Past exam of the options I course of the University of Oxford / 2008 / D (Integration)

A null set has Lebesgue outer measure zero: for every $\epsilon >0$ it can be covered by countably many open intervals with total length below $\epsilon$. If $E_j$ are null, cover $E_j$ by intervals of total length below $\epsilon 2^{-j}$, for $j\ge 1$. The union of all these covers is countable and covers ${\bigcup }_j{E}_j$ with total length below $\epsilon$. As $\epsilon$ is arbitrary, the countable union is null.

Let $A_n=\{x:f(x)\ge 1/n\}$. To avoid presupposing finite measure of $A_n$, first truncate it: $A_{n,k}=A_n\cap [-k,k]$ is measurable and has finite measure. Its indicator divided by $n$ is integrable and is bounded above by $f$. The allowed monotonicity property givesThus every $A_{n,k}$ is null. Their countable union $A_n$ is null by part (a), and so is $E={\bigcup }_n{A}_n$. Every positive value of $f$ is at least $1/n$ for some integer $n$, hence $\{f>0\}=E$. Consequently $f=0$ outside a null set.

The integrand is positive and continuous on the interval in question. Substitute $t=\log (1+x)$ to obtainIf $\alpha \ge 1$, the exponential factor is at most one and the comparison integral of $t^{-2}$ converges. If $\alpha <1$, exponential growth dominates $t^2$, so the integrand eventually exceeds one and the integral diverges. ThusThe logarithmic square is what makes the boundary value $\alpha =1$ integrable.

The function is measurable and $0\le {\cos }^2(1/x)\le 1$. The interval has finite measure, so its integral is at most one. Hence it is integrable, despite its oscillations near zero.

As $x\to \infty$, ${\cos }^2(1/x)\to 1$. It is therefore at least $1/2$ on some infinite tail. The integral of that lower bound is infinite, so it is not integrable on this interval.