 Solution

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

False. Take

k (x) = 1/ (1 + ∣ x ∣)

. This is bounded and measurable on

R

, and tends to zero as

x \to \infty

, 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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