# Countable products of total measures

Suppose $\kappa = \mathfrak{c} =2^{\aleph_0}$ is a real valued measurable cardinal with a witnessing measure $m:\mathcal{P}(\kappa) \to [0, 1]$ - So $m$ is a diffused (points have zero measure) $\kappa$-additive measure on $\mathcal{P}(\kappa)$. Let $\mu = \otimes_{n < \omega} m$ be the countble product measure - So $\mu$ is defined on the sigma algebra generated by the family $\{\Pi_{n < \omega} X_n : (\forall n)(X_n \subseteq \kappa)\}$. Let $\mathcal{N}$ be the sigma ideal of $\mu$-null subsets of $\kappa^{\omega}$.

What can be say about the cardinal invariants associated with the ideal $\mathcal{N}$? For example, can either one of $add(\mathcal{N}), non(\mathcal{N}), cov(\mathcal{N})$ be $\aleph_1$?

Definitions: $non(\mathcal{N})$ is the least size of a set not in $\mathcal{N}$, $add(\mathcal{N})$ is the least size of a family $F \subseteq \mathcal{N}$ whose union is not in $\mathcal{N}$ and $cov(\mathcal{N})$ is the least size of a family of members of $\mathcal{N}$ whose union is $\kappa^{\omega}$.

• For some background, what can be said about cardinal characteristics of a product of two total measures? Commented May 2, 2017 at 20:17
• For finite products, all invariants equal $\kappa$ (by Fubini's theorem). Commented Jul 14, 2017 at 6:12

• $non(\mathcal{N})=\kappa$.
If $S \subseteq \kappa^\omega$ is of size $<\kappa$, then all projections $\pi_i(S)\subseteq \kappa$ are also of size $<\kappa$, hence have measure $0$. But then $S\subseteq \prod_{i<\omega} \pi_i(S)$ also has measure $0$.
(It seems to me that it would be enough that one of the projections is null.)

• $cov(\mathcal{N})$ is bounded by the covering number for the (Borel) null ideal on $2^\omega$, which I will call $\nu$. [EDITED. I do not know if $cov(\mathcal{N})$ can be less than $2^{\aleph_0}$. The "standard" model for real valued measurable is obtained by adding $\kappa$ random reals to a model where $\kappa$ is measurable, so in that model we have $\nu=2^{\aleph_0}$, in which case my claim $cov(\mathcal{N})\le \nu$'' is irrelevant.]

• Let $f:\kappa \to 2$ be such that $f^{-1}(\{i\})$ has measure $\frac12$ for $i=0,1$.
• Define $F:\kappa^\omega\to 2^\omega$ by $F(x)(n) = f(x(n))$ for all $x\in \kappa^\omega$.
• Then $F^{-1}(U) \subseteq \kappa^\omega$ has the same measure as $U$ for all open sets $U\subseteq 2^\omega$; hence $F^{-1}(N)$ is a null set whenever $N$ is.
• Let $(N_j:j<\nu)$ be a family of null sets covering $2^\omega$.
• Define $M_j = F^{-1}(N_j) \subseteq \kappa^\omega$. Then $(M_j:j<\nu)$ is a cover of $\kappa^\omega$ by null sets.
• I now think that $cov(\mathcal N)=\nu=2^{\aleph_0}$ always holds if $\mathfrak c$ is rvm, but I don't have a proof yet. Commented Jul 14, 2017 at 5:17
• You can do this by constructing an inverse measure preserving map $f: \kappa \to 2^{\omega}$ ($\kappa$ is any atomlessly rvm). This is also how we lift the measure on $\kappa$ to a total extension of Leb. measure. Commented Jul 14, 2017 at 5:45
• If there is an isomorphism between the given measure algebra (call it $(B_m,m)$) on $\kappa$ and the Lebesgue measure on $2^\omega$ (call it $(B_\lambda,\lambda)$, then of course the measure algebra $(B_\mu,\mu)$ will also be isomorphic. But I cannot show that $B_m$ has Maharam weight $\aleph_0$. ($m$ might not be an extension of Lebesgue measure.) Commented Jul 14, 2017 at 13:37
• If $\kappa$ is rvm and $m$ is a witnessing Maharam homogeneous measure on $\kappa$, then forcing with the associated measure algebra is same as adding $\theta$ random reals for some $\theta \geq \kappa^{+}$. In particular, the measure algebra of $m$ cannot be isomorphic to the standard Lebesgue measure algebra. This is a theorem of Gitik and Shelah. Commented Jul 14, 2017 at 16:15