Extending a function that gives a value to convex functions to a measure I am wondering if such a result exists (or similar) and or if there is a "simple" proof.
Let $\mathcal X$ be a bounded and closed subset of a topological vector space, let $\Sigma$ be the Borel $\sigma$-algebra associated and let $\Gamma$ be the set of bounded convex continuous function on $\mathcal X$, suppose that the topology on $\mathcal X$ is the weakest topology such that all functions in $\Gamma$ are continuous. Suppose we have $\mu:\Gamma\to\mathbb R$ that is such that
\begin{align*}
\mu(1)&=1\\
\forall f,g\in\Gamma \text{ such that }f\leq g,~~~\mu(f)&\leq \mu(g)\\
\sum_{i\in\mathcal I}a_i\mu(f_i) &=\mu\left( f \right)
\end{align*}
Whenever $\sum_{i\in\mathcal I} a_i f_i=f\in\Gamma$ for $f_i$ in $\Gamma$, $a_i \geq 0$ and $\mathcal I$ some countable set. Here $\left(\sum_{i\in\mathcal I} a_i f_i\right) (x)=\sup\left\{ \sum_{j\in \mathcal J} a_j f_j(x) : \mathcal J\subseteq \mathcal I,|\mathcal J|<\infty\right\}$

There exists a unique probability measure $\mu:\Sigma\to\mathbb R$ such that when seen as a function from $\Gamma$ to $\mathbb R$ through Lebesgue integration, $\mu$ matches with the previous function.

The reason why I think this may be true is that it is known that the Choquet ordering is a partial order on a convex closed bounded set (see Definition 2.9 and Proposition 2.10 in this), this would give uniqueness so I think we have to argue about existence. We may need to add extra assumption, my goal here is to try to find necessary and sufficient condition for extitence of such a probability measure.

Maybe one way to go would be to try to use Caratheodory's extension theorem on a pre-measure that we define on the topology of $\mathcal X$. For instance I am thinking that it could be that for a closed set $A$, $\mu(A)=\inf\{ \mu(f)-\mu(g) : f,g\in\Gamma ~s.t.~f-g\geq \mathbf 1_A  \}$ where $\mathbf 1_A$ denotes the indicator function of $A$. If this is a good candidate (maybe we have to change into $\inf$) then we can define the measure on open sets $A$ as $\mu(A)=1-\mu(A^c)$ and then extend this pre-measure to the Borel $\sigma$-algebra using Caratheodory's extension theorem. We then have to show that this measure integrates to the right value on $\Gamma$ and conclude using Proposition 2.10 of the previous document.

An argument proposed by @gerw is that we can extend $\mu$ to be a linear functional on the closure of $\Gamma-\Gamma$ which may match with the set of measurable functions (or at least contain all $1_A$ for $A$ open or close) and then by the Riesz representation theorem there is a probability measure that represents that functional.

I added the condition that if $f\leq g$ are in $\Gamma$, then $\mu(f)\leq \mu(g)$. this will indeed be true if $\mu$ comes from a probability measure and it is possible to build values for $\mu(f)$ for any $f\in\Gamma$ such that the other two axioms are satisfied but not this one. An example is when $\mathcal X=[0,1]$, then all convex function are countable sums of the functions of the type $h_p:x\to |x-p|$ and adding the $1$ function, but if we fix any probability measure $\mu$ and take the mapping $\mu:\Gamma \to \mathbb R$ and define it to be the same on $h_p$ except for instance for $h_{1/2}$ then extend the values to $\Gamma$ linearly, we have $\mu(1)=1$ and $\sum_{i\in\mathcal I} a_i \mu(f_i)=\mu(f)$ but for sure there is no probability measure.
 A: Here is an attempt when $\mathcal X$ is compact, in order to generalize to closed and bounded, the only details missing are to show that $f'$ is continuous whenever $f$ is continuous and bounded as well as the fact that Riesz-Markov-Kakutani's theorem doesn't apply (there may be a generalization which I am not aware of). All of this is probably very related to order dual and order bound, not sure if there is a way to express it in this framework. Any improvement of this answer or other answer would be most appreciated.
We can extend $\mu:\Gamma\to\mathbb R$ to $\mu:\Gamma-\Gamma\to\mathbb R$ trivially as $\mu(f_1-f_2)=\mu(f_1)-\mu(f_2)$. Now the function is linear (and even countably linear) and satisfy $\mu(f)\geq 0$ for all $f\in\Gamma-\Gamma$ such that $f\geq 0$. Of course we still have $\mu(1)=1$.
The next step is to extend $\mu:C_c(\mathcal X)\to\mathbb R$, we can do this by using Hahn-Banach separation theorem (here the sublinear function we use is important later on to prove we have a probability distribution), It is clear that $\Gamma-\Gamma$ is a subspace of $C_c(\mathcal X)$ and we can use the sublinear function $p:f\to \mu(f')$ where $f'(p)=\sup \{ \nu(f):\nu\sim x \}$ where the supremum is taken over all Radon measure that averages to $x$ in the sense of the Bochner integral. This function is clear concave (it is actually the concave envellope on compact sets) and I think it should be continuous but couldn't prove it (if any one knows how to do that I would really appreciate it). The thing is that $p$ is really defined on $C_c(\mathcal X)$ and is sublinear such that $\mu(f)\leq p(f)$ for any $f\in\Gamma-\Gamma$ (because $f\leq f'$). Applying the Hahn-Banach theorem gives us a linear function $\mu:C_c(\mathcal X)\to\mathbb R$  such that it matches with the previous one on $\Gamma-\Gamma$ and $\mu\leq p$. If $f\in C_c(\mathcal X)$ is such that $-f\geq 0$, then $\mu(f')= p(f)\geq \mu(f)$ but clearly $f'\leq 0$ since $f\leq 0$ and so $f\geq 0$ implies $\mu(f)\geq$. By Riesz-Markov-Kakutani's theorem we get a a unique positive Radon measure $\nu$ on $\mathcal X$ such that $\nu(f)=\mu(f)$ for all $f\in C_c(\mathcal X)$, but we still have that $\nu(1)=\mu(1)=1$ and so this is our final goal.
