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Let $X$ be a topological space with a Borel $\sigma$-algebra $\mathcal B(X)$. There is a family of probability measure on $X$, which is denoted as $P:X\times \mathcal B(X)\to[0,1]$.

I would like to consider a family with a following property:

For any non-empty open set $B$ there exists $x\in X$ such that $P(x,B)>0$.

Roughly speaking, the support of family of measures $\{P(x,\cdot)\}_{x\in X}$ is the whole space $X$. I wonder if it can be said formally. It is not irreducibility of a Markov chain, since the chain can admit this property being reducible and may not admit this property being irreducible.

The support of measure is defined as $$ \operatorname{supp}\mu = \{x\in X:\mu(U(x))>0\forall \text{ open neighborhood }U(x) \}. $$

I guess that the support of family of measures should be defined like this $$ \operatorname{supp}P:=\overline{\bigcup\limits_{x\in X}\operatorname{supp}P(x,\cdot)} $$ but first, it's not clear if $\operatorname{supp}P=X$ is equivalent to my condition, second, I had no chance to find this definition of $\operatorname{supp}P$ in the literature and finally this definition seems a bit weird cause there is an infinite union of closed sets.

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Just for a fixed $x$, so you have a single probability measure $Q=P(x,\cdot)$, what do you mean by the support? If you let $S$ be the intersection of all closed sets $E$ with $Q(E)=1$, then I don't think it follows that $Q(S)=1$. So, there does not have to be such a thing as the minimal closed set on which $Q(S)=1$, which is how you might want to define the support. There does, if you require $Q$ to be inner regular on open sets. –  George Lowther Sep 24 '11 at 1:44
    
But, if $P(x,\cdot)$ does have a support for every $x$, then there will be a minimal closed set $S$ on which $P(x,S)=1$ for all $x$, which is exactly what you wrote for ${\rm supp}P$. Equivalently, $S$ is closed such that $P(x,S)=1$ for all $x$ and, for every open $U$ with $U\cap S\not=\emptyset$, there is an $x$ with $P(x,U\cap S) > 0$. Your condition is equivalent to $S=X$. Well, it would be if you had required $B$ to be a nonempty open set. –  George Lowther Sep 24 '11 at 1:53
    
@George: Thanks for the correction and for the attention - I fixed the non-emptyness of $B$, also added the definition of support of measure I meant. I didn't understand your first sentence in the last comment: which support do you mean? –  Ilya Sep 24 '11 at 10:57
    
I was thinking of the support of a probability measure as the smallest closed set with probability 1. This does not exist in general. You could define the support as the intersection of all closed sets with probability 1 - as (in different words) Wikipedia does. In general, though, this does not have to have probability 1. –  George Lowther Sep 27 '11 at 23:40
    
Regards your question, "support of a family of measures" seems like a good description of what you define. I'm not aware of the term being used elsewhere though. –  George Lowther Sep 27 '11 at 23:43

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