# Probability measures and convex combination

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Let $\mathcal B(\Omega)$ be the Borel algebra of $\Omega$ (the σ-algebra of Borel sets in $\Omega$). I understand that if we define a "convex combination" of probability measures by $$\bigg(\sum_{k=1}^n w_k\mu_k\bigg)(E)=\sum_{k=1}^n w_k\mu_k(E),$$ then every convex combination of probability measures is a probability measure. I'm particularly interested in the probability measures defined in the following way: For each $s\in\Omega$, we define $$\mu_s(E)=\chi_E(s)=\begin{cases}1 & \text{ if }s\in E\\ 0 & \text{ if }s\notin E\end{cases}$$
Let $S_0$ be the set of all probability measures defined this way, and let $S$ be the set of all convex combinations of members of $S_0$. I have found that S is closed under convex combinations.

My question is: Are there any probability measures on $\mathcal B(\Omega)$ that aren't members of $S$?

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Not sure if I understand your question. The $\mu_s$ that you define are the Dirac measures, and you want to know if there are probability measures not in the convex hull of the Dirac measures, right??

Well, to give you a counterexample: let $\Omega=[0,1]$ equipped with the Borel-sigma-algebra. Then the Lebesgue measure is an example of measure that is not in your convex hull.

Your post raises interesting questions though, so I will go into some more depth:

Let $\Omega$ be a locally compact Hausdorff space. Then there is an interesting result known as the Riesz representation theorem. This says that the set of all regular Borel measures on $\Omega$ is bijective with the positive bounded linear functionals on $\mathcal{C}_0(\Omega)$.

That is, given $\mu$ a regular Borel measure, then

$$\mathcal{C}_0(\Omega)\rightarrow \mathbb{R}:f\rightarrow \int fd\mu$$

is a positive bounded linear functional. (positive just means that a positive function f is sent to a positive number in $\mathbb{R}$).

So to study all the (regular) measures, it suffices to study the positive functionals on $\mathcal{C}_0(\Omega)$. And to study the probability measures, it suffices to study the positive functionals with norm 1. These functionals are called states.

Now, your Dirac measures correspond to the functionals

$$ev_s:\mathcal{C}_(\Omega)\rightarrow \mathbb{R}:f\rightarrow f(s)$$

These are not only linear functionals, but also multiplicative! That is, they satisfy $ev_s(fg)=ev_s(f)ev_s(g)$. Such a functionals are called pure states.

Note that we can put a topology on $(\mathcal{C}_0(\Omega))^\prime$: the weak*-topology!

Now, one can prove (use the Krein-Milman theorem), that the set of all states is the weak*-closure of the the convex hull of all the pure states.

That is: gives your Dirac probability measures and take the convex hull. Then you have almost all the probability measures. You only need to take the weak*-closure of them.

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Edit: I wrote this before I saw your second post. I will read the second post now, and reply later (maybe tomorrow).

Not sure if I understand your question. The $\mu_s$ that you define are the Dirac measures, and you want to know if there are probability measures not in the convex hull of the Dirac measures, right??

Well, to give you a counterexample: let $\Omega=[0,1]$ equipped with the Borel-sigma-algebra. Then the Lebesgue measure is an example of measure that is not in your convex hull.
Thanks. Yes, that's what I meant. I guess it was much simpler than I thought. (I did expect the answer to be "yes". I just wanted to make sure).

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