Extending Rowbottom's Theorem to Multiple Measures Simultaneously? Rowbottom's Theorem states that if $\kappa$ is measurable and $U$ is a normal measure on $\kappa$, then for every $f:[\kappa]^{<\omega}\to \tau$, $\tau<\kappa$, there is an $H\in U$ such that for each $n$, $f\upharpoonright [H]^n$ is constant.
I'm hoping to extend Rowbottom's theorem to multiple measures simultaneously, in the following sense:
Conjecture:
Let $\vec{U}=\langle U_\alpha \mid \alpha<\lambda\rangle$ be a Mitchell order-increasing system of normal measures on $\kappa$ and let $f:[\kappa]^{<\omega}\to \tau$ with $\tau<\kappa$. 
Then there is an $H\in \bigcap_{\alpha<\lambda} U_\alpha$ such that for all $n<\omega$, $f\upharpoonright [H]^n$ is constant.
Is this true?
 A: The conjecture as stated is false, since the function $\alpha \mapsto o^{\vec{U}}(\alpha)$ is a counterexample. Yet, sometimes it is the only counterexample. 
Theorem: Let $f\colon \kappa^{<\omega} \to \tau$ and let $\langle U_\alpha \mid \alpha < \lambda\rangle$ be a sequence of normal measures on $\kappa$, $\lambda < \kappa$. Pick a function $h \colon \kappa \to \lambda$ such that for all $\alpha < \lambda$, $h^{-1}(\alpha) \in U_\alpha$. 
Then, there is a set $H \in \bigcap_{\alpha < \lambda} U_\alpha$ and a function $g \colon \lambda^{<\omega} \to \tau$ such that for all $\alpha_0 < \dots < \alpha_{n-1} \in H$, $f(\alpha_0, \dots, \alpha_{n-1}) = g(h(\alpha_0), \dots, h(\alpha_{n-1}))$.
Proof: By induction on $n$. For $n = 1$, this is clear: find a measure one set in $U_\alpha$ of $\alpha$ with $h(\alpha) = \zeta$ for each $\zeta < \lambda$ with a fixed value. Take their union.
For $n = 2$, for each value for $\alpha_0$ there is a set $A_{\alpha_0}$ in the filter and a function $g_{\alpha_0} \colon \lambda \to \tau$ such that for all $\alpha_1 \in A_{\alpha_0}$, $f(\alpha_0, \alpha_1) = g_{\alpha_0}(h(\alpha_1))$. Using the inductive hypothesis (with $\tau^\lambda <\kappa$ many colors), we can make $g_{\alpha_0}$ depend only on $h(\alpha_0)$. This defined a single function $g \colon \lambda^2 \to \tau$. Using the normality of the filter we can take a diagonal intersection of the $A_{\alpha_0}$ and obtain a single large set $A$. Intersecting it with the large set on which $g(h(\alpha), \bullet) = g_\alpha(\bullet)$, we get that for all $\alpha < \beta$ in $A$, $f(\alpha,\beta) = g(h(\alpha), h(\beta))$. Continuing this way for $\omega$ many steps and using the $\sigma$-completeness of the filter, we obtain the (somewhat) homogeneous set $H$. QED 
If we take longer sequences of measures, even Mitchell increasing, then we run into a problem. For example, for $\lambda = \kappa$, we can take $f(\alpha, \beta)$ to be $1$ iff $o(\beta) = \alpha$. In this case, there is no set $H$ in the intersection filter such that the value of $f$ on $[H]^2$ depends only in their Mitchell order.
