Let A,B ⊆ R such that A ⊂ B. If A /∈ L, show that m∗(B \A) > 0.
Since A is not measurable, we know that m*(A) < ∞ and there exists a measurable set E such that E contains A and m*(E \ A) > 0 (by definition of non-measurable set).
Since A ⊂ B, we have B ⊆ E. Therefore,
(B \ A) ⊆ (E \ A)
This implies that
m*(B \ A) ≤ m*(E \ A)
However, we know that m*(E \ A) > 0. Therefore, m*(B \ A) must also be greater than 0, as otherwise we would have m*(E \ A) = 0.
Hence, we have shown that if A /∈ L, then m∗(B \A) > 0.
To show that m*(B \ A) > 0, we can prove this by contradiction.
1. Assume that m*(B \ A) = 0.
2. Since m* represents the Lebesgue outer measure, m*(B \ A) = inf{∑(i=1 to ∞) |B_i|}, where B_i are intervals covering B \ A.
3. As m*(B \ A) = 0, it implies that for any ε > 0, there exists a countable collection of intervals B_1, B_2, ..., such that ∑(i=1 to ∞) |B_i| < ε.
4. Since A ⊂ B, we can write B + (B \ A) ⊇ A, where B + (B \ A) represents the union of B and (B \ A).
5. By monotonicity of the outer measure, we have m*(B + (B \ A)) ≥ m*(A).
6. Since m*(B + (B \ A)) = m*(B) + m*(B \ A) and m*(A) = m*(B), this simplifies to m*(B) + m*(B \ A) ≥ m*(B).
7. By subtracting m*(B) from both sides of the inequality, we obtain m*(B \ A) ≥ 0.
8. This contradicts our assumption that m*(B \ A) = 0.
9. Therefore, our initial assumption is false, and we can conclude that m*(B \ A) > 0.
Hence, if A ⊆ B and A ∉ L, it implies that m*(B \ A) > 0.