For a Hilbert space $H$, a locally compact (Hausdorff) group $G$ and its (left) Haar measure $\mu$, it follows by the Riesz representation theorem that we may identify an operator $\int_G f(s)d\mu(s) \in B(H)$ for any $f\in C_c(G,H)$ so that for every $\xi,\eta\in H$ $$\biggl<\bigg(\int_G f(s)d\mu(s)\bigg)\xi, \eta\biggr>_H=\int_G\langle f(s)\xi,\eta\rangle_Hd\mu(s)$$
Now, if we assume that a $C^*$-algebra $A$ is faithfully and nondegenerately represented on $H$ and that $f\in C_c(G,A)$, I want to realize that $\int_G f(s)d\mu(s)\in A$.
How do I do this? My guess: To approximate $f$ uniformly by elements in the algebraic tensor product $C_c(G)\odot A$ (with the obvious inclusion $C_c(G)\odot A\subseteq C_c(G,A)$ given by $f(s)\otimes a\mapsto f(s)a$) and then realize that for every $g\in C_c(G)\odot A$ we automatically have $\int_G g(s)d\mu(s)\in A$. Even though this is most likely trivial, I am having trouble convincing myself of this last part.
All hints are appreciated.
Prove that for $g=\sum_{i=1}^{n} g_i \otimes a_i\in C_c(G)\odot A$ you have $\int_G g d \mu=\sum_{i=1}^{n} (\int_G g_i d \mu) a_i\in A$
Note: you should also argue that you indeed can find such a uniform approximation of any $C_c(G,A)$ by elements from $C_c(G)\odot A$.