Rig module (?) of measures with scalar multiplication given by Lebesgue integration Let $R$ be a rig. A rig module $M$ on $R$ is a module on $R$ except there may be no additive inverses, yet $0\cdot v = 0$.
Let $(E, \mathcal E)$ be a measurable space and $\mathcal B$ be the Borel-$\sigma$-Algebra of the extended reals $[-\infty\, ..\infty]$. Now let $R$ be the set of all measurable, nonnegative functions $(E,\mathcal E) \to ([-\infty\, ..\infty], \mathcal B$). Equipped with usual structure this is a rig. 
Thinking naively, a measurable space seems to yield a rig module of measures in the following way:
Let $M$ be the set of all measures on $(E,\mathcal E)$ equipped with usual addition and scalar multiplication $\cdot : R\times M \to M$ given by the Lebesgue integral:
$$f\cdot \mu := \int\limits \mathbb{1}_{\_}\cdot f\,d\mu$$
($\mathbb{1}_{\_}$ maps measurable sets to their indicator function; this is known to be a measure again).
Then indeed $(M,+)$ is a commutative monoid and:


*

*$(f+g)\cdot \mu = f\cdot \mu + g\cdot \mu$ (since integrals are linear)

*$f\cdot(\mu + \nu) = f\cdot \mu + f\cdot \nu$ (by this)

*$(f\cdot g)\cdot \mu = f\cdot (g\cdot \mu)$ (by a theorem concerning "indefinite integrals", I don't know the name)

*$1\cdot \mu = \mu$

*$0\cdot \mu = 0$


i.e. $M$ is an $R$-Rig module. 

Is this really true or am I missing something here?

 A: The reason why you cannot answer the question yourself is the fact that you have artificially restricted the objects that you are working with to only object that have an intuitive meaning. Mathematics always evolves by extending the conceptual framework to include less intuitive objects, with the advantage that this makes certain results conceptually very elegant (think about jumping from $\Bbb R$ to $\Bbb C$ or from the affine space to the projective one).
In our case, your question admits a very natural answer provided that we accept to change a few things.


*

*Since we want a module, and the multiplication of a measure by a negative function is no longer a measure, we'll have to use signed measures; if you want to work with complex-valued functions, then you'll obviously have to work with complex measures.

*Working with signed objects raises a problem: what should be the result of multiplying the function $x \mapsto \begin{cases} -1, & x \le 0 \\ 2, & x > 0 \end{cases}$ with the usual Lebesgue measure on $\Bbb R$? How would you define $\infty - \infty$? In order to avoid these annoying infinities, we restrict ourselves to working with only finite measures.

*Finally, we want to come up with a space of functions that should be integrable with respect to every such measure as chosen above. If you take $E$ to be a topological space, and if you choose $\mathcal E$ to be the $\sigma$-algebra generated by the topology, in order to make all the continuous functions measurable, then the space $C_b (E)$ of continuous, bounded functions on $E$ is the natural choice, because all its elements are automatically integrable, with finite integrals.
Call the resulting space of measures $\mathcal M (E)$ - it can be shown to be a Banach space over $\Bbb R$ (or $\Bbb C$, depending on what you work with).
Now, the multiplication
$$(f \mu) (S) = \int _S f \ \Bbb d \mu = \int 1 _S f \ \Bbb d \mu$$
is perfectly well defined and has all the properties that you ask for. In the complex case, defining $\mu ^* (S) = \overline {\mu (S)}$ will make $\mathcal M (E)$ a $C_b (E)$- $*$-module.
If you want to have even more fun, take $E$ to be a topological group, in which case you may also defined a multiplication of measures (called "convolution") by
$$(\mu * \nu) (S) = \int \left( \int 1_S (xy) \ \Bbb d \mu (x) \right) \Bbb d \nu (y) .$$
This will make $\mathcal M (E)$ into a $C_b (E)$- $*$-algebra.
The important thing to remember here is that sometimes, in order to answer a mathematical question, one has to change the conceptual framework, by intelligently extending some concepts while, at the same time, restricting others, until a natural mathematical construction is obtained.
A: Every nonnegative measurable function is a monotone limit of simple functions. By monotone convergence it follows that
$$
\int f\,d(g\cdot \mu)=\int fg\,d\mu
$$
and similarly
$$
\int f\,d(\mu+\nu)=\int f\,d\mu+\int f\,d\nu
$$
for all $\mu,\nu\in M$, $f,g\in R$ (both equalities hold by definition for indicator functions $f$).
Now you can easily verify all bullets by integration over nonnegative measurable functions. For example
$$
\int\phi\,d((fg)\cdot\mu)=\int \phi f g\,d\mu=\int \phi f\,d(g\cdot\mu)=\int\phi\,d(f\cdot(g\cdot\mu))
$$
for all $f,g,\phi\in R$, $\mu\in M$ etc.
