Within the monotone coupling, $I_p$ increases with $p$ and
$$
\bigcup_{p'<p}I_{p'}=\{M<p\}.
$$
Taking a countable cofinal sequence $p'\uparrow p$ and using <continuity from below of a measure> gives
$$
\lim_{p'\uparrow p}\theta(p')=\mathbb P(M<p).
$$
Since $\{M<p\}\subseteq I_p$, subtraction from $\theta(p)=\mathbb P(I_p)$ proves
$$
\boxed{\theta(p)-\lim_{p'\uparrow p}\theta(p')=\mathbb P(I_p\cap\{M=p\})}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 204 / 2 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-204/2/e/solution