I am interested in studying the following problem:
\begin{align}
\sup\_{\mu \in \mathcal{D} } \int\_{\mathbb{R}} f(x) d\mu(x)
\end{align}
where $\mu$ is a probability measure. Assume that $\mathcal{D}$ is closed, covex and compact (in weak^{\*} topology). 


We know that if $f(x)$ is continuous and bounded then $\mu \to \sup\_{\mu \in \mathcal{D} } \int\_{\mathbb{R}} f(x) d\mu(x)$ is a continous functional and 
\begin{align}
\sup\_{\mu \in \mathcal{D} } \int\_{\mathbb{R}} f(x) d\mu(x)&=\max\_{\mu \in \mathcal{D} } \int\_{\mathbb{R}} f(x) d\mu(x), \text{ this steps is due to contintinuity,}\\
&=\max\_{\mu \in \text{Extrem Points of} \mathcal{D} } \int\_{\mathbb{R}} f(x) d\mu(x), \text{ this step is due to linearity of the functional,}\\
\end{align}


**My question:** Now, assumes that $f(x)$ is continuous and positive but not bounded from above. So, we can not claim that $\mu \to \sup\_{\mu \in \mathcal{D} } \int\_{\mathbb{R}} f(x) d\mu(x)$ is a continous functional.


However, can we still say that 


\begin{align}
\sup\_{\mu \in \mathcal{D} } \int\_{\mathbb{R}} f(x) d\mu(x)=\sup\_{\mu \in \text{Extrem Points of} \mathcal{D} } \int\_{\mathbb{R}} f(x) d\mu(x),
\end{align}


from the linearity?


The example I have in mind is $f(x)=x^2$. 



Maximizing linear function (not necessarily continuous) over a compact, closed and convex domain
The answer is yes. 
Call ${\mathcal E}$ the set of extreme points of ${\mathcal D}$.
It is a consequence of [Choquet's theorem](https://en.wikipedia.org/wiki/Choquet_theory) that for every $\mu\in{\mathcal D}$ there exists a probability measure $\nu=\nu\_\mu$ on ${\mathcal E}$ such that $\mu=\int\_{\mathcal E}\rho d\nu(\rho)$ and in particular
$$\int\_{\mathbb R}f(x)d\mu(x)=\int\_{\mathcal E}\left(\int\_{\mathbb R}f(x)d\rho(x)\right)d\nu(\rho)\leq\sup\_{\rho\in{\mathcal E}}\int\_{\mathbb R}f(x)d\rho(x)$$


