Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / D / 2 (D2)

On $(0,1)$, $|g(x)|\le 1$, so ${\int }_0^1|g|\le 1$. For the second function,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.

For $x\ge 1$, the bound $|\sin u|\le |u|$ gives $|g(x)|\le x^{-3/2}$, whose tail integral is finite. Thus $g{1}_{[1,\infty )}$ is integrable.

Near one, $h(x)=[(x-1)(x+1){]}^{-1/2}$ has an integrable inverse-square-root singularity. At infinity, however, $h(x)\ge 1/x$ for $x>1$, and the harmonic tail diverges. Thus $h{1}_{[1,\infty )}$ is not integrable. Assigning $h(1)=0$ changes neither conclusion, since one point has Lebesgue measure zero.