Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / D / 2 / 1

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- Solution 1

Solution

On

(0, 1)

,

∣ g (x) ∣ \leq 1

, so

\int_{0}^{1} ∣ g ∣ \leq 1

. For the second function,

\int_{0}^{1} h (x) d x = \int_{0}^{1} \frac{d x}{1 - x ^{2}} = \frac{π}{2} < \infty.

Both functions are measurable, and therefore both restrictions are Lebesgue integrable. The oscillation of

g

near zero does not affect absolute integrability on this finite interval.

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