Let $X,Y \in L_{2} ( \Omega, P)$. Then $ | \mathbb{E} (XY) | \leq \sqrt{\mathbb{E} X^{2} \mathbb{E} Y^{2}}$ I have trouble with understanding proof of next theorem:

Let $X,Y \in L_{2} ( \Omega, P)$. Then
$$ | \mathbb{E} (XY) | \le \sqrt{\mathbb{E} X^{2} \mathbb{E} Y^{2}} .$$
Proof:
Let $\Omega = \{ \omega_{n} \colon n \in\mathbb{N} \}$. Then, for every $n \in\mathbb{N}$
$$  | \mathbb{E} (XY) |  \le  \mathbb{E} | XY |  = \sum_{k=1}^{n} | X (\omega_{k}) | \cdot | Y (\omega_{k}) | \cdot P(\{\omega_{k}\})$$
Futher, using Cauchy Schwartz inequality, and tending $n \to \infty$ proof is done.

This
$$ | \mathbb{E} (XY) | \le  \mathbb{E} | XY | = \sum_{k=1}^{n} |X (\omega_{k})| \cdot |Y(\omega_{k})| P(\{\omega_{k}\})$$ is what is bothering me. How can we observe $
\mathbb{E} | XY | $ as a sum of finite numbers?
 A: This is not really an answer to your question, but is too much for a comment.
I think it is better just to drop this proof (which is dubious and at least not general).

Observe that for every $t\in\mathbb R$ we have:$$\mathbb E(tX-Y)^2\geq0$$ or equivalently:$$t^2\mathbb EX^2-2t\mathbb EX\mathbb EY+\mathbb EY^2\geq0\tag1$$
This indicates that the discriminant here is not positive:$$4(\mathbb EX\mathbb EY)^2-4\mathbb EX^2\mathbb EY^2\leq0$$
Hence:$$|\mathbb EX\mathbb EY|\leq\sqrt{\mathbb EX^2\mathbb EY^2}$$

P.S. If $\mathbb EX^2=0$ then we are not dealing with a quadratic equation in $(1)$ but that case is trivial because then automatically also $\mathbb EX=0$.
A: Step with equality $|\mathbb{E} (XY) | \leq \mathbb{E} | XY | = \sum_{k=1}^{n} ( | X (\omega_{k}) | | Y (\omega_{k}) | P({\omega_{k}}))$ isn't correct, since there (can be) more than $n$ elements in $\Omega$ for which $X,Y$ attain nonzero values with nonzero probabilities (so you have to take limit there or go different way).
I would do it as follows. Let $\Omega = \{ \omega_n : n\in \mathbb N \} $. Let $p_k = \mathbb P(\omega_k), k\in \mathbb N $. Given that, we can define an inner product:
$  \rho(X,Y)  = \sum_{k=1}^\infty X(\omega_k)Y(\omega_k)p_k = \sum_{k=1}^{\infty} (XY)(\omega_k)p_k = \mathbb E[XY] $
By Cauchy-Schwarz :
$ |\mathbb E[XY] |^2 \leq \mathbb E[|X||Y|]^2 = \rho(|X|,|Y|)^2 \leq \rho(|X|,|X|)\rho(|Y|,|Y|) = (\sum_{k=1}^\infty |X|^2(\omega_k)p_k)(\sum_{k=1}^\infty |Y|^2(\omega_k)p_k) = (\sum_{k=1}^\infty X^2(\omega_k)p_k)(\sum_{k=1}^\infty Y^2(\omega_k)p_k) = \mathbb E[X^2] \mathbb E[Y^2]$
Taking square-root, you get what you wanted.
