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).
Let A,B ⊆ R. Show that |m∗(A) − m∗(B)| ≤ m∗(A B), where m∗(A B) = (A\B)∪(B\A)
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