Question

This is a theorem on Lebesgue outer measure. Please provide the proof for this theorem. Also, does this theorem have a name?

Theorem 2.3: Let {Aₙ: n ∈ ℕ} be a collection of subsets of ℝ. Then m*(⋃ₙ=₁ᵢⁿ Aₙ) ≤ ∑ₙ=₁ᵢⁿ m*(.

Answer 1

The theorem is the Countable Subadditivity axiom of Lebesgue outer measure, stating the measure of a union is at most the sum of the measures. The proof uses the definition of outer measure and converging sums of lengths of intervals that cover the sets.

Step-by-step explanation:

Proof of the Theorem on Lebesgue Outer Measure

The theorem you're asking about states that the Lebesgue outer measure of the union of a countable collection of sets is at most the sum of their individual Lebesgue outer measures. This is known as the Countable Subadditivity axiom of Lebesgue outer measure. First, we note that the Lebesgue outer measure m* is defined for all subsets A of the real numbers ℝ as the infimum of the sum of the lengths of intervals that cover A.

1. For each natural number n, let {A_n: n ∈ ℕ} be our collection of subsets of ℝ.
2. The union ⋃ₙ=1ᵢₙ A_n can be covered by a countable union of intervals since each A_n can be individually covered.
3. By the definition of outer measure, the measure of each A_n is the infimum of sums of the lengths of the covering intervals.
4. Thus, for any ε > 0, we can find a covering of each A_n with total length at most m*(A_n) + ε/2^n.
5. The union of these coverings for all n covers ⋃ₙ=1ᵢₙ A_n, and the sum of the lengths is at most ∑ₙ=1ᵢₙ (m*(A_n) + ε/2^n), which converges to ∑ₙ=1ᵢₙ m*(A_n) + ε.
6. Because ε was arbitrary, we take the limit as ε approaches 0 and obtain m*(⋃ₙ=1ᵢₙ A_n) ≤ ∑ₙ=1ᵢₙ m*(A_n), proving the theorem.