This is meant as note.
Given a measure space and a Hilbert space.
Then there's an identification: $$\mathcal{L}^2(\mu)\hat{\otimes}\mathcal{H}\cong\mathcal{L}^2_\mathcal{H}(\mu):\quad \varphi\otimes\eta\leftrightarrow\eta\varphi$$
In particular, this yields: $$\mathcal{L}^2(\Omega;\mu)\hat{\otimes}\mathcal{L}^2(\Omega';\mu')\cong\mathcal{L}^2(\Omega\times\Omega';\mu\otimes\mu')$$ How to prove this carefully?
Please, if you find a flaw let me know.
First by Pettis' measurability criterion one has: $$l(\eta\varphi)=l(\eta)\varphi\in\mathcal{M}(\mu)\implies\eta\varphi\in\mathcal{M}(\mu)$$ or alternatively one can check directly: $$\sigma_n\in\mathcal{S}(\mu):\quad\sigma_n\to\varphi\implies\eta\sigma_n\to\eta\varphi$$ So the identification is well-defined as: $$\int\|\eta\varphi\|^2\mathrm{d}\mu=\|\eta\|^2\int|\varphi|^2\mathrm{d}\mu<\infty$$
Next by monotone convergence it is an embedding: $$\sum_{\varepsilon\delta}\kappa_\varepsilon\lambda_\delta\delta\varepsilon=0\implies0=\|\sum_{\varepsilon\delta}\kappa_\varepsilon\lambda_\delta\delta\varepsilon\|^2=\sum_{\varepsilon\delta}|\kappa_\varepsilon|^2|\lambda_\delta|^2\|\delta\|^2\int|\varepsilon|^2\mathrm{d}\mu\implies\sum_{\varepsilon\delta}\kappa_\varepsilon\lambda_\delta\varepsilon\otimes\delta=0$$
Finally characteristic functions write as: $$X=\eta\chi\hat{=}\chi\otimes\eta$$
But simple functions are dense: $$F\in\mathcal{L}^2_\mathcal{H}(\mu):\quad\int\|F-S_n\|^2\mathrm{d}\mu\to0\quad(S_n\in\mathcal{S}_\mathcal{H}(\mu))$$ So the image is indeed all.