Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-08 (Codex)
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Closure properties of measurable functions used by the integral

Statement

Let (X,A,μ) be a measure space.

  1. If f,g:X→R‾ are measurable and f+g is defined pointwise, then f+g is measurable.
  2. If c∈R and f:X→R is measurable, then cf is measurable.
  3. If f:X→R is measurable, then f+=max⁡{f,0}, f−=max⁡{−f,0}, and ∣f∣=f++f− are measurable.
  4. If E∈A and f:X→[0,+∞] is measurable, then fχE is measurable, where this function equals f on E and 0 off E (in particular, 0⋅(+∞)=0 here).
  5. If (fn) is a sequence of measurable functions X→R‾, then inf⁡nfn is measurable; if moreover fn↑f pointwise, then f is measurable.

Facts & Assumptions

Given: A measure space (X,A,μ) and functions or sets as in the relevant clause.

[L1]

A function h:X→R‾ is measurable exactly when {h>a}∈A for every real a (Extended-real-valued measurable functions).

[L2]

A sigma-algebra contains X and ∅ and is closed under complements and countable unions; countable intersections follow by taking complements (Sigma-algebras).

Proof

technique · direct
1.1L1L2L3

For any extended-real-valued h, the identities {h<a}=⋃q∈Q, q<a(X∖{h>q}) and {h>a}=⋃q∈Q, q>a(X∖{h<q}) follow from rational density, including when h(x) is infinite. Thus [L2] and [L3] show that measurability of all strict sublevels is equivalent to measurability of all strict superlevels.

1.2givenL1L2

Put h=fχE as defined in clause 4. For a<0, {h>a}=X, since h≥0. For a≥0, {h>a}=E∩{f>a}. Both sets are measurable, including at a=0, so clause 4 follows.

1.3L1L2L3given

For a pointwise-defined sum, {f+g>a}=⋃q∈Q({f>q}∩{g>a−q}). Indeed, if both summands are finite and their sum exceeds a, choose a rational strictly between a−g(x) and f(x). If one summand is +∞, the other is not −∞, and a rational meeting the two inequalities still exists; if a summand is −∞, the defined sum cannot exceed a. The reverse inclusion follows by adding the inequalities. The union is countable and measurable, proving clause 1.

2.1L1L2step 1.1

For c>0, {cf>a}={f>a/c}; for c<0, {cf>a}={f<a/c}. If c=0, each superlevel is either X or ∅. Step 1.1 and [L1] therefore prove clause 2.

2.2givenL1L2step 1.1

Put u=inf⁡nfn and v=sup⁡nfn. The defining order properties of infimum and supremum give {u<a}=⋃n{fn<a} and {v>a}=⋃n{fn>a}, including infinite values. By step 1.1 and [L2] these sets are measurable, so u and v are measurable by [L1]. If fn↑f, then f=v pointwise. This proves clause 5.

3.1L1L2step 1.3step 2.1

The function −f is measurable by step 2.1. For any real-valued measurable h, the superlevel of max⁡(h,0) is X when a<0 and {h>a} when a≥0. Apply this to h=f and h=−f to obtain measurable f+ and f−. They are finite-valued, so step 1.3 makes ∣f∣=f++f− measurable, proving clause 3.

4.1step 1.3step 2.1step 3.1step 1.2step 2.2∎

Clauses 1–5 follow respectively from steps 1.3, 2.1, 3.1, 1.2, and 2.2.

Depends on

Used by

Dependency tree · two levels

32 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