3

We know that probability measures are tight if the metric space is separable and complete.

Here tight means there exists a compact set in that metric space say $K$ such that $P(K) > 1- \epsilon$.

I want to create a probability measures which is not tight. For that we have to violates the separability or completeness condition. Suppose we violates separability. And consider the space $l_{\infty}$ with respect to supremum norm. We know that this space is not separable. But how to construct an probability measures here?

If we violates completeness then also this holds. As a example $c_{00}$ space is not complete. Then how to construct probability measures there?

Any kind of simple examples are appreciated.

I know there is some explanation and example available in the stack exchange and those are talking about 'left limit topology' kind of things. I need an simple example and construction not that much advance that's why asked this question.

Cantor_Set
  • 1,058
  • I don't think a simple example exists. – Kavi Rama Murthy Apr 24 '22 at 07:19
  • 1
    It is consistent with the usual axioms on set theory that every Borel probability measure is supported on a separable set. Under some set-theoretic axioms (existence of a measurable cardinal), it is consistent that there exists a probability measure on the Borel $\sigma$-algebra generated by the discrete metric that assigns measure zero to singletons. Such a measure would not be regular. It is then also consistent that there exists a counterexample in every uncountable complete metric space. – Michael Greinecker Apr 24 '22 at 07:52

2 Answers2

4

Separable (but not complete) metric space with a measure that is not tight.

Perhaps Murthi is right, that there is no simple example. So I will provide a not-so-simple example.

Let $B \subseteq [0,1]$ be a Bernstein set. This means: $$ \text{for every uncountable closed set $F \subseteq [0,1]$, we have both $F\cap B \ne \varnothing$ and $F\setminus B \ne \varnothing$.} \tag1$$ (A Bernstein set $B$ exists according to ZFC.) Our separable metric space is $B$ with the restriction of the usual metric $|x-y|$ of $[0,1]$.

Let $\lambda^*$ Lebesgue outer measure on $[0,1]$. The Bernstein set $B$ has $\lambda^*(B) = 1$, so defining $\mu$ by $$ \mu(E) = \lambda^*(E)\quad\text{for all } E \in \operatorname{Borel}(B) \tag2$$ gives us a probability measure $\mu$ on $B$. But $\mu$ is not tight ... Indeed, let $K \subseteq B$ be compact. Then by $(1)$, $K$ is countable, so by $(2)$, $\mu(K) = 0$.

GEdgar
  • 111,679
3

$c_{00}$ is an increasing union of subsets $K_n$ where $K_n$ consists of sequences where only the first $n$ elements are allowed to be nonzero. Thus for every probability measure on $c_{00}$ and every $\epsilon>0$, there is some $n$ such that $P(K_n)>1-\epsilon$. So that space will not yield an example.

Every Borel probability measure on a metric space has a separable support, see Lemma 2.1 in https://arxiv.org/pdf/1612.03213.pdf But whether this support has full measure is more delicate.

Yuval Peres
  • 21,955