Finite-set homogeneity for a normal measure

First fix a colouring $c:[\kappa {]}^m\to \lambda$, where $\lambda <\kappa$. If $m=0$, its singleton domain makes it constant. If $m=1$ and no colour class belongs to $U$, the complement of every colour class belongs to $U$. Their intersection belongs to $U$ by $\kappa$-completeness, but it is empty, a contradiction. Thus some measure-one set is homogeneous in the unary case. Notice also that every tail $\kappa \setminus (\alpha +1)$ is in $U$: intersect the complements of its fewer than $\kappa$ singleton points.

Induct on $m$. Assume the result for $m$ and consider $c:[\kappa {]}^{m+1}\to \lambda$. For every $\alpha <\kappa$, extend the tail colouring $t\mapsto c(\{\alpha \}\cup t)$ from $[\kappa \setminus (\alpha +1){]}^m$ to all of $[\kappa {]}^m$ by assigning one fixed value of $\lambda$ off the tail. Apply the induction hypothesis to this total extension, obtaining a homogeneous $H_{\alpha }\in U$, and put $B_{\alpha }=H_{\alpha }\cap (\kappa \setminus (\alpha +1))\in U$. Its restriction to the tail is the original colouring, so $B_{\alpha }$ is homogeneous for that colouring; call its constant value $i_{\alpha }<\lambda$. Using Choice, make these selections simultaneously. The diagonal intersection $D=\{\beta <\kappa :(\mathrm{\forall }\alpha <\beta )\beta \in B_{\alpha }\}$ belongs to $U$.

Apply the unary case to $\alpha \mapsto i_{\alpha }$ and take $X\in U$ on which it has constant value $i$. Put $A=D\cap X$. If ${\alpha }_0<\cdots <{\alpha }_m$ lie in $A$, then ${\alpha }_j\in B_{{\alpha }_0}$ for every $0<j\le m$, by the definition of $D$. Consequently $c(\{{\alpha }_0,\dots ,{\alpha }_m\})=i_{{\alpha }_0}=i$. This proves the fixed-arity claim for every finite $m$.

For every $n<\omega$, use the fixed-arity claim to choose $A_n\in U$ on which $c_n$ is constant. Since $\omega <\kappa$, countable completeness gives $A={\bigcap }_{n<\omega }{A}_n\in U$. Restricting a constant colouring remains constant, so this $A$ works for every $n$, including $n=0$. Choice selects the family $⟨B_{\alpha }:\alpha <\kappa ⟩$ at each induction stage and the countable family $⟨A_n:n<\omega ⟩$; the filter calculations after those selections are choice-free. [F1, F3, step 3.1, discharge-induction] $\mathrm{\square }$