Past exam of the integration course of the University of Oxford / 2015 / 1

A measurable space is a pair $(\Omega ,\mathcal{F})$ where $\mathcal{F}$ is a sigma-algebra of subsets of $\Omega$: it contains $\Omega$, is closed under complements relative to $\Omega$, and is closed under countable unions. A measure on it is a function $\mu :\mathcal{F}\to [0,\infty ]$ with $\mu (\varnothing )=0$ and countable additivity:whenever the measurable sets $A_n$ are pairwise disjoint. Infinite values are allowed. The triple $(\Omega ,\mathcal{F},\mu )$ is a measure space.

Because $E\in \mathcal{F}$, the family ${\mathcal{F}}_E$ contains $E$. If $A\in {\mathcal{F}}_E$, then $E\setminus A=E\cap (\Omega \setminus A)$ belongs to $\mathcal{F}$ and is a subset of $E$, so it belongs to ${\mathcal{F}}_E$. If each $A_n\in {\mathcal{F}}_E$, their countable union belongs to $\mathcal{F}$ and remains a subset of $E$. Thus ${\mathcal{F}}_E$ is a sigma-algebra on $E$, with relative complements taken in $E$, not in $\Omega$.

Equivalently,the trace sigma-algebra on the measurable set $E$. The measurability of $E$ is what makes this equal to the family in the question.

The function $\nu$ takes values in $[0,\infty ]$ and has $\nu (\varnothing )=0$. For any pairwise disjoint $A_n\in {\mathcal{F}}_E$, countable additivity of $\mu$ givesTherefore $\nu$ is a measure on $(E,{\mathcal{F}}_E)$: it is the restriction of a measure to a measurable subset.

For the increasing family, define the measurable setsThey are pairwise disjoint, with $A_N={\bigcup }_{n=1}^N{D}_n$ and ${\bigcup }_n{A}_n={\bigcup }_n{D}_n$. By countable additivity,The nonnegative series is defined by its increasing partial-sum limit, so this proof also covers an infinite value. This is continuity from below of a measure.

For the decreasing family, put $C_n=B_1\setminus B_n$. These measurable sets increase, andSince $\mu (B_1)<\infty$, finite additivity gives legitimate finite subtractions:Apply the first result to $C_n$ and subtract from $\mu (B_1)$ to obtainThis is continuity from above of a measure. Its finiteness hypothesis matters: for Lebesgue measure and $B_n=[n,\infty )$, every $B_n$ has infinite measure but their intersection is empty. The proof must not subtract $\infty -\infty$.

For $x<y$, finite additivity on the disjoint pieces of the half-line givesBy monotonicity of a measure and the Lebesgue measure of an interval,Exchanging the two points if necessary provesThus the cumulative measure of a finite-measure set is a Lipschitz function with Lipschitz constant at most one, hence is uniformly continuous: given $\epsilon >0$, the choice $\delta =\epsilon$ works simultaneously at every point.

To locate its endpoint limits, the measurable sets $E\cap (-\infty ,N]$ increase to $E$, so continuity from below of a measure gives $h(N)\to m(E)$. The measurable sets $E\cap (-\infty ,-N]$ decrease to the empty set and all have finite measure, so continuity from above of a measure gives $h(-N)\to 0$. Since $h$ is nondecreasing,For $0<c<m(E)$, choose $R>0$ with $h(-R)<c<h(R)$. The intermediate value theorem on $[-R,R]$ gives $t$ with $h(t)=c$. ThenThe endpoint values $c=0,m(E)$ can also be achieved by $A=\varnothing ,E$. This proves divisibility of Lebesgue measure for arbitrary finite-measure Lebesgue measurable sets on the real line, including unbounded ones. The non-atomic measure structure permits prescribed fractions of measure; a measure concentrated at one point would not have this property.