Past exam of the options I course of the University of Oxford / 2007 / D / D1

The monotone convergence theorem states that if measurable nonnegative functions satisfy $f_n\uparrow f$ almost everywhere on a measure space, thenallowing either side to be $+\infty$.

Use the nonnegative increasing truncations $f_n=x^{-2}{1}_{[1/n,1]}$. They increase almost everywhere to the specified function. On each compact interval the continuous function has equal Riemann and Lebesgue integrals, so the fundamental theorem of calculus givesMonotone convergence yields $\int f=+\infty$. Since a nonnegative function is Lebesgue integrable exactly when its integral is finite, this function is not Lebesgue integrable.

The truncations $g_n=x^{1/2}{1}_{[1/n,1]}$ are measurable, nonnegative, and increase almost everywhere to $g$. As before their integrals agree with the corresponding Riemann integrals, givingMonotone convergence therefore gives the finite value $\boxed{\int g=2/3}$. Nonnegativity makes this the integral of the absolute value as well, proving Lebesgue integrability.