Math and science::Analysis::Tao, measure::01. Jordan measure

Proof outline

An elementary cover $C$ of a set $E$ can be broken into two disjoint elementary sets, $A$ and $B$ which union to $C$. This can be done by taking the intersection and difference of $C$ with some other elementary set $F$. $A$ will be a cover for $E\setminus F$ and $B$ a cover for $E\cap F$. As $A$ and $B$ are disjoint, we have $m(A\cup B)=m(A)+m(B)$. $A$ and $B$ will have an elementary measure larger than the infimum representing $m^{\ast ,(J)}(E\setminus F)$ and $m^{\ast ,(J)}(E\cap F)$ respectively, however, they will have a summed measure smaller than that of $C$ which can be made arbitrarily close to $m^{\ast ,(J)}(E)$.

Implication

In this formulation, the Carathéodory type property can be thought of as being one answer to the question: under what conditions do we have finite additivity of outer Jordan measure (not mere sub-additivity)? 

This question is especially pertinent when considering the fact that Jordan outer measure, while additive for Jordan measurable sets, is not additive in general: the outer measure of the union of the bullets and the bullet-riddled square is 1 (as union is $[0,1]$), yet the outer measure of each set individually is also 1.

The answer is that if a set $E$ is split by an elementary set into $A$ and $B$, then any cover of $E$ can be split (without repeat) and cover $A$ and $B$. This makes $m^{\ast ,(J)}(A)+m^{\ast ,(J)}(B)\le m^{\ast ,(J)}(E)$. The reverse inequlity is simply the characteristic of sub-additivity, which is indeed possessed by outer measure.

Proof

For any real $\delta >0$ there exists an elementary set $C$ covering $E$ with elementary measure $m(C)\le m^{\ast ,(J)}(E)+\delta$. $C$ can be broken into two disjoint elementary sets $A=C\setminus F$ and $B\cap F$ for some elementary set $F$, and we will have $m(C)=m(A)+m(B)$. This means that $m(C)=m(A)+m(B)\le m^{\ast ,(J)}(E)+\delta$. It is true (can be checked with some set algebra) that $A$ is a cover for $E\setminus F$ and $B$ a cover for $E\cap F$. Consequently, $m(A)\ge m^{\ast ,(J)}(E\setminus F)$ and $m(B)\ge m^{\ast ,(J)}(E\cap F)$. Substituting this back into the previous inequality we see that:

$\delta$ was arbitrary so we can say:

That proves the harder of the two inequalities needed! Paired with the sub-additivity of outer Jordan measure, we have equality: