No. For each realized sample let $A_n=\{X_1,\ldots,X_n\}$. This is a finite Borel set, and continuity of $F$ makes $P(A_n)=0$ almost surely, while $P_n(A_n)=1$. Therefore
$$
\sup_{A\in\mathcal B(\mathbb R)}|P_n(A)-P(A)|=1
$$
almost surely for every $n$, so the class of all Borel-set indicators cannot satisfy a ULLN.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 210 / 1 / c / Solution

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