Let $H$ be a Hilbert space and $(X,\mu,\mathcal{A})$ a $\sigma$-finite, complete measure space. My textbook states that $L_2(X,\mu,H)$ together with $$\langle\cdot,\cdot\rangle_2\colon L_2(X,\mu,H)^2\to\mathbb{K},\quad (f,g)\mapsto\int_X\langle f,g\rangle_H\, d\mu$$ is a hilbert space and that this would be obvious using some result $(\star)$ and Hölder's inequality.
This is not obvious to me as I fail to see how, given $f=[f'],g=[g']\in L_2(X,\mu,H)$, it is$\langle f',g'\rangle_H\in\mathcal{L}_0(X,\mu,\mathbb{K})$: I know that $(f',g')\in\mathcal{L}_0(X,\mu,H^2)$. The result $(\star)$ states that
In separable Banach spaces $E$, $F$, given $u\in\mathcal{L}_0(X,\mu,E)$, $v\in\mathcal{C}(f(X), F)$, it is $u\circ v\in\mathcal{L}_0(X,\mu,F)$.
However, $E=H^2$ is not assumed to be separable, so how is this result applicable?
I have looked at the method used by Folland to prove that $fg$ is measurable if $f$ and $g$ are, and think that the same method can be applied to this case.
Let $f,g : X \to H$ be fixed and let $h : X \to \mathbb K$ be defined by $h(x) = \langle f(x), g(x) \rangle$. Then we can write $h = \langle \cdot, \cdot \rangle \circ p$ where $p: X \to H^2$ is given by $p(x) = (f(x), g(x))$.
Now, $\langle \cdot, \cdot \rangle : H^2 \to \mathbb K$ is continuous and therefore measurable $(\mathcal B_{H^2}, \mathcal B_{\mathbb K})$-measurable. Here $\mathcal B_Y$ denotes the Borel-sets on $Y$. Also, $p$ is $(\mathcal B_X, \mathcal B_{H^2})$-measurable (here $\mathcal B_X = \mathcal A$ according to your notation) as a tuple of $(\mathcal B_X, \mathcal B_{H})$-measurable functions. Then the composition $\langle \cdot, \cdot \rangle \circ p$ is $(\mathcal B_X, \mathcal B_{\mathbb K})$-measurable.