Is integration on manifolds unique? Suppose $M$ is a smooth oriented compact connected $m$-dimensional manifold and let $A^m(M)$ denote the set of exterior differential $m$-forms on $M$. We can integrate members of $A^m(M)$, and the mapping $\phi\mapsto\int_M\phi$ has several properties:


*

*It's not identically $0$.

*It's linear.

*It's symmetric.


By "it's symmetric", I mean that if $f:M\to M$ is an orientation-preserving diffeomorphism, then $\int_M \phi = \int_M f^*\circ \phi$. I think that's true and expresses the idea of symmetry I have in mind. Basically, all points on $M$ are equivalent as far as integration is concerned.
My question is: Do these properties uniquely determine the space $A^m(M)$ and integration? More precisely, is there some smooth compact connected $m$-manifold $M$, a space $S$ of smooth sections of the tensor bundle of $M$, and a linear functional $\lambda:S\to\mathbb{R}$ that obeys the three properties above but is not just integration of exterior $m$-forms, up to constant scale factor?
 A: This is a special case of a result from analysis and measure theory. The following two theorems are quoted from section 2.5 of Federer "Geometric Measure Theory." Theorem 2 is a consequence of Theorem 1. The spaces $L^p(X,\mu)$ are the "Lp spaces" of measure theory.
Theorem 1
Let $X$ be a measurable space and let $L$ be a set of measurable functions $X\to\mathbb{R}$ such that $c\in[0,\infty)$ and $f,g\in L$ implies $f+g\in L$ and $cf\in L$ and $\inf\{f,g\}\in L$ and $\inf\{f,c\}\in L$, and provided $f\leq g$, also $g-f\in L$.
(For example, if $X$ is a smooth compact manifold, then $L$ could be the space of continuous functions.)
If $\lambda:L\to\mathbb{R}$ satisfies


*

*$\lambda(f+g)=\lambda(f)+\lambda(g)$ for any $f,g\in L$

*$\lambda(cf)=c\lambda(f)$ for any $c\in[0,\infty)$ and any $f,g\in L$

*$f\geq g$ implies $\lambda(f)\geq\lambda(g)$

*if $f_n$ is pointwise increasing and convergent to $g$, then $\lambda(f_n)$ is increasing and convergent to $\lambda(g)$.


then there is a measure $\mu$ on $X$ such that $\lambda(f)=\int_X f\,d\mu$ for all $f\in L$.
Theorem 2
Let $(X,\mu)$ be a measure space, which is the union of countably many measurable subsets with finite measure. Let $\lambda:L^1(X,\mu)\to\mathbb{R}$ be a linear map such that
$$\sup\Big\{|\lambda(f)|:f\in L^1(X,\mu)\text{ s.t. }\int_X |f|\,d\mu=1\Big\}<\infty.$$
Then there exists $g\in L^{\infty}(X,\mu)$ such that
$$\lambda(f)=\int_X fg\,d\mu\qquad\forall f\in L^1(X,\mu).$$
If $g'$ is any other such function then $g=g'$ almost everywhere with respect to $\mu$.
Application of Theorem 2.
Let $M$ be a smooth compact oriented $n$-manifold, and let $T:\Omega^n_{C^\infty}(M)\to\mathbb{R}$ be a linear map. Let $\omega$ be a smooth nonvanishing $n$-form on $M$. Suppose that
$$\underbrace{\sup\Big\{|T(\nu)|:\nu\in\Omega^n_{C^\infty}(M)\text{ s.t. }\int_M \Big|\frac{\nu}{\omega}\Big|\omega\leq 1\Big\}}_{c}<\infty.$$
Define a measure $\mu$ on $M$ by $\mu(A)=\int_A\omega$, and define $\lambda:C^\infty(M)\to\mathbb{R}$ by $\lambda(f)=T(f\omega).$ If $f\in C^\infty(M)$ has $\int_M |f|\omega=1$ then one has $|T(f\omega)|\leq c$. So (see wikipedia "continuous linear extension") $\lambda$ extends to a map $L^1(M,\mu)\to\mathbb{R}$, and
$$\sup\Big\{|T(f)|:f\in L^1(M,\mu)\text{ s.t. }\|f\|_{L^1(M,\mu)}\leq 1\Big\}<\infty.$$
According to Theorem 2, there exists $g\in L^\infty(M,\mu)$ such that $\lambda(f)=\int_M fg\omega$ for all $f\in L^1(M,\mu)$. So, for any smooth $n$-form $\nu$, one has
$$T(\nu)=T\Big(\frac{\nu}{\omega}\omega\Big)=\int_M \frac{\nu}{\omega}g\omega=\int_M g\nu.$$
By taking $\nu$ to be supported in arbitrarily small regions, it should be an easy exercise to show that diffeomorphism-invariance of $T$ implies that $g$ is constant.
On the extra condition. Note that if $\omega'$ is another smooth nonvanishing $n$-form with $\frac{\omega'}{\omega}$ positive everywhere, then
$$\Big|\frac{\nu}{\omega'}\Big|\omega'=\Big|\frac{\nu}{\omega'}\Big|\frac{\omega'}{\omega}\omega=\Big|\frac{\nu}{\omega}\Big|\omega$$
and so the sup-condition on $T$ is actually of a topological character, and not dependent on the selection of $\omega.$ I think it is plausible that the sup-condition follows from the diffeomorphism-invariance of $T$ anyway. The idea of a proof could be to use a triangulation to decompose $M$ into small disjoint regions which are mutually diffeomorphic to each other, and to write $\nu$ as the sum of its restrictions to each region. Then $\nu$, using $\omega$ as a reference, could be well-approximated by a step function with supports given by the small regions, with $T$ applied to such a form being nicely comparable with the integral of the form, using linearity and diffeomorphism-invariance
