Past exam of the real and harmonic analysis course of the University of Oxford / 2007 / 2

An outer measure on $X$ is a function $\mu :\mathcal{P}(X)\to [0,\infty ]$ satisfying $\mu (\varnothing )=0$, monotonicity, and countable subadditivity:Its measurable sets are those satisfying Caratheodory measurability criterion:The test runs over all subsets $E$, including nonmeasurable ones.

The Caratheodory measurability criterion yields a complete sigma-algebra ${\mathfrak{M}}_{\mu }^{\ast }$, and $\mu$ restricted to it is a countably additive measure. Completeness means that every subset of a measurable null set is measurable.

For the requested algebra proof, $X$ and $\varnothing$ plainly satisfy the criterion. Replacing $A$ by $X\setminus A$ merely exchanges the two terms, proving closure under complements. If $A,B$ are measurable, apply their splitting identities successively to any $E$:Subadditivity applied to the first two pieces impliesThe reverse inequality is outer subadditivity. Hence $A\cup B$ is measurable, proving closure under finite unions and therefore the algebra of sets property. The argument never subtracts infinite values.

A metric outer measure on a metric space $(X,d)$ is an outer measure satisfyingHere $d(A,B)=\inf \{d(a,b):a\in A,b\in B\}$. Every Borel set is measurable for a metric outer measure. Consequently its measurable sigma-algebra contains the Borel sigma-algebra; it may be strictly larger.

For $E\subseteq {\mathbb{R}}^n$, define the Lebesgue outer measure bywhere the boxes are bounded and $\mathrm{vol}(Q_j)={\prod }_r(b_{jr}-a_{jr})$. The empty covering handles the empty set. Monotonicity follows by inclusion of admissible covers. Taking covers within $\epsilon 2^{-j}$ of the outer measures of a sequence of sets proves countable subadditivity; when the sum is infinite the inequality is automatic. Lebesgue measure is the restriction of ${\mathcal{L}}^{\ast }$ to its Caratheodory-measurable sets.

If $d(A,B)=\delta >0$, subdivide each box of any countable cover of $A\cup B$ into finitely many half-open boxes of diameter smaller than $\delta$, preserving its total volume. No small box can meet both $A$ and $B$. The boxes meeting $A$ cover $A$, and the disjoint family meeting $B$ covers $B$, so the original cover costs at least ${\mathcal{L}}^{\ast }(A)+{\mathcal{L}}^{\ast }(B)$. Taking the infimum and then using subadditivity givesThus the outer measure underlying Lebesgue measure is a metric outer measure, and in particular every Borel set is Lebesgue measurable.

A regular outer measure has, for every $E\subseteq X$, a measurable hull $H\supseteq E$ with $\mu (H)=\mu (E)$.

Suppose $\mathcal{C}$ is an algebra containing ${\mathfrak{M}}_{\mu }^{\ast }$ and $\mu$ is finitely additive on $\mathcal{C}$. Fix $A\in \mathcal{C}$ and an arbitrary $E$ of finite outer measure. Take a measurable hull $H$ of $E$. Both $H\cap A$ and $H\setminus A$ belong to $\mathcal{C}$, soOuter subadditivity gives equality. If $\mu (E)=\infty$, outer subadditivity already forces the sum on the right to be infinite, again giving equality. Therefore $A$ satisfies the Caratheodory criterion. We have proved $\mathcal{C}\subseteq {\mathfrak{M}}_{\mu }^{\ast }$, soThis proves the maximal additivity algebra of a regular outer measure without any sigma-finiteness assumption.