 Solution

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

Let

H_{0} = \int_{0}^{π} ∣ h (u) ∣ d u < \infty

. On each period, substitute

x = jπ + u

\int_{jπ}^{(j + 1) π} e^{- β x} ∣ h (x) ∣ d x = e^{- β jπ} \int_{0}^{π} e^{- β u} ∣ h (u) ∣ d u \leq e^{- β jπ} H_{0} .

Summing over

j \geq 0

gives the convergent geometric bound

\int_{0}^{\infty} e^{- β x} ∣ h (x) ∣ d x \leq \frac{H _{0}}{1 - e ^{- β π}} < \infty.

Measurability is preserved under multiplication by the exponential, so this proves Lebesgue integrability.

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