I don't know a reference, but I have some thoughts. I'm going to tackle this using just homology. I'll ignore gradings in the name of legibility.
There are three kinds of homology in this story: $H(M)$, $H(M,b M)$, and $H(b M)$. The last of these is self-dual; the other two are dual to each other. What I just said is literally true over $Q$, with 'dual to' meaning 'canonically isomorphic to the vector-space dual of'. Over $Z$ of course it's only true in a derived sense. One consequence of the derived statement is that, of the free abelian groups $H(M)/{torsion}$, $H(M,b M)/{torsion}$, and $H(b M)/{torsion}$, the last one is self-dual in the $Hom(-,Z)$ sense while the other two are dual to each other. Another is that the torsion part of $H(b M)$ is self-dual in the $Hom(-,Q/Z)$ sense and the torsion parts of the others are dual to each other in the same sense.
Then there are the other three players: the images of the three maps $H(M)\rightarrow H(M,b M)\rightarrow H(b M)\rightarrow H(M)$. Call them $A$, $B$, and $C$. (Each of the three can also be described as a kernel, or as a cokernel.) Here the rational story is very nice: $A$ is self-dual (more precisely the perfect pairing between $H(M)$ and $H(M, b M)$ when restricted to be a pairing between $H(M)$ and $A$ yields a pairing between $H(M)/ker=A$ and $A$ which is perfect). And $B$ and $C$ are dual to each other (more precisely, in the perfect self-pairing of $H(b M)$ the subspace $B$ is its own orthogonal complement, so that $B$ and $C=H(b M)/B$ are dual).
Over $Z$ things become murkier for $A$, $B$, and $C$. It's true that we get a pairing of $A$ with itself, and that this yields a nondegenerate pairing between the free abelian group $A/{torsion}$ and itself. But it's not perfect; it just injects $A$ into $Hom(A,Z)$. The same goes for pairing $B/{tors}$ with $C/{tors}$. And as for pairings of the torsion parts into $Q/Z$, the images just don't seem to behave well in general.
Update in response to Greg's update to the question: I think the following is basically what you're saying, but it can be said just algebraically and has nothing to do with choosing a splitting (i. e. identifying a quotient of $H(M)/tors$ with a subgroup). The sequence of free abelian groups $H(M)/tors\to H(M,bM)/tors\to H(bM)/tors$ is not exact. The image of $H(M)/tors\to H(M,bM)/tors$, which is the same as my $A/tors$, is not a summand in general; the kernel of $H(M,bM)/tors\to H(bM)/tors$ is a summand and consists of all elements such that some multiple is in $A/tors$; call this $A'$. So $A'$ contains $A/tors$ with finite index. And we get a perfect Z-pairing between $A/tors$ and $A'$. Likewise we get $B'$ and $C'$ containing $B/tors$ and $C/tors$ with finite index, and Z-dual to $C/tors$ and $B/tors$. I don't know what to say about all of this in relation to pairings into $Q/Z$.
Best Answer
Here is my own favorite construction of the (Lebesgue) integral.
Suppose M is an arbitrary smooth manifold. Denote by Or(M) the orientation line bundle of M. This bundle is equipped with a canonical Riemannian metric. Vectors of length 1 in the fiber of Or(M) over a point p∈M correspond canonically to the two orientations of the tangent space at the point p. The manifold M is orientable if and only if the bundle Or(M) is trivializable. Choosing an orientation of M amounts to choosing an isometric trivialization of Or(M).
The bundle Or(M) together with its natural metric is flat. Hence we can twist the de Rham complex Ω^0(M)→⋯→Ω^n(M) by Or(M) and obtain the following twisted de Rham complex: Ω^0(M)⊗Or(M)→⋯→Ω^n(M)⊗Or(M). (Here by a complex I mean a complex of sheaves.) The line bundle Ω^n(M)⊗Or(M) is called the bundle of densities and is denoted by Dens(M). This bundle has a canonical orientation (hence it is trivializable), but does not have a canonical metric or a canonical trivialization.
The cohomology of the twisted de Rham complex (with compact support) is called the twisted de Rham cohomology (with compact support). We have a canonical map C^∞_cs(Dens(M))→H^n_cs(M,Or(M)). Here C^∞_cs is the space of global sections of a vector bundle with compact support and H^n_cs denotes the nth cohomology with compact support.
The Poincaré duality gives us a canonical isomorphism H^n_cs(M,Or(M))→H_0(M). Finally, the map from M to the point induces a map in homology H_0(M)→H_0(∙)=R.
The composition of maps C^∞_cs(Dens(M))→H^n_cs(M,Or(M))→H_0(M)→H_0(∙)=R gives us a map ∫: C^∞_cs(Dens(M))→R, which is the integration map. Note that the actual integration (over each connected component) happens in the first map. The second map is an isomorphism and the third map simply sums integrals over individual connected components.
The map f∈C^∞_cs(Dens(M))→∫|f|∈[0,∞) is a norm on C^∞_cs(Dens(M)). Completing C^∞_cs(Dens(M)) in this norm yields L_1(M)(=L^1(M)), which can be identified with the space of finite complex-valued measures on M.
The space of bounded measurable functions on M (=L_0(M)=L^∞(M)) can be constructed by completing C^∞(M) in the σ-weak topology induced by L_1(M). Other L_p spaces can be constructed in a similar way to L_1(M) by completing sections of the bundle of p-densities instead of the bundle of 1-densities Dens(M).
The development of the remainder of measure theory in this approach largely parallels the one explained in one of my previous answers.
I want to stress that these constructions do not rely on any existing integration theory. In fact, they can be used to build integration theory on smooth manifolds from scratch without ever referring to the usual measure theory with its lengthy and technical proofs.
Added later: In the middle of the proof, we relied on Poincaré duality, which can be most easily established using sheaf cohomology. To this end, one must show that the de Rham cohomology is isomorphic to the sheaf cohomology with real coefficients. This boils down immediately to the Poincaré lemma.
The simplest way to establish the Poincaré lemma is as follows. The de Rham complex of a finite-dimensional smooth manifold M is the free C^∞-dg-ring on the C^∞-ring C^∞(M). If M is the underlying smooth manifold of a finite-dimensional real vector space V, then C^∞(M) is the free C^∞-ring on the vector space V* (the real dual of V). Thus, the de Rham complex of a finite-dimensional real vector space V is the free C^∞-dg-ring on the vector space V*. This free C^∞-dg-ring is the free C^∞-dg-ring on the free cochain complex on the vector space V*. The latter cochain complex is simply V*→V* with the identity differential. It is cochain homotopy equivalent to the zero cochain complex, and the free functor from cochain complexes to C^∞-dg-rings preserves cochain homotopy equivalences. Thus, the de Rham complex of the smooth manifold V is cochain homotopy equivalent to the free C^∞-dg-ring on the zero cochain complex, i.e., R in degree 0.