Closures, interiors and Jordan measure
- Proof outline: see other side
- Proof outline: see other side
- Proof outline: see other side
- Proof outline: see other side
(4), shows that Jordan outer measure (and inner measure) do not possess finite additivity for non-measurable sets. For the outer measure case, the question of what conditions might be needed to imply fintie additivity is partially answered by the Caratheodory type property.
The proofs below rely on the monotonicity of the Jordan inner and outer measure—two properties that Tao doesn't introduce explicitly. Tao does state the existence of monotonicity of both when the set being measured is Jordan measurabled.
1. and the closure of have the same Jordan
outer measure.
Proof. Claim 1:
Claim 1, Jordan outer measure of a set is not greater than the outer measure
of the set's closure. Proof. This follows from monotonicity of Jordan outer
measure: any elementary set that contains
Claim 2, Jordan outer measure of a set's closure is not greater than the
outer measure of the set itself. Proof. Consider any almost-disjoint elementary
cover
Any elementary set can be expressed as a finite union of disjoint boxes, so to consider the measure of an arbitrary disjoint cover is to consider the measure of any arbitrary cover.
2. and the interior of have the same
Jordan inner measure.
The proof of this has a symmetry with the previous proof.
Proof. Claim 1:
Claim 1, Jordan inner measure of a set is not less than the inner measure of
the set's interior. Just like the proof above, the first claim follows from
monotonicity, this time the monotonicity of Jordan inner measure. As
Claim 2, Jordan inner measure of a set is not greater than the inner measure of the set's interior. Consider any almost-disjoint set of boxes contained
within
3. is Jordan measurable iff the topological boundary
of has Jordan outer measure zero.
Forward implication proof. Assume that
The inner equality is true by the definition of Jordan measurability, and the two outer equalities hold by the two statements that were proven above.
Applying monotonicity of inner and outer measure we conclude that both
Reverse implication proof. Assume that
Lemma.
So, I'm stuck on this proof.
Visually, monotonicity and sub-additivity gives us the following situation:
And the ability to convert covers to be open or closed gives us the following situation:
Using the assumption about zero outer measure for the boundary, we can apply sub-aditivity of the outer and inner measures again. This squashes all outer measures together and all inner measures together:
But I'm not sure how to get to this:
4. Jordan measure of the bullet-riddled square and the set of bullets
The closure of the rationals is the reals, and the closure of the reals minus the rationals is the reals. So the outer measure of both sets is the reals, in this case just the subset
The Jordan outer measure of both sets is equal to the outer measure of