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.
Arithmetic and lattice operations preserve measurability whenever they are defined
Statement
Let be a measurable space and let be measurable. Then:
- is measurable for every real scalar ;
- , , , , and are measurable;
- if is pointwise defined, then is measurable;
- with the convention of The convention is used only for pointwise products of measurable functions, the pointwise product is measurable.
Facts & Assumptions
Given: A measurable space and measurable functions .
Extended-real measurability is equivalent to measurability of the threshold sets . (Threshold characterisations of real-valued and extended-real-valued measurability)
The positive and negative parts are and . (The positive and negative parts of a function)
In this proof, the pointwise product uses the page convention .
Proof
Scalar multiples are measurable. If , then [given, L1] ; if , then ; and if , the function is constant. So [L1] gives measurability of , and in particular of .
The threshold identities
show via [L1] that and are measurable. By [L2], this proves measurability of and ; replacing by also gives . [step 1.1, L1, L2]
Assume is pointwise defined. For every real ,
The inclusion from right to left is immediate. For the converse, if then either , in which case any rational works, or is finite and one may choose a rational with . Thus [L1] gives measurability of . [step 2.1, L1]
Suppose first that are nonnegative and measurable. [step 3.1, L1, A1] If , then . If , then
Again the inclusion from right to left is immediate. For the converse, if , choose a rational with and ; this is possible because either is finite positive and the rationals are dense, or , in which case any sufficiently large positive rational works. Hence nonnegative products are measurable by [L1]. [step 3.1, L1, A1]
For general measurable and , step 2.1 gives measurable nonnegative [step 2.1, step 3.1, step 4.1, L2, A1] functions . By step 4.1 the four products are measurable. Put
At each point, at least one of and is zero, because at least one of and at least one of is zero. So the difference is pointwise defined without the forbidden form, and step 3.1 makes it measurable. By the usual sign decomposition, , with the convention [A1] at the points. [step 2.1, step 3.1, step 4.1, L2, A1]
Steps 1.1 through 5.1 prove all four claims.
Depends on
Used by
Dependency tree · two levels
7 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
- Sheldon Axler, Measure, Integration and Real Analysis, Section 2B (standard reference, not scraped)
- John K. Hunter, Measure Theory, Definition 3.3 (standard reference, not scraped)