# Finitely additive measure on Cartesian square of countable set

Let $$\mu$$ be a probability measure on $$(\omega, 2^\omega, \mu)$$ measure space which is finitely additive and $$\mu(A)=0$$ for finite sets. We can define as usual $$\mu^2$$ on semiring $$\mathcal{G}=\{A\times B~|~A,B\in 2^\omega\}\subset2^{\omega\times\omega}$$: $$\mu^2(A\times B) = \mu(A)\cdot\mu(B)$$ and extend it to the ring $$\langle\mathcal{G}\rangle$$ of subsets of $$2^{\omega\times\omega}$$ generated by $$\mathcal{G}$$. Then we can do Peano-Jordan extension, i.e. define: $$\mathcal{U} = \{X\in 2^{\omega\times\omega}~|~\inf_{X\subset S\in\langle\mathcal{G}\rangle}\mu(S)=\sup_{X\supset S\in\langle\mathcal{G}\rangle}\mu(S)\}$$ The question is: what we can say about $$\mathcal{U}$$? How can we describe it in set-theoretic terms. Is it, for example, ring of subsets isomorphic to $$2^\omega$$ or it coincides with $$2^{\omega\times\omega}$$ (which is isomorphic to $$2^\omega$$) or something else? Are $$(\omega, 2^\omega, \mu)$$ and $$(\omega\times\omega,\mathcal{U},\mu^2)$$ isomorphic?

• It is not $2^{\omega\times\omega}$. Indeed, $\{(m,n)\mid m\leq n\}\notin\mathcal{U}$. For if $A\times B\subseteq \{(m,n)\mid m\leq n\}$, then $A$ must be finite and therefore $\mu^2(A\times B)=0$. A similar argument applies to the complement $\{(m,n)\mid m> n\}$. – Michael Greinecker Mar 18 at 13:28

## 1 Answer

Here is something that might be of interest to you, unfortunatley this is too long to be a comment.

As Michael Greinecker has said, $$\mathcal U\neq 2^{\omega\times\omega}$$ and thus $$(\omega, 2^\omega, \mu)$$ and $$(\omega\times\omega, \mathcal U, \mu^2)$$ cannot be isomorphic. However, in the case where $$\mu$$ comes from a non-principal ultrafilter on $$\omega$$ (i.e. $$\mu(A)\in\{0, 1\}$$ for all $$A\subseteq\omega$$), $$\mu^2$$ can always be extended to $$\mu^2\subseteq\lambda$$ so that in fact $$(\omega, 2^\omega, \mu)\cong(\omega\times\omega, 2^{\omega\times\omega}, \lambda)$$.

To see this, let $$\mathcal F$$ be the ultrafilter derived from $$\mu$$. It is straightforward to show that $$\mathcal U=\{X\subseteq\omega\times\omega\mid \exists A, B\in\mathcal F\ (A\times B\subseteq X\vee A\times B\subseteq \omega\setminus X)\}$$ and for $$X\in\mathcal U$$, $$\mu^2(X)=1$$ iff $$\exists A, B\in\mathcal F\ A\times B\subseteq X$$.

We now start to define the bijection $$f:\omega\times\omega\rightarrow\omega$$ that will induce the isomorphism in the end.

Claim 1 There is a partition $$(P_n)_{n\in\omega}$$ of $$\omega$$ into infinite $$\mu$$-nullsets so that $$\{n\in\omega\mid n\in P_n\}\in\mathcal F$$.

Proof: Let $$D$$ be any infinte, co-infinite set in $$\mathcal F$$ and choose $$(P_n)_{n\in\omega}$$ to be any partition of $$\omega$$ with $$P_n\cap D=\emptyset$$ for $$n\notin D$$ and $$n\in P_n$$ for $$n\in D$$. $$\square$$

Next we choose for each $$n$$ an enumeration $$P_n=\{m^n_k\mid k\in\omega\}$$ so that if $$n\in P_n$$ then $$n=m^n_n$$. Define $$f$$ as $$f(n, k)=m^n_k$$. This is a bijection and for $$\mathcal F$$-almost all $$n$$ we have $$f(n, n)=n$$. Now let $$\lambda(X)=\mu(f[X])$$ for $$X\subseteq\omega\times\omega$$ be the pullback of $$\mu$$ via $$f$$.

Claim 2 $$\lambda$$ extends $$\mu^2$$.

Proof: Assume $$X\in\mathcal U$$ and $$\mu^2(X)=1$$. Then there are $$A,B\in\mathcal F$$ with $$A\times B\subseteq X$$. Let $$D=\{n\in\omega\mid n\in P_n\}=\{n\in\omega\mid n=m^n_n\}\in\mathcal F$$. We have $$(D\cap A\cap B)^2\subseteq X$$ and thus $$f[X]\supseteq f[(D\cap A\cap B)^2]\supseteq D\cap A\cap B\in\mathcal F$$ which gives $$\lambda(X)=1$$. If otherwise $$\mu^2(X)=0$$ then the argument above shows $$\lambda(\omega\setminus X)=1$$, hence $$\lambda(X)=0$$. $$\square$$

This at least gives an embedding $$(\omega\times\omega, \mathcal U,\mu^2)\rightarrow (\omega\times\omega, 2^{\omega\times\omega}, \lambda)\cong (\omega, 2^\omega, \mu)$$.

• Thank you. I can add that in case of selective ultrafilter your embedding becomes isomorphism. Unfortunately in my case the measure is not ultrafilter. – ar.grig Mar 19 at 4:58