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Let $\Omega$ be a locally compact topological space. Let us denote $B_{\infty}(\Omega)$, by the set of all uniformly bounded measurable functions.

Let $H$ be a separable Hilbert space. We also denote $M(\Omega,B(H))$, by the set of all bounded linear maps $\mu:C_0(\Omega)\to B(H)$. It is clear that any two vectors $\zeta,\eta$ in $H$ induce the Radon Measure $\mu_{\zeta,\eta}$ given by $$\mu_{\zeta,\eta}(f):=\langle\mu(f)\zeta,\eta\rangle$$

We now obtain the following linear map

$$\gamma:B_{\infty}(\Omega)\otimes M(\Omega,B(H))\to B(H) : f\otimes\mu\to \int f d\mu $$ where $\langle(\int f d\mu)~\zeta,\eta\rangle:=\int f d\mu_{\zeta,\eta}$.

Questions:

1- Does there exist any tensor product $\otimes_{\alpha}$ under which, $$\gamma\left(B_{\infty}(\Omega)\otimes_{\alpha} M(\Omega,B(H))\right)=B(H) ?$$

2- What is the best tensor product $\otimes_{\alpha}$ under which, $B_{\infty}(\Omega)\otimes_{\alpha} M(\Omega,B(H))$ is much more similar to $B(H)$?. I mean $\gamma$ will be isometric or at least invective, ...

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Let $T\in B(H)$. Fix some nonzero function $f_T\in C_0(\Omega)$. Consider a bounded linear functional $\varphi:C_0(\Omega)\to\mathbb C$ such that $\varphi(f_T)=1$. Define $$ \mu_T(g)=\varphi(g)\,T. $$ Then $\mu_T\in M(\Omega,B(H)$ and $\mu_T(f_T)=T$. Then $$ \gamma(f_T\otimes\mu_T)=\mu_T(f_T)=T. $$ So $\gamma$ is always surjective from the algebraic tensor product.