Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / D (Integration)

A Lebesgue null set $E\subset \mathbb{R}$ can, for every $ϵ>0$, be covered by countably many open intervals whose total length is less than $ϵ$. Equivalently, its Lebesgue outer measure is zero.

If $E_j$ is null, choose for it such a cover of total length below $ϵ2^{-j}$, $j\ge 1$. The union of these countably many countable covers covers ${\bigcup }_j{E}_j$ and has total length below $ϵ$. Since $ϵ$ is arbitrary, a countable union of null sets is null.

For positive integers $n,m$, let $E_{n,m}=\{x:f(x)\ge 1/n\}\cap [-m,m]$. This measurable set has finite measure, and $f\ge n^{-1}{1}_{E_{n,m}}$. Monotonicity of the Lebesgue integral givesso every $E_{n,m}$ is a Lebesgue null set. The set where $f>0$ is the countable union of these sets over $n,m$. Part (a) makes it null, henceThe bounded truncation ensures that only integrable indicator functions are used with the stated comparison rule.

There is no singularity at the lower endpoint one. Put $t=\log (1+x)$, so $dx=e^{t}dt$. The nonnegative integral becomesFor $\alpha =1$, it converges as a reciprocal-square integral. For $\alpha >1$, exponential decay makes it converge. For $\alpha <1$, its integrand eventually grows without bound, so its integral diverges. ThusThis is the logarithmic borderline in a power-tail integrability test.

The function is measurable and $0\le {\cos }^2(1/x)\le 1$. Since $(0,1)$ has finite Lebesgue measure, comparison gives ${\int }_0^{1}g\le 1$. Therefore $g$ is integrable on $(0,1)$, despite oscillating infinitely often near zero.

For $x>1$, $1/x\in (0,1)$ and ${\cos }^2(1/x)\ge {\cos }^2(1)>0$. Its integral over the infinite interval is bounded below by the integral of a positive constant. Therefore $g$ is not integrable on $(1,\infty )$. In fact it tends to one at infinity.