The Grothendieck Group of the Monoid of Measures

Let $f:(X,\mathrm{\Sigma })\to (Y,\mathrm{\Omega })$$f:(X,\Sigma)\to (Y,\Omega)$ be a morphism in $\mathrm{Meas}$$\mathrm{Meas}$.

We could define a functor $\mathcal{M}:\mathrm{Meas}\to \mathrm{Mon}$$\mathcal M:\mathrm{Meas}\to\mathrm{Mon}$ as follows:

For a measurable space $(X,\mathrm{\Sigma })$$(X,\Sigma)$, we define $\mathcal{M}(X,\mathrm{\Sigma })$$\mathcal M(X,\Sigma)$ be all the measure on it. It forms a monoid.

For $\mu ,{\mu }^{\prime }\in \mathcal{M}(X,\mathrm{\Sigma })$$\mu,\mu'\in\mathcal M(X,\Sigma)$, we define $(\mu +{\mu }^{\prime })(E)=\mu (E)+{\mu }^{\prime }(E)$$(\mu+\mu')(E)=\mu(E)+\mu'(E)$.

Easy to check that $\mu +{\mu }^{\prime }\ge 0,(\mu +{\mu }^{\prime })(\mathrm{\varnothing })=0$$\mu+\mu'\ge 0,(\mu+\mu')(\empty)=0$, also we have

$$(\mu +{\mu }^{\prime })\left(\coprod _{i=1}^{\mathrm{\infty }}{E}_i\right)=\mu \left(\coprod _{i=1}^{\mathrm{\infty }}{E}_i\right)+{\mu }^{\prime }\left(\coprod _{i=1}^{\mathrm{\infty }}{E}_i\right)=\sum _{i=1}^{\mathrm{\infty }}\mu (E_i)+\sum _{i=1}^{\mathrm{\infty }}{\mu }^{\prime }(E_i)=\sum _{i=1}^{\mathrm{\infty }}\mu (E_i)+{\mu }^{\prime }(E_i)=\sum _{i=1}^{\mathrm{\infty }}(\mu +{\mu }^{\prime })(E_i)$$

For $\mathcal{M}(f):\mathcal{M}(X,\mathrm{\Sigma })\to \mathcal{M}(Y,\mathrm{\Omega })$$\mathcal M(f):\mathcal M(X,\Sigma)\to\mathcal M(Y,\Omega)$, it is defined by

$$\mu ⟼\mu \circ f^{-1}$$

Easy to see that $(\mu +{\mu }^{\prime })\circ f^{-1}=\mu \circ f+{\mu }^{\prime }\circ f$$(\mu+\mu')\circ f^{-1}=\mu\circ f+\mu'\circ f$.

Then we could compose it with Grothendieck Group functor.

We get a functor from $\mathrm{Meas}\to \mathrm{Ab}$$\mathrm{Meas}\to\mathrm{Ab}$.

We could talk about

$$0⟶K_0(\mathcal{M}(X,\mathrm{\Sigma }))⟶K_0(\mathcal{M}(Y,\mathrm{\Omega }))⟶K_0(\mathcal{M}(Z,\mathrm{\Delta }))⟶0$$

is exact or not.

For which measurable space $(X,\mathrm{\Sigma })$$(X,\Sigma)$, $K_0(M(X,\mathrm{\Sigma }))$$K_0(M(X,\Sigma))$ will be a free/projective/injective/flat... Module?