Probabilities Measures

Question

Consider then a convex subset of probability measures $A$ that is closed relative to the set of all probability measures. I'm wondering if $\forall \mu \in$ closure($A$), does there exists a positive measure $\nu$ such that $\mu + \nu \in A$?
It is well known that $C^\*_0(\mathbb{R})$ (the continuous dual space of $C_0(\mathbb{R})$, which is all continuous functions on $\mathbb{R}$ that vanish at $\pm \infty$) can be identified with the space of all regular signed measures. Equip this space with the weak-* topology, i.e. where $\mu_n$ converges weakly to $\mu$ if $\int f d\mu_n \rightarrow \int f d\mu$ for all $f \in C_0(\mathbb{R})$. I'm looking at the set of all probability measures in this space (positive measures for which $\mu(\mathbb{R}) = 1$). This set is not closed in the weak-* topology, since sequences such as $\delta_n \to 0$ (zero measure) as $n \to \infty$.

Consider then a convex subset of probability measures $A$ that is closed relative to the set of all probability measures. I'm wondering if $\forall \mu \in$ closure($A$), does there exists a positive measure $\nu$ such that $\mu + \nu \in A$?

Answer 1

This is false. Here's a counterexample.

For $n \geq 2$, let $\mu_n$ be the probability measure $\mu_n = (1/n)\delta_0 + (1/2 - 1/n)\delta_1 + (1/2)\delta_n$. The convex hull of $\{\mu_n: n \geq 2\}$ is contained in the set $P$ of probability measures; let $A$ be its closure in $P$ for the relative weak* topology. Then I claim that (1) the measure $\mu = (1/2)\delta_1$ is in the weak* closure of $A$, but (2) there is no positive measure $\nu$ such that $\mu + \nu \in A$.

Well, (1) is clear; $\mu$ is the weak* limit of the sequence $(\mu_n)$. For (2), suppose you had a net $(\mu_\alpha)$ of convex combinations of the $\mu_n$ which converged weak* to a probability measure of the form $\mu + \nu$ with $\nu$ positive. Then by examining the coefficients of $\delta_1$, we can show that mass has to be escaping to infinity as we take the limit in $\alpha$, contradicting the assumption that $\mu + \nu$ is a probability measure. That's the idea; I think I'll leave it to you to work out the details.

Answer 2

Let A be the set of all probability measures supported in (0,1). Then $\delta_1$ (unit mass at 1) is in the weak-* closure of A, but there is no measure in A that is greater or equal to $\delta_1$.