 Solution

Let $r$ be the number at risk just after $a$, and suppose the $d$ events are grouped into tie sizes $d_1,\dots ,d_m$ at ordered times in $(a,b]$. With no intervening censoring or entry, their risk counts are $r$, $r-d_1$, $r-d_1-d_2$, and so on. Hence the Kaplan–Meier telescoping across a censor-free interval givesOnly the total $d$ appears. Consequently the order of the events and their grouping into ties do not affect the estimate at $b$. Censoring exactly at $b$ does not change this conclusion when it is processed after the events at that time, as in the standard convention. The calculation assumes $r>0$; no estimator beyond an exhausted risk set is being inferred from unobserved follow-up.