Past exam of the real and harmonic analysis course of the University of Oxford / 2008 / 2

The Lebesgue outer measure iswhere $|Q|$ is the product of its side lengths. The infimum ranges over countable covers; allowing other standard endpoint conventions gives the same outer measure by arbitrarily small enlargement. Lebesgue measure is its restriction to Lebesgue measurable sets.

Enlarged unit cubes give ${\mathcal{L}}^{\ast }([0,1{]}^n)\le 1$. For any countable open-box cover of this compact cube choose a finite subcover. A finite grid formed from its box endpoints and the cube endpoints partitions the cube into rectangular cells. The sum of the volumes of covering boxes is at least the sum of these cell volumes, which is one. Boundary faces have elementary zero volume and do not affect the finite volume comparison. Thus every covering sum is at least one, proving

A set $H$ is Lebesgue measurable when the Caratheodory measurability criterion holds:Subadditivity always gives the inequality with left side at most right side. For $H=\{x_i>t\}$, split each covering box $Q_j$ across the plane $x_i=t$. Its two pieces have total volume $|Q_j|$. The piece with a closed cut face can be enlarged by arbitrarily little into an open box; hence its outer measure is at most its corresponding volume, just as for the other piece. Subadditivity applied separately on the two sides consequently givesFor finite outer measure take a covering sum arbitrarily close to ${\mathcal{L}}^{\ast }(E)$; infinite outer measure follows already from subadditivity. Thus every coordinate half space is measurable.

Carathéodory's theorem says that the measurable sets form a sigma-algebra and the outer measure restricted to it is a complete measure. Coordinate half spaces generate all open boxes and then all open sets, using countably many rational-endpoint boxes. They therefore generate the Borel sigma-algebra. Hence every Borel set is Lebesgue measurable.

A regular outer measure has a measurable hull of equal outer measure for every set: for each $A$ there is measurable $G\supseteq A$ with ${\mathcal{L}}^{\ast }(G)={\mathcal{L}}^{\ast }(A)$. Lebesgue outer measure has even a Borel hull. If its value on $A$ is finite, choose open-box covers of total cost below ${\mathcal{L}}^{\ast }(A)+1/j$, and let $U_j$ be their open unions. Then $G={\bigcap }_j{U}_j$ is Borel andFor infinite outer measure the whole space is a suitable hull. Thus Lebesgue outer measure is regular.

If $A$ is measurable, additivity immediately gives the requested equality. Conversely suppose the equality holds. Choose a measurable hull $G$ of $A$ and put $B=G\cap M$. Then $A\subseteq B\subseteq M$ and $\mathcal{L}(B)={\mathcal{L}}^{\ast }(A)$. Since $B$ is measurable, its splitting of $M\setminus A$ givesThe assumed equality and $\mathcal{L}(M)=\mathcal{L}(B)+\mathcal{L}(M\setminus B)$ imply ${\mathcal{L}}^{\ast }(B\setminus A)=0$. These subtractions are legitimate because $\mathcal{L}(M)<\infty$. Outer-measure-null sets are measurable by the Caratheodory criterion. Hence $A=B\setminus (B\setminus A)$ is measurable, proving the equivalence.