 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-4/25j/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Take the truncated dyadic approximation

g_{n} (x) = min {n, 2^{- n} ⌊ 2^{n} f (x)⌋} .

It has finitely many values and measurable level sets, so is simple. Downward rounding and truncation ensure

0 \leq g_{n} \leq f

. For a fixed finite

f (x)

, eventually

n > f (x)

, and then

0 \leq f (x) - g_{n} (x) < 2^{- n}

. Thus

g_{n} (x) \to f (x)

everywhere. If extended value infinity is permitted, define

g_{n} = n

there, also giving the required limit.

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