Setup: Let $(M,g)$ be a (possibly non-compact) Riemannian manifold with volume density $d_gV$. Then one may think of $(M,g)$ as a measure space $(\Omega,\mathcal{A},\mu)$, where $\Omega:=M$, $\mathcal{A}:=\sigma(\tau_M)$ is the $\sigma$-Algebra generated by the topology $\tau_M$ of $M$ and for any $A \in \mathcal{A}$, $\mu(A):=\int_M{\chi_A d_gV}$, where $\chi_A:M \to [0,1]$ is the characteristic function of $A$. We obtain $\int_{M}{f d\mu} = \int_M{f d_gV}$, where the left hand side is understood to be an integral in the measure theoretic sense and the right hand side is an integration of a density. This enables us to define the space $L_p(\mu)$ with norm $\|f\|_{L_p(M)}^p = \int_{M}{|f|d_gV}$ on a manifold and apply all the results from integration theory to it, e.g. that it is a Banach space and so on.

My question is: Does this work in the following more general setup: Extend the Riemannian metric on $M$ to a fibre metric in $\bigwedge^k T^{\;*}M$, $0 \leq k \leq m$, (as described in the paragraph below). Then one may define $L_p$-spaces of differential forms by setting $\|\omega\|_{L_p(M)}^p := \int_{M}{|\omega|^pd_gV}$ and setting $L_p^k(M)$ to be the space of all measurable $k$-forms on $M$ (i.e. with Lebesgue measurable coefficient functions in any chart) such that $\|\omega\|_{L_p(M)}<\infty$. Is it possible to construct a measure space $(M,\mathcal{A},\mu)$ such that $L_p^k(M)$ may be thought of as an $L_p(\mu)$ as well?. The problem obviously is the range of a differential form. Formally it is a map $\omega\colon M \to \bigwedge^k T^*M$, i.e. it takes values in the vector bundle $\bigwedge^kT^*M$. Even if integration theory is available for functions on measure spaces with values in Banach spaces, this does not help since the bundle itself is not a vector space. I am interested in this question, because otherwise I see no alternative but to establish all the results about integration theory for $L_p^k(M)$ again, i.e. that it is a Banach space, Lebesgue Dominated Convergence Theorem, Fubini/Tonelli etc. That seems a bit exaggerated since intuitively this space is not so fundamentally different.

Construction of the fibre metric: For any $0 \leq k \leq m$ the Riemannian metric may be extended canonically to differential forms in $\Omega^k(M)$ in the following way: For one forms $\omega,\eta \in \Omega^1(M)$ define $g(\omega,\eta):=g(\omega^\sharp, \eta^\sharp)$, where $\sharp:T^*M \to TM$ is the sharp operator with respect to $g$. Then define $g$ on decomposable forms by $g(\omega^1 \wedge \ldots \wedge \omega^k, \eta^1 \wedge \ldots \wedge \eta^k):= \det(g(\omega^i, \eta^j))$.