 Solution

Since there are finitely many arrival times and each is strictly below one almost surely, $U_{\ast }:={\max }_i{U}_i<1$ almost surely. For every $t\ge U_{\ast }$ with $t<1$, all weights have arrived and $M_t=0$. Therefore $M_t\to 0$ almost surely as $t\uparrow 1$.

Nevertheless $\mathbb{E}{M}_t=S>0$ for every $t<1$. If the family were uniformly integrable, its almost-sure convergence along any deterministic sequence tending to one would imply convergence in $L^1$, and hence convergence of expectations to zero. This is impossible. Thus the martingale is not uniformly integrable. Its constant expected mass is carried by increasingly rare, late arrivals.