If $E_n\downarrow E$ and $\mu(E_1)<\infty$, then $\mu(E_n)\downarrow\mu(E)$. Apply <continuity from below of a measure> to the increasing differences $E_1\setminus E_n$.

Continuity from above of a measure

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

If $E_n\uparrow E$, then <countable additivity> gives $\mu(E_n)\uparrow\mu(E)$, even when the limiting measure is infinite.

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