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Borel representatives make the convolution integrand Borel measurable
Statement
Let be Borel measurable functions. Then
is Borel measurable on . In particular, for each fixed , the section is measurable.
Facts & Assumptions
Given: Borel measurable functions on .
The functions and are Borel measurable by hypothesis.
The Borel product on is the Euclidean Borel sigma-algebra on (The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}).
Composition with Borel functions preserves measurability, and measurable arithmetic operations preserve measurability (Composition with a Borel measurable outer map preserves measurability, Arithmetic and lattice operations preserve measurability whenever they are defined).
Sections of product-measurable functions are measurable (Every section of a product-measurable function is measurable).
Proof
The map given by [L2, L3, given, construct] is continuous, hence Borel measurable. Since and are Borel measurable on by [L2] and [L3], the functions and are Borel measurable.
Multiplication on is continuous, so [L3] makes [L3, L4, step 1.1]
Borel measurable on . Then [L4] gives measurability of each section .
Depends on
- Convolution of two functions on $\mathbb{R}^n$
- A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra
- The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}
- Composition with a Borel measurable outer map preserves measurability
- Arithmetic and lattice operations preserve measurability whenever they are defined
- Every section of a product-measurable function is measurable
Used by
- FALSE: the Borel-representative discipline in convolution is unnecessary because continuous precomposition always preserves Lebesgue measurability False statement
- Convolution on L¹(ℝⁿ) is independent of the chosen Borel representatives Lemma
- Convolution with a mollifier is smooth, and derivatives pass under the integral sign Theorem
- If f,g ∈ L¹(ℝⁿ), then f*g exists almost everywhere, belongs to L¹, and ‖f*g‖₁ ≤ ‖f‖₁ ‖g‖₁ Theorem
Dependency tree · two levels
26 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
- Walter Rudin, Real and Complex Analysis, 3rd ed. (standard reference, not scraped)