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?

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    $\begingroup$ 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 $\{(m,n)\mid m> n\}$. $\endgroup$ – Michael Greinecker Mar 18 at 13:28

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 $(\omega\times\omega, \mathcal U,\mu^2)\rightarrow (\omega\times\omega, 2^{\omega\times\omega}, \lambda)\cong (\omega, 2^\omega, \mu)$.

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

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