# Measurability of $f\otimes g:(X\times Y,\mathcal{B}(X)\otimes \mathcal{B}(Y)) \to (H_1 \otimes H_2,\mathcal{B}(H_1\otimes H_2))$

Assume that $(X,d)$ and $(Y,\rho)$ are metric spaces, and that $f:X \to H_1$ and $g:Y\to H_2$ are isometries (especially Borel measurable) into Hilbert spaces. Now define the mapping $$f\otimes g:X\times Y \to H_1 \otimes H_2 \quad \quad \text{by} \quad f\otimes g(x,y) = \iota\circ f(x)\otimes g(y),$$ where $H_1 \otimes H_2$ is the completion of the tensor product of $H_1$ and $H_2$ (i.e. a Hilbert space) and $\iota$ is the embedding into the completion.

I need to show that $f\otimes g$ is $\mathcal{B}(X)\otimes \mathcal{B}(Y))/\mathcal{B}(H_1\otimes H_2)$-measurable or at the very least that $h^* \circ f\otimes g$ is measurable for every $h^* \in (H_1 \otimes H_2)^*$, i.e. $f\otimes g$ is weakly measurable.

Can anybody show me how to proceed? For example can we find a generator for $\mathcal{B}(H_1\otimes H_2)$ which allows for a smooth proof.