Past exam of the probability, statistics and options I course of the University of Oxford / 2007 / D / 1

The monotone convergence theorem states that if measurable functions $f_n$ are nonnegative and $f_n\uparrow f$ pointwise, then $f$ is measurable andallowing either side to be infinite. The same conclusion holds under almost-everywhere monotone convergence after changing values on a null set. No assumption that the integrals stay bounded is required.

Use the increasing nonnegative measurable functions $f_n(x)=x^{-2}{1}_{[1/n,1]}(x)$. Their pointwise limit is the given function except at irrelevant endpoint choices, and the fundamental theorem of calculus on $[1/n,1]$ givesmonotone convergence therefore gives $\int f=\infty$. Since $f\ge 0$, this is also $\int |f|$, and $f$ is not Lebesgue integrable. Measurability alone does not make a finite integral.

Similarly set $g_n(x)=x^{1/2}{1}_{[1/n,1]}(x)$. These nonnegative measurable functions increase to $g$ almost everywhere. The fundamental theorem of calculus givesBy monotone convergence,The integral is finite and $g\ge 0$, so $g$ is Lebesgue integrable. Identifying Lebesgue and ordinary integrals of continuous functions on compact intervals justifies each truncated calculation.