TheoremBase

Integrable Functions and Their Lebesgue Integral

A real-valued measurable function is integrable when the integrals of its positive and negative parts are finite; L¹(μ) is the set of integrable functions, and the integral of such a function is the difference of those two integrals.

Statement

In the setting of Measure Spaces, Borel Sets and Lebesgue Measure on the Real Line: Standing Notation, let (X,F,μ)(X,\mathcal{F},\mu) be a measure space. Let Meas(X,F;R,B(R))\mathrm{Meas}(X,\mathcal{F};\mathbb{R},\mathcal{B}(\mathbb{R})) be the set of measurable functions X→RX\to\mathbb{R} of Measurable Map, and the Set of Measurable Maps (ii), let M+=M+(X,F)\mathcal{M}^{+}=\mathcal{M}^{+}(X,\mathcal{F}) be the set of measurable functions into [0,∞][0,\infty], and for g∈M+g\in\mathcal{M}^{+} let ∫Xg dμ∈[0,∞]\int_{X}g\,d\mu\in[0,\infty] be its integral. For f:X→Rf:X\to\mathbb{R}, f+f^{+} and f−f^{-} are its positive part and negative part.

Integrable functions. A function f∈Meas(X,F;R,B(R))f\in\mathrm{Meas}(X,\mathcal{F};\mathbb{R},\mathcal{B}(\mathbb{R})) is integrable with respect to μ\mu if

∫Xf+ dμ<∞and∫Xf− dμ<∞\int_{X}f^{+}\,d\mu<\infty\qquad\text{and}\qquad\int_{X}f^{-}\,d\mu<\infty

(here f+,f−∈M+f^{+},f^{-}\in\mathcal{M}^{+} by Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i) and Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)).

The set L1(μ)\mathcal{L}^{1}(\mu). L1(X,F,μ)\mathcal{L}^{1}(X,\mathcal{F},\mu), also written L1(μ)\mathcal{L}^{1}(\mu), is the set of those f∈Meas(X,F;R,B(R))f\in\mathrm{Meas}(X,\mathcal{F};\mathbb{R},\mathcal{B}(\mathbb{R})) that are integrable with respect to μ\mu (a set by The Functions f ↦ f⁺ and f ↦ f⁻ on the Real-Valued Functions on a Set, and a Predicative Form of g = f⁺ and g = f⁻ (ii) and Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices (iv)).

The integral. For f∈L1(μ)f\in\mathcal{L}^{1}(\mu), the integrals ∫Xf+ dμ\int_{X}f^{+}\,d\mu and ∫Xf− dμ\int_{X}f^{-}\,d\mu are finite, hence real numbers (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (ii)). The integral of ff with respect to μ\mu is the real number

∫Xf dμ=∫Xf+ dμ−∫Xf− dμ,\int_{X}f\,d\mu=\int_{X}f^{+}\,d\mu-\int_{X}f^{-}\,d\mu,

also written ∫f dμ\int f\,d\mu; the function f↦∫Xf dμf\mapsto\int_{X}f\,d\mu from L1(μ)\mathcal{L}^{1}(\mu) to R\mathbb{R} is the integral with respect to μ\mu (it exists by The Functions f ↦ f⁺ and f ↦ f⁻ on the Real-Valued Functions on a Set, and a Predicative Form of g = f⁺ and g = f⁻ (iii)). A function in both M+\mathcal{M}^{+} and L1(μ)\mathcal{L}^{1}(\mu) thus has two integrals; where it matters, the text names which one is meant.

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…