Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / D / 1

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.