Let $\alpha$ be a countable ordinal. Is there necessarily a family $(\epsilon_{\beta,n})_{n\in\mathbb{N},\,\beta<\alpha}$ such that $\epsilon_{\beta,n}>0$, $\sum_n \epsilon_{\beta,n} = 1$, and $\left(\sum_{\beta<\gamma<\alpha}\epsilon_{\gamma,n}\right)/\left(\sum_{\beta\le\gamma<\alpha}\epsilon_{\gamma,n}\right)\to0$ uniformly in $\beta$ as $n\to\infty$?
(I have a family of probability distributions $P_\beta$ on pairwise disjoint $\Omega_\beta$ and would like to find a sequence of distributions $Q_n := \sum_{\beta<\alpha} \epsilon_{\beta,n} P_\beta$ on $\bigcup_\beta\Omega_\beta$ such that $Q_n(A|\bigcup_{\beta\le\gamma<\alpha}\Omega_\gamma)\to P_\beta(A\cap\Omega_\beta)$ uniformly in $A$ and $\beta$.)
No.
Suppose there were, for $\alpha>\omega$, and let $S_{\beta,n} := \sum_{\beta\le\gamma<\alpha}\epsilon_{\gamma,n}$. Then the assumption is that $S_{\beta+1,n}/S_{\beta,n}\to0$ uniformly in $\beta$. Thus, there is some $n$ such that $S_{\beta+1,n}/S_{\beta,n} < \frac12$ for all $\beta$. For finite $\beta = k$, this implies that $S_{k,n}\to0$ for $k\to\infty$ (holding $n$ fixed). But $0 < \epsilon_{\omega,n} \le S_{k,n}$, contradiction.