Lebesgue Integral - Homework Problem 3 April 1975 APM404

This was a favorite homework problem I kept around for a long time in a folder so I thought I'd include it here for posterity.

Let (X,M,μ) = ([a,b], Lebesgue sets, Lebesgue measure) and α(x) = x ∀ x ∈[a,b]. Show that (i) ƒ is μ measurable if and only if ƒ ∈ H₂^α and (ii) if ∫ ƒ dμ < ∞ ( or ∫ ƒ dα < ∞) then ∫ ƒ dμ = ∫ ƒ dα

1. Let ƒ ∈ H₂^α. Then there exist continuous (and therefore μ measurable) functions ƒ_n which converge to ƒ ae (almost everywhere). Since μ was obtained from an outer measure, (X,M,μ) is a complete space and ƒ is μ measurable.[(i)

2. Assume E ⊂ [a,b], E ∈ M. Then there exists a measurable set C = ∪_n=1,∞ C_n, where the C_n are closed, C ⊂ E and μ(E ∩ C^∼) = 0.

• Let B_m = ∪_n=1,m C_n. Then B_m is closed (a finite union of closed sets) and ∪_m=1,∞ B_m = C
• Thus the characteristic function X_Bm ∈ H₂^α (since B_m complement - B_m^∼ - is open; a result established on APM403 exam) and X_Bm = X_{∪n=1,m Cn} ≥ X_{∪n=1,m-1 Cn} = X_Bm-1
• Also, |X_Bm| ≤ X_[a,b] ∈ L₁ and ∫ X_Bm dα < ∫ X_[a,b] dα = b-a and X_Bm ↑ X_E ae, since μ(E ∩ C^∼) = μ(E ∩ (∪_m=1,∞ B_m )^∼) = 0 so by the Lebesgue Dominated Convergence Theorem, X_E ∈ L₁
• Finally, μ(E) ≡ ∫ X_Eμ(dx) = ∫ X_E dx ≡ α(E) by our choice of μ and α. Thus, if E ∈ M, E ⊂ [a,b] is compact, then μ(E) = α(E) ≤ b-a.

3. Let ƒ be simple, say ƒ(x) = ∑_i=1,n α_iX_Ei(x) where E_i ∩ E_j = ∅ for i ≠ j and [a,b] ⊃ E_i ∈ M, ∀_i.

• Since H₂^α is a vector space, ƒ ∈ H₂^α and
  ∫ ƒ dμ = ∑_i=1,n α_iμ(E_i) = ∑_i=1,n α_iα(E_i) = ∫ ƒ dα
• The integrals make sense since ∀_i, α_i < ∞ and μ(E_i) = α(E_i) ≤ b-a.

4. Let ƒ be bounded and measurable (non-negative), say ƒ ≤ N, ∀ x ∈ [a,b]. There exist simple functions φ_n ↑ ƒ ae (φ_n ∈ H₂^α, ∀ n by (3) above) and

• ∫ φ_n dα = ∫ φ_n dμ ≤ ∫ ƒ dμ ≤ N(b-a)
• so φ_n ∈ L₁, φ_n ↑ ƒ ae, |φ_n| ≤ N, ∫_a,bN dx < ∞ implies f ∈ L₁ and ∫ φ_n converge to ∫ ƒ
• Also, by the Monotone Convergence Theorem, ∫ φ_n dμ → ∫ ƒ dμ
• Therefore, ∫ ƒ dα = lim ∫ φ_n dα = lim ∫ φ_n dμ = ∫ ƒ dμ

5. Let ƒ be measurable, non-negative and consider g_N = ƒ ∧ N as N → ∞

• g_N ∈ H₂^α (by (4)) and g_N ↑ ƒ and ∫ g_N dα = ∫ g_N dμ ≤ ∫ ƒ dμ
• If ∫ ƒ dμ < ∞, then ƒ ∈ H₂^α and ∫ ƒ dμ = lim ∫ g_N dμ = lim ∫ g_N dα = ∫ ƒ dα

6. Let ƒ be measurable (μ). ∫ ƒ dμ exists if and only if ∫ ƒ⁺ dμ < ∞ and ∫ ƒ^- dμ < ∞

• From (5), ∫ ƒ⁺ dμ < ∞ and ∫ ƒ^- dμ < ∞ so ƒ⁺ and ƒ^- are in H₂^α and
• ∫ ƒ dμ = ∫ ƒ⁺ dμ - ∫ ƒ^- dμ = ∫ ƒ⁺ dα - ∫ ƒ^- dα = ∫ ƒ dα

Thus, ƒ ∈ H₂^α implies ƒ is measurable (μ) and if ƒ is μ-integrable then ƒ ∈ L₁^α and ∫ ƒ dμ = ∫ ƒ dx.

Corollary: Let ƒ ∈ H₂^α and c ∈ R^*. If E = {x ∈ [a,b]: ƒ(x) > c} then X_E ∈ H₂^α.

1. If c = ∞ then E = ∅ and X_∅ = 0 ∈ H₂^α
2. If c = -∞ then E = [a,b] so X_E ∈ H₂^α [X_E = | 1 - X_E^∼| ∈ H₂^α ]
3. Assume c ∈ R. ƒ ∈ H₂^α ⇒ ƒ is μ-measurable (Lebesgue) ⇔ E is μ-measurable (Lebesgue) and by step (2) of the problem, X_E ∈ H₂^α