Uniform-arrival residual mass martingale ($M_t={\sum }_i{\theta }_i{1}_{\{U_i>t\}}/(1-t)$)

For finitely many independent uniform arrival times on $[0,1]$, and fixed positive weights, normalize the unarrived mass by the remaining time. Conditional on not having arrived by $s$, each arrival remains uniform on $(s,1)$, giving the martingale identity. The process is càdlàg, nonnegative, and has constant mean equal to the total weight. It vanishes eventually before time one on each typical path, so it is not uniformly integrable when the total weight is positive.

Let $U_1,\dots ,U_n$ be independent uniform random variables on $[0,1]$, let ${\theta }_i>0$, and set $S={\sum }_i{\theta }_i\in (0,1)$ and $X_t={\sum }_i{\theta }_i{1}_{\{U_i\le t\}}$. ThenIndeed, the complementary crossing is that of level one by the uniform-arrival residual mass martingale $M_t=(S-X_t)/(1-t)$, with $M_0=S$. Its jumps are downward, so its first crossing of one is continuous. An arrival coincides with a possible crossing time with probability zero: the possible times belong to the finite deterministic set obtained from subsets of the weights. Stop at that crossing or at $r<1$. The stopped values are in $[0,1]$, and as $r\uparrow 1$ they converge to the crossing indicator because all arrivals occur before one almost surely. The optional stopping theorem and dominated convergence theorem give crossing probability $S$. For $S=1$, the strict displayed inequality already fails at time zero.