Let A,B ⊆ R. Show that |m∗(A) − m∗(B)| ≤ m∗(A B), where m∗(A B) = (A\B)∪(B\A)
We can prove this using the triangle inequality for the outer measure.
First, we note that A and B can overlap, so we cannot simply calculate the difference between their measures. Instead, we need to take into account their union and intersection.
Let C = A ∩ B and D = A ∪ B. Then,
m*(A) = m*(D) - m*(D \ A)
m*(B) = m*(D) - m*(D \ B)
Adding these equations, we get
m*(A) + m*(B) = 2m*(D) - [m*(D \ A) + m*(D \ B)]
Now, we consider the set E = (A \ B) ∪ (B \ A). Since E is the disjoint union of D \ A and D \ B, we have
m*(E) = m*(D \ A) + m*(D \ B)
Combining the two previous equations, we get
m*(A) + m*(B) = 2m*(D) - m*(E)
Rearranging, we get
m*(A) - m*(B) = 2m*(D) - m*(E) - 2m*(B)
Since m*(E) ≥ 0 and m*(B) = m*(C) by subadditivity, we can further simplify this to
|m*(A) - m*(B)| ≤ 2m*(D) - 2m*(C) = m*(D) + m*(D) - m*(C) - m*(C) = m*(D) - m*(C) ≤ m*(E)
Thus, we have shown that |m*(A) - m*(B)| ≤ m*(A B).
To prove that |m*(A) - m*(B)| ≤ m*(AB), where m*(AB) = (A\B)∪(B\A), we can break it down into two cases:
Case 1: If m*(A) ≤ m*(B), then we have:
m*(A) = m*((A\B)∪(A∩B)) (since A = (A\B)∪(A∩B))
≤ m*((A\B)∪(B\A)) (since A∩B ⊆ B\A)
≤ m*(AB) (since (A\B)∪(B\A) ⊆ AB)
Since |m*(A) - m*(B)| = m*(B) - m*(A) ≥ 0, it follows that |m*(A) - m*(B)| ≤ m*(AB).
Case 2: If m*(A) > m*(B), then we can interchange A and B in Case 1 to get:
m*(B) = m*((B\A)∪(B∩A)) (since B = (B\A)∪(B∩A))
≤ m*((B\A)∪(A\B)) (since B∩A ⊆ A\B)
≤ m*(AB) (since (B\A)∪(A\B) ⊆ AB)
Again, |m*(A) - m*(B)| = m*(B) - m*(A) ≤ 0, so |m*(A) - m*(B)| ≤ m*(AB).
Therefore, in both cases, we have shown that |m*(A) - m*(B)| ≤ m*(AB).