Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / D (Integration)

The Lebesgue outer measure isThe Lebesgue inner measure is $m_{\ast }(E)=\sup \{m(K):K\subseteq E,K\text{ compact}\}$, with $m_{\ast }(\varnothing )=0$. For a bounded set inside a finite interval $I$, this is equivalently $|I|-m^{\ast }(I\setminus E)$.

A set is Lebesgue measurable if it satisfies Carathéodory's criterion:For finite outer measure this is equivalent to equality of inner and outer measure. For a general unbounded set, an equivalent local criterion is $m_{\ast }(E\cap [-n,n])=m^{\ast }(E\cap [-n,n])$ for every positive integer $n$. Merely writing $m_{\ast }(E)=m^{\ast }(E)=\infty$ would not establish measurability.

A null set has Lebesgue outer measure zero. Equivalently, for every $\epsilon >0$ it can be covered by countably many open intervals whose total length is less than $\epsilon$. Such a set is automatically Lebesgue measurable and has Lebesgue measure zero.

If $A\subseteq E$ and $E$ is a null set, every interval cover of $E$ also covers $A$. Thus monotonicity of Lebesgue outer measure gives $0\le m^{\ast }(A)\le m^{\ast }(E)=0$. Hence every subset of a null set is null, and in particular is Lebesgue measurable.

A real measurable function is one whose inverse image of every open subset of $\mathbb{R}$ is Lebesgue measurable. Equivalently,These upper rays generate the Borel sigma-algebra, making this equivalent to requiring measurable inverse images of all Borel sets.

If $a\ge 0$, the set $\{f>a\}$ is contained in $E$, since $f=0$ outside $E$. It is therefore a null set and is Lebesgue measurable. If $a<0$, its complement $\{f\le a\}$ is contained in $E$ and is measurable for the same reason. Complements of measurable sets are measurable. Thus every upper level set is measurable, proving $f$ is a measurable function.

Let $d=h-g$. It vanishes outside the null set $E$, so part (e) proves $d$ is a measurable function, regardless of its values on $E$. Since the sum of two measurable functions is measurable, $h=g+d$ is measurable. This is the completeness of Lebesgue measure under changes on null sets.

For sigma-finite measures, Tonelli's theorem states that the double integral of a nonnegative measurable function equals either iterated integral, allowing $+\infty$. Fubini's theorem states that if a measurable function is absolutely integrable, its sections are integrable almost everywhere, both iterated integrals exist, and they equal the double integral.

Assign any value at $(0,0)$, which does not affect Lebesgue integrability. Near that corner set $s=x+y$ and $t=(x-y)/(x+y)$. Then $x=s(1+t)/2$, $y=s(1-t)/2$, and the absolute Jacobian determinant is $s/2$. On $0<s<1$, the full range is $-1<t<1$, soThe remaining part of the unit square has $x+y\ge 1$ and is harmless. The improper power integral is finite precisely when $\alpha <3$. Under the given restriction,For those values, the change $(x,y)\mapsto (y,x)$ preserves the square and changes the sign of $F$. Absolute integrability justifies this change of variables, givingFor $\alpha \ge 3$ the positive and negative parts each diverge, so symmetry does not define a Lebesgue integral.