I am attempting to prove the finite additivity of the Jordan Measure (meaning if A and B Jordan Measurable and are disjoint then the Jordan Measure of A ∪ B equals the Jordan Measure of A + the Jordan Measure of B).
To clarify, my definition of Jordan Measurable is that the characteristic function defined on the set is Riemann Integrable.
If A and B are separable, it is clear as there will be a partition fine enough that each cube in the partition only intersects A or B (but not both).
However, I am having trouble when A and B are not separable. You should be able to make the volume of the cubes that intersect both A and B arbitrarily small but I am having trouble formally articulating this
Thanks
I am going to first state an equivalent criterion for Jordan measurability and then work with this criterion.
Call a set elementary if it is a finite union of boxes. Clearly, elementary sets are Jordan measurable. Suppose $E$ is Jordan measurable. The measure of each elementary set contained in $E$ is less than the lower integral of the characteristic function of $E$. So, the supremum of the measures of all elementary sets contained in $E$ (called the Inner Jordan measure of $E$) is less than the lower integral of the characteristic function of $E$. Similarly, the infimum of the measures of all elementary sets containing $E$ (called the Outer Jordan measure of $E$) is greater than the upper integral of the characteristic function of $E$. So, if the inner and outer Jordan measure of $E$ are equal then $E$ is Jordan measurable.
Conversely, if $E$ is Jordan measurable, then we have a partition for which the upper and lower sums of the characteristic function of $E$ are arbitrarily close. Let $T$ be the union of the boxes of this partition which intersect $E$, and let $S$ be the union of the boxes of this partition which are contained in $E$. The measures of $T$ and $S$ are nothing but the upper and lower sum respectively, so the measure of $T$ and measure of $S$ are arbitrarily close, which means the inner Jordan measure and the outer Jordan measure are equal, and equal to the Jordan measure of $E$.
Coming back to our disjoint Jordan measurable sets $A$ and $B$, we see that the Jordan outer measure of $A\cup B$ is not more than the sum of Jordan measures of $A$ and $B$, since we can have elementary sets containing $A$ and $B$ whose measures are arbitrarily close to the Jordan measures of $A$ and $B$, and the union of these elementary sets has measure less than or equal to the sum of their measures.
We can also have elementary sets contained in $A$ and $B$ whose measures are arbitrarily close to the Jordan measures of $A$ and $B$, and the union of these elementary sets has measure equal to the sum of their measures (here we use the fact that $A$ and $B$ are disjoint, and that disjoint elementary sets are seperable). So, the inner Jordan measure of $A \cup B$ is not less than the sum of Jordan measures of $A$ and $B$.
The inner Jordan measure of any set is less than or equal to its outer Jordan measure, and so we have the result.