I'm afraid I don't have a reference at hand. However, I can sketch out a proof that the two definitions are equivalent.
The hard direction of implication is showing that if $\text{Span}(\{ x \otimes y : x \in H_1, y \in H_2 \})$ is dense in $H$, then $\text{Span}(\{e_k \otimes f_l : k, l \in \mathbb N \})$ is dense in $H$.
Step 1: Show that for any $x \in H_1$ and $y \in H_2$, there exists a sequence $(z_n)_{n \in \mathbb N}$ with terms in $\text{Span}(\{e_k \otimes f_l : k, l \in \mathbb N \})$, such that $\lim_{n \to \infty} z_n = x \otimes y$.
It easy to see how to construct $z_n$.
- Since $\text{Span}(\{ e_k : k \in \mathbb N \})$ is dense in $H_1$, there exists a sequence $(x_n)_{n \in \mathbb N}$ with terms in $\text{Span}(\{ e_k : k \in \mathbb N \})$ such that $\lim_{n \to \infty} x_n = x$.
- Since $\text{Span}(\{ f_l : l \in \mathbb N \})$ is dense in $H_2$, there exists a sequence $(y_n)_{n \in \mathbb N}$ with terms in $\text{Span}(\{ f_l : l \in \mathbb N \})$ such that $\lim_{n \to \infty} y_n = y$.
Now define $ z_n := x_n \otimes y_n$, for each $n \in \mathbb N$.
It's clear that $z_n \in \text{Span}(\{e_k \otimes f_l : k, l \in \mathbb N \})$. With a little work, you can verify that $\lim_{n \to \infty} z_n = x \otimes y$.
[The technique here is to write \begin{align}\left\| z_n - x \otimes y \right\|_H^2 & \leq \left\| z_n - x \otimes y_n \right\|_H^2 + \left\| x \otimes y_n - x \otimes y \right\|_H^2 \\ & = \left\| x_n - x\right\|_{H_1}^2 \left\| y_n \right\|_{H_2}^2 + \left\| x\right\|_{H_1}^2 \left\| y_n - y \right\|_{H_2}^2 \end{align}
and then show that the limit of the right hand side as $n \to \infty$ is zero.]
Step 2: Deduce that for any $z \in \text{Span}(\{ x \otimes y : x \in H_1, y \in H_2 \})$, there exists a sequence $(z_n)_{n \in \mathbb N}$ with terms in $\text{Span}(\{e_k \otimes f_l : k, l \in \mathbb N \})$, such that $\lim_{n \to \infty} z_n = z$.
This follows almost immediately from the result of Step 1.
Step 3: Conclude that $\text{Span}(\{e_k \otimes f_l : k, l \in \mathbb N \})$ is dense in $H$.
From Step 2, we know that
$$ \text{Span}(\{ x \otimes y : x \in H_1, y \in H_2 \}) \subset \overline{\text{Span}(\{e_k \otimes f_l : k, l \in \mathbb N \})} \subset H.$$
Hence
$$ H = \overline{\text{Span}(\{ x \otimes y : x \in H_1, y \in H_2 \})} \subset \overline{\text{Span}(\{e_k \otimes f_l : k, l \in \mathbb N \})} \subset H. $$
This shows that $\text{Span}(\{e_k \otimes f_l : k, l \in \mathbb N \})$ is dense in $H$.
The set $\{e_k \otimes f_l : k, l \in \mathbb N \}$ is clearly orthonomal, and we've just shown that its span is dense in $H$. So $\{e_k \otimes f_l : k, l \in \mathbb N \}$ is an orthonomal basis for $H$.
Best Answer
Bochner integrable r.v.'s take values inside a separable subspace , so we can certainly assume separability of $H$. Let $(e_n)$ be an orthonormal basis for $H$. We have to show that $E \sum \langle X, e_n \rangle \langle Y, e_n \rangle = \langle EX, EY \rangle$. Note that $EX=\sum E\langle X, e_n \rangle e_n$ $\cdots$ (1). So $\langle EX, EY \rangle=\sum E\langle X, e_n \rangle E\langle Y, e_n \rangle $. The proof is now clear by indepndence provide interchange of sums and expectations i have used can be justified. But the justification is easy using Fubini's Theorem: $E\sum Y_n=\sum EY_n$ if $\sum E|Y_n| <\infty$.
(1): $\langle EX, e_n \rangle =E\langle X, e_n \rangle$ (since Bochner integral coincides with Pettis integral). Hence $EX =\sum \langle EX, e_n \rangle e_n =\sum E\langle X, e_n \rangle e_n$.