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Absolute continuity of the integral

Statement

Let $f \in L^{1} (μ)$ and let $ε > 0$ . Then there is $δ > 0$ such that for every measurable $E$ , $μ (E) < δ ⟹ \int_{E} ∣ f ∣ d μ < ε .$

Facts & Assumptions

Given: An integrable function $f$ and a real number $ε > 0$ .

[L1]

The truncations $∣ f ∣ \land n$ increase pointwise to $∣ f ∣$ , so their integrals converge to $\int ∣ f ∣ d μ$ by monotone convergence (Monotone convergence for the integral).

[L2]

The integral over a measurable set is defined by $\int_{E} h d μ = \int h χ_{E} d μ$ (Integral over a measurable subset).

[L3]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L4]

Integrability means $\int ∣ f ∣ d μ < + \infty$ (Integrable real and complex functions, and their integrals).

[L5]

The nonnegative integral is additive (Additivity of the nonnegative Lebesgue integral).

Proof

technique · direct

1.1L1L4L5choose

Put $h_{n} = ∣ f ∣ - ∣ f ∣ \land n \geq 0$ . The pointwise identity $∣ f ∣ = (∣ f ∣ \land n) + h_{n}$ and [L5] give $\int h_{n} = \int ∣ f ∣ - \int (∣ f ∣ \land n)$ ; the subtraction is valid because both integrals are finite by [L4]. By [L1] choose $n$ so large that $\int h_{n} d μ < ε / 2$ , and put $δ : = ε / (2 n + 1)$ .

2.1step 1.1L2L3L5algebra∎

If $μ (E) < δ$ , then [L2], [L3], and [L5] give $\int_{E} ∣ f ∣ d μ = \int_{E} (∣ f ∣ \land n) d μ + \int_{E} h_{n} d μ \leq n μ (E) + \int h_{n} d μ < n δ + ε / 2 < ε$ . The last inequality follows from $n / (2 n + 1) < 1 / 2$ .

Depends on

Dependency tree · 2 levels

Sources

John K. Hunter, Measure Theory Notes, Proposition 4.16