Stefan).
