Past exam of the options II course of the University of Oxford / 2012 / D / D3

The monotone convergence theorem states that if measurable functions $0\le f_n\uparrow f$ pointwise on a measure space, then $\int f_n\uparrow \int f$, allowing the limit to be infinite.

Extend each restriction to $I_n$ by zero on $I\setminus I_n$. The functions $f{1}_{I_n}$ are nonnegative and increase pointwise to $f$ because the intervals are increasing and exhaust $I$. HenceFor nonnegative $f$, integrability means finiteness of this integral, soNo boundedness or closedness of the intervals is required.

The function is nonnegative, measurable, and even. Direct integration givesThus it is integrable, with integral $\boxed{2}$.

On finite intervals the function is continuous, butConsequently its nonnegative integral on the half-line is infinite: it is not integrable. The divergence occurs at infinity, not at the origin.

For every real $x$, the numerator is at most $2e^{|x|}$, while the denominator is at least $e^{2|x|}$. ThereforeThe upper bound is integrable by part (a), so the comparison theorem shows that the function is integrable on $\mathbb{R}$.

We prove absolute integrability. On $(0,1]$, $|\sin (1/x)|\le 1$ and the denominator is at least one, so $|f(x)|\le 1$. On $[1,\infty )$, the standard inequality $|\sin t|\le |t|$ givesSubstitution $t=\log x$ yieldsThus ${\int }_0^{\infty }|f(x)|dx\le 1+\pi /2<\infty$, and the function is integrable despite its oscillation near zero.