Lower Expectation Let $X$ be, for simplicity, a finite set (with the discrete topology).
Denote with $M(X)$ the set of probability measures on $X$ endowed with the weak topology.
For $\mu\in M(X)$  and a (necessarily measurable) function $f:X\rightarrow[-1,1]$ denote with $E_{\mu}(f)$ the expected value of $f$.
For a (closed) set $A\subseteq M(X)$, define the lower expectation of $A$, denoted by $E(A)$, as the following functional of type $(X\rightarrow [-1,1])\rightarrow [-1,1]$:
$$E(A)(f)= \displaystyle \inf \{ E_{\mu}(f) \ | \  \mu \in A \}.$$
For a (closed) set $A\subseteq M(X)$ denote with $H(A)$ its convex hull of $A$, defined as expected.
It is easy to see that:
Proposition: For all $A,B\subseteq M(X)$, if $H(A)=H(B)$ then $E(A)=E(B)$.
Now my question is about the inverse direction of the previous statement.
QUESTION 1: Is it true that for $A,B\subseteq M(X)$, if $E(A)=E(B)$ then $H(A)=H(B)$?
QUESTION 2: What about restricting attention to functions $f$ of type $X\rightarrow [0,1]$?
Remark: note that, restricting even further to characteristic functions $f:X\rightarrow\{0,1\}$, the statement of QUESTION $1$ is not true anymore and the following is an example:
Example. Consider $X=\{a,b,c\}$, $\mu_{1}= \{ a\mapsto 0.3, b\mapsto 0.3, c\mapsto 0.4\}$, $\mu_{2}=\{ a\mapsto 0.4, b\mapsto 0.3, c\mapsto 0.3\}$ and $\mu_{3}=\{a\mapsto 0.5, b\mapsto 0.4, c\mapsto 0.1 \}$. Now consider $A=\{\mu_{1},\mu_{2}\}$ and $B=\{\mu_{1},\mu_{2},\mu_{3}\}$. Now $H(A)\neq H(B)$ because $\mu_{3}$ is not a convex combination of $\mu_{1}$ and $\mu_{2}$. Yet, for every set $Y\subseteq X$ (i.e., function $f:X\rightarrow\{0,1\}$, it holds that $E(A)(Y)=E(B)(Y)$.
 A: There is a one-to-one correspondence between convex sets of probability distributions and affinely superadditive lower expectations (or lower previsions in Walley's terminology). You should check http://sites.poli.usp.br/p/fabio.cozman/research/credalsetstutorial/introduction/node5.html
A: The first question has already been answered in the comments and by @Kenny: yes! Taking the convex closure of the set of probability measures has no influence on the lower expectation, so  $E(A)=H(A)$ (and also $E(B)=H(B)$). Therefore, $E(A)=E(B)$ implies $H(A)=H(B)$ because the functionals $E$ and $H$ are identical.
As for the second question, the answer is yes too.
To see why, choose any variable $g\colon X\to\mathbb{R}$. Note that $g$ is bounded, and in particular has a minimum and a maximum, because $X$ was assumed to be finite.
Let \begin{align}
\alpha&:=\min g \\
\beta&:=\max g - \min g \\
f&:=\begin{cases}
\frac{g-\alpha}{\beta} & \text{if $\beta>0$} \\
0 & \text{if $\beta=0$}
\end{cases}
\end{align}
Note that $f\colon X\to[0,1]$ and $g=\alpha+\beta f$ with $\beta\ge 0$.
Also,
$$E(A)(g)=E(A)(\alpha+\beta f)=\alpha +\beta E(A)(f)$$
In other words, $E(A)$ is completely determined by its restriction to variables of the form $f\colon X\to [0,1]$.
Now assume $E(A)(f)=E(B)(f)$ for all $f\colon X\to [0,1]$. Then for any $g\colon X\to\mathbb{R}$, by the previous part, we know that there are $f\colon X\to [0,1]$, $\alpha\in\mathbb{R}$ and $\beta\in\mathbb{R}_{\ge 0}$ (as constructed before) such that $g=\alpha+\beta f$, and
$$E(A)(g)=\alpha +\beta E(A)(f)=\alpha +\beta E(B)(f)=E(B)(g)$$
Note that we never used countable additivity. So these arguments also extend to infinite $X$, bounded variables (using $\sup$ and $\inf$ instead of $\max$ and $\min$), and closed sets of finitely additive probability charges (under an appropriate topology).
