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 $\kappa^{\omega}$.
 A: *

*$non(\mathcal{N})=\kappa$.
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.  


