$\mathcal F$ is set of events. Can you give an example of some algebra $\mathcal A$ over $\mathbb N$ and a non-zero finitely additive measure $\mu$ on this algebra, which has a countably additive extension to the $\sigma$-algebra generated by this algebra, and moreover, such that when shifting any set $A ∈ \mathcal F$ by an integer $n$ (calling the resulting set $A + n$), the following equation is fulfilled: $A + n ∈ \mathcal{A}$, $\mu (A + n) = \mu (A)$?
3
-
2$\begingroup$ There is only one translation-invariant countably additive measure on $\mathbb{N}$, up to scaling: the counting measure. Additionally, why not just take $\mathcal{A}$ as the full $\sigma$-algebra itself? $\endgroup$– user44191Commented Dec 3, 2018 at 10:10
-
$\begingroup$ Are you requiring $\mu({\bf N})=1$, i.e.\ some kind of left-invariant mean? $\endgroup$– Yemon ChoiCommented Dec 4, 2018 at 0:18
-
1$\begingroup$ If you consider the counting measure on $\mathbf{Z}/k\mathbf{Z}$ and consider its inverse image on $\mathbf{Z}$ (defined on the $\sigma$-algebra $\mathcal{A}$ consisting of subsets invariants by $\pm k$), then you get such a measure, but on $\mathbf{Z}$. After intersecting with $\mathbf{N}$, you get a measure almost as required, except the last axiom. Indeed, writing $T(n)=n+1$, your last axiom is that $A$ measurable implies $T(A)$ measurable and $\mu(T(A))=\mu(A)$. What holds here is that $A$ measurable implies $T^{-1}(A)$ measurable and $\mu(T^{-1}(A))=\mu(A)$. $\endgroup$– YCorCommented Dec 5, 2018 at 20:45
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
There is no such measure.
Assume there is. Then we must have that $\mathbb{N} \in \mathcal{A}$. We can translate by $1$ to get $\mathbb{N} + 1 \in \mathcal{A}$. But $\mathbb{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. By translation again, $\mu(\{n\}) = \mu(\{1\})$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \mu(\{1\}) \sum_{a \in A} 1 = |A| \mu(\{1\})$. Therefore, any measure that is translation-invariant and countably additive must be the counting measure (up to scaling). Further, if this measure is finite (that is, if $\mu(\mathbb{N}) < \infty$, then it must be zero.
$\begingroup$
$\endgroup$
3
Here is an example. Let $\mu:{\cal P}(\mathbb{N})\to[0,1]$ be defined by $$\mu(A) := \lim\inf_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} \text{ , for any }A\subseteq \mathbb{N}.$$
answered Dec 3, 2018 at 9:24
-
5$\begingroup$ Your algebra $\mathcal{A}$ is $\mathcal{P}(\mathbb{N})$, right? But this is already a $\sigma$-algebra, so the extension of $\mu$ is $\mu$ itself which is not countable additive, or do I miss something? $\endgroup$– DirkCommented Dec 3, 2018 at 9:36
-
4$\begingroup$ For the record, $\mu$ is not finitely additive. We can build a subset $A$ of $\mathbb{N}$ such that the limsup is strictly larger than the liminf. Then, because $1-x$ is an order-reversing isomorphism of $[0,1]$, $$\mu(\mathbb{N}\setminus A) = 1 - \limsup_{n \to \infty} \frac{|A \cap \{1,\ldots,n\}|}{n} < 1 - \mu(A)$$. $\endgroup$ Commented Dec 4, 2018 at 17:16
-
3$\begingroup$ Why don't you correct, amend or delete your post, since it's been mentioned that it's wrong? $\endgroup$– YCorCommented Dec 5, 2018 at 11:08