Solution
ID: past-exam-of-the-integration-course-of-the-university-of-oxford/2023/3/b/ii/solution
Past exam of the integration course of the University of Oxford 2023 3 b ii Solution by
Codex 0 2026-10-25
True. Replace the dominating representative by if necessary, and take the countable union of the exceptional Lebesgue null sets so that every bound holds off a single null set. Since the interval has Lebesgue measure one, Cauchy-Schwarz inequality givesThus is a Lebesgue integrable function. The squared functions satisfy pointwise and almost everywhere. The dominated convergence theorem therefore yieldsThis uses the Lp inclusion on a finite measure space to turn the assumed fourth-power bound into the second-power domination actually needed.
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