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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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Bochner integrability criterion

Statement

Let f:Ω→X be strongly measurable. Then f is Bochner integrable if and only if

∫Ω∥f∥ dμ<∞.

Here an a.e.-defined scalar function is integrated through any measurable representative supplied by the strong simple approximation. Moreover the Bochner integral is independent of the approximating sequence in its definition.

Facts & Assumptions

[L1]

Bochner integrability means L1 approximation by integrable simple functions and defines the integral as the norm limit of their integrals (Bochner-integrable function).

[L2]

Nonnegative integrals are monotone and positively homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral) and additive (Additivity of the nonnegative Lebesgue integral).

[L3]

Pointwise limits and countable suprema of measurable scalar functions are measurable (Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable).

[L4]

Scalar dominated convergence gives convergence in L1 (Dominated convergence); its nonnegative foundation is monotone convergence (Monotone convergence for the integral).

Proof

technique · direct

Given: A strongly measurable f and the conventions in the Statement.

1.1givenL1L2

Prove necessity of scalar norm integrability. [given, L1, L2] Suppose first that f is Bochner integrable and choose (sn) as in [L1]. For some n, ∫∥f−sn∥<∞, while integrability of the simple function gives ∫∥sn∥<∞. Since ∥f∥≤∥f−sn∥+∥sn∥, [L2] gives ∫∥f∥<∞.

1.2givenL1

Prove approximation independence. [given, L1] If (sn) and (tn) are any two defining approximations, the simple norm inequality gives ∥∫sn−∫tn∥≤∫∥sn−tn∥, which is at most ∫∥sn−f∥+∫∥f−tn∥. Both terms tend to zero, so the two norm limits coincide.

1.3givenL3construct

Construct dominated simple approximants for sufficiency. [given, L3, construct] Conversely assume ∫∥f∥<∞. On the exceptional measurable null set of a strong approximation, replace both f and every approximant by zero (and call the representative again f). Thus measurable simple functions un converge pointwise to f; [L3] makes ∥f∥ measurable. Define sn=un1{∥un∥≤2∥f∥}. Then sn is simple and measurable, ∥sn∥≤2∥f∥, and sn→f pointwise: when f(ω)≠0, the inequality defining the retained part holds eventually, while at a zero of f either retained values tend to zero or the replacement is zero.

2.1L2step 1.3

Verify that the constructed simple functions are integrable. [L2, step 1.3] Each sn is integrable. Indeed, if a nonzero value x occurs, its level set is contained in {2∥f∥≥∥x∥}, whose measure is at most 2∫∥f∥/∥x∥<∞ by [L2].

3.1L1L4step 1.3step 2.1

Obtain convergence in L1. [L1, L4, step 1.3, step 2.1] The pointwise convergence in step 1.3 and ∥f−sn∥≤3∥f∥ allow [L4] to be applied. Hence ∫∥f−sn∥→0. Together with step 2.1 this is exactly the approximation required in [L1].

4.1L1step 1.1step 1.2step 3.1∎

Step 1.1 proves necessity, step 3.1 proves sufficiency, and step 1.2 proves that the resulting integral is approximation-independent. The zero function, the empty measure space, and a single simple function are included by taking the constant zero or constant simple approximation.

Depends on

Used by

…and 20 more results.

Cited to discharge well-definedness by Bochner-integrable function.

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources