 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-31/1/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Take

m = ⌈(b - a) / ε ⌉

. The ceiling, as printed in the PDF, makes the intervals cover the bounded range of f. If

a = b

, use the single set

X

instead, since f is constant. By continuity their preimages are closed. On every nonempty preimage

C_{i}

, the oscillation of f is at most epsilon. Hence for every

y \in C_{i}

sup_{C_{i}} f \leq f (y) + ε, sup_{C_{i}} f - I (y) \leq sup_{x \in X} (f (x) - I (x)) + ε .

Taking the supremum over y in the second expression gives the bound for

sup_{C_{i}} f - in f_{C_{i}} I

; if that infimum is infinite, the bound is automatic. Apply the finite-cover estimate to obtain

n lim sup n^{- 1} lo g E e^{n f (X_{n})} \leq x sup (f (x) - I (x)) + ε .

This is the exponential-integral upper estimate in the bounded form of Varadhan lemma.

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