About $\alpha$-indecomposability. One defines $\alpha$-indecomposability, for ultrafilters $\mathcal{D}\subset\mathcal{P}(\kappa)$, as: for every $\langle I_\beta:\beta<\alpha\rangle$ with $\bigcup_{\beta<\alpha}I_\alpha=\kappa$, one can find $s\in[\alpha]^{<\alpha}$ with $\bigcup_{\beta\in s} I_\beta\in\mathcal{D}$.
Now, i want to prove: if $\mathcal{D}\subset\mathcal{P}(\kappa)$ is $\mu^{(n)}$-indecomposable, for every $n\geq1$, then:
\begin{equation}
\prod\mu^{(n)}/\mathcal{D}=\bigcup_{\nu<\mu^{(n)}}\prod\nu/\mathcal{D}.
\end{equation}
Here $\mu^{(n)}$ is the $n$th succesor of $\mu$: $\mu^{(0)}=\mu$, $\mu^{(n+1)}=(\mu^{(n)})^+$.
 A: The reversed inclusion is obvious, so I will just give a proof of the inclusion
\begin{equation}
\prod\mu^{(n)}/\mathcal{D}\subseteq\bigcup_{\nu<\mu^{(n)}}\prod\nu/\mathcal{D}.
\end{equation}
Assume that $f : \kappa \to \mu^{(n)}$ is a function. We shall check that it is bounded almost everywhere with respect to $\mathcal{D}$; that is, we have some $\nu<\mu^{(n)}$ such that
$$\{\xi < \kappa : f(\xi)<\nu\}\in\mathcal{D}.$$
Assume that we know $f$ is bounded almost everywhere with an upper bound $\nu$. Then we have a bounded function $g$ such that $[f]=[g]$ and $g(\xi)\le \nu$ for all $\xi$. (How? You can find such $g$ if you are familiar with ultrfilters.) Hence $[f]$ is an element of $\prod \nu/\mathcal{D}$.
Take $X_\alpha = \{\xi<\kappa : f(\xi) < \alpha\}$. Then $\bigcup_{\alpha<\mu^{(n)}} X_\alpha = \kappa$. By $\mu^{(n)}$-indecomposability, there is a $s\subset \mu^{(n)}$ such that $|s|<\mu^{(n)}$ and $\bigcup_{\alpha\in s} X_\alpha \in\mathcal{D}$. We know that $\mu^{(n)}$ is regular for each $n\ge 1$, so $\nu := \sup s < \mu^{(n)}$. Moreover, if $\alpha\in s$ then $X_\alpha \subseteq X_\nu$. 
Therefore $X_\nu = \{\xi<\kappa : f(\xi) < \nu\} \in \mathcal{D}$, and so $f$ is bounded almost everywhere.
