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Assume that $\mu$ is a probability measure on $\mathbb Z$ with some given properties. I now want to show that $\mu$ is strictly positive.

I found a proof which starts like this:

Assume $\mu$ is not strictly positive. Then there exists a $x_0$ such that $\mu(x_0)=0$ and $\mu(x_0+1)+\mu(x_0-1)>0$

M question: a measure is called strictly positive if every non-empty open subset has strictly positive measure (Wikipedia). As we are on $\mathbb Z$ every set should be open. So is "strictly positive" here just another way of saying "takes only positive values" (except for empty set)?

And I don't understand how they concluded the second part, why can't $\mu(x_0+1)$ and $\mu(x_0-1)$ have measure $0$?

PeterGarder
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    It would be helpful to state what your "given properties" are, otherwise we don't really have much hope of working out why such an $x_0$ exists. – Jason Jun 07 '17 at 21:25
  • @Jason You are right,but it's from a bigger proof I didn't know how to state them compactly without a lot of other unnecessary informmation. I mostly wanted to check if these conclusions follow diretly from the definition of "not strictly positive" – PeterGarder Jun 07 '17 at 21:32
  • I misread the condition initially - there is indeed enough information. – Jason Jun 07 '17 at 21:33

2 Answers2

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Maybe, they mean the open set $(x_0-1,x_0+1)\cap \mathbb Z$ which is basically a singleton, and must have positive measure. This would give a contradiction.

Juanito
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They said "there exists $x_0$". First since the measure is not zero, then there exists $y$ with positive measure and we assume there exists $z$ with zero measure. now take the largest $x_0$ such that $\mu (x_0)=0$ and $x_0$ is between y and z this $x_0$ must satisfy the condition.

Yanko
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    This is not quite correct - it is possible that $\mu$ is both positive and zero infinitely many times. To get around this, note that there exists $x_1,x_2$ such that $\mu(x_1)=0$ (by assumption) and $x_2$ such that $\mu(x_2)>0$ (otherwise $\mu\equiv0$, which is not a probability measure). Now your trick works if you restrict attention to those integers between $x_1$ and $x_2$, inclusive. – Jason Jun 07 '17 at 21:40