 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2011/d/d3/a/i/solution

False. Take the bounded measurable function

k (x) = {(1 + x)^{- 1}, 0, x > 0, x \leq 0.

It tends to zero at positive infinity, but

\int_{0}^{R} ∣ k (x) ∣ d x = lo g (1 + R) ⟶ \infty.

Thus decay to zero alone does not imply Lebesgue integrability.

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