# Supremum of the integral over the compact and convex set belongs the extreme set

Let $$T:X \to X$$ be a continuous function on a compact metric space. We say that $$\mu$$ is $$T-$$invariant if $$\mu(T^{-1}(A))=\mu(A)$$ for all $$A \in \mathbb{B}_{X}$$. We denote by $$M(X,T)$$ the space of all $$T-$$invariant measure which is a nonempty, compact and convex in weak* topology. Let $$f:X \to \mathbb{R}$$ be a continuous function. Denote $$G(f)=\sup_{\mu \in M(X,T)} \int f d\mu$$.

$$\textbf{Question}:$$ Why the supremum is always attained by a measure that belongs to the extreme set $$M(X,T)$$?

My attempt: I think that is related to the fact $$M(X,T)$$ is compact and convex in weak* topology, but I don't know why the measure must belong the extreme set $$M(X,T)$$?

• What do you know about extreme sets? Have you tried using the definition? Apr 23, 2021 at 13:22
• @supinf : Let $\mu=\lambda \mu_{1}+(1-\lambda)\mu_{2}.$ Take integral $\int f \mu= \lambda \int f d\mu_{1}+(1-\lambda)\int f\mu_{2}$, and then take supremum over $M(X,T)$. I don't know what I can get now.
Apr 23, 2021 at 13:35
• @supinf : regarding extreme sets, I know a couple of theorems and definitions. I feel I should use Krein–Milman, but I don't know where I missed it. I would appreciate it if you could help me.
Apr 23, 2021 at 13:43

First, you are right about the compactness. A continuous function over a (nonempty) compact set always obtains a maximum and a minimum. In this case, the set is compact in the weak-$$*$$ topology, and the function $$h:C(X)\to\Bbb R,\qquad f\mapsto \int f\,\mathrm d\mu$$ is continuous in the weak-$$*$$ topology, so this fits together.
Consider the subset of $$M(X,T)$$ where the maximum is obtained. This subset is again weak-$$*$$ compact and convex. Pick an extremal point of this subset (here you have to use Krein-Milman, to argue that such an extremal point exists). This should also be an extremal point of $$M(X,T)$$.