Past exam of the options I course of the University of Oxford / 2009 / D (Integration)

The Lebesgue outer measure can be defined byA null set has $m^{\ast }(E)=0$, equivalently it admits such a cover of total length below every positive $\epsilon$.

The usual Cantor set is the intersection of the sets obtained from $[0,1]$ by repeatedly removing every open middle third. At stage $n$ there are $2^n$ closed intervals of length $3^{-n}$. Enlarge them slightly to open intervals whose total added length is below $\epsilon$. Thus $m^{\ast }(C)\le (2/3{)}^n+\epsilon$ for every $n$ and $\epsilon >0$, proving $m^{\ast }(C)=0$.

For the different deletion rule in this problem, let $L_n$ be the total length of the stage-$n$ intervals. Removing the stated fraction from each component givesFor $n\ge 1$ the telescoping product simplifies toBecause $K\subseteq K_n$, the same finite interval-enlargement argument gives $m^{\ast }(K)\le L_n$. Taking the limit provesOnly interval covers and their lengths were used; no continuity theorem for integrals is needed.

One version of dominated differentiation under the integral sign is the following. Let the parameter range be open. Suppose $f(\cdot ,y)$ is integrable, ${\partial }_{y}f(x,y)$ exists for almost every $x$, and near each parameter value its absolute value is bounded by an integrable function of $x$ independent of $y$. ThenThis follows by applying the dominated convergence theorem to difference quotients, bounded by that same integrable function through the mean value theorem.

On any compact parameter interval $0<y\le B$, the original integrand is bounded by $B/(1+x^4)$, using $|\sin (x^{2}y)|\le B{x}^2$. Its first and second parameter derivatives areTheir absolute values are bounded respectively by $(1+x^4{)}^{-1}$ and $x^2/(1+x^4)$, both integrable on $(0,\infty )$. Applying the theorem twice proves that $F$ is twice differentiable and givesThis last integral is absolutely convergent: its integrand is bounded near zero and bounded by $x^{-2}$ at infinity. Substituting $t=x\sqrt{y}$ yieldsThe expression for $c$ is finite and independent of $y$, which is all that is required.