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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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Borel representatives make the convolution integrand Borel measurable

Statement

Let f~,g~:RnC be Borel measurable functions. Then

H(x,y):=f~(xy)g~(y)

is Borel measurable on R2n. In particular, for each fixed xRn, the section yH(x,y) is measurable.

Facts & Assumptions

Given: Borel measurable functions f~,g~ on Rn.

[A1]

The functions f~ and g~ are Borel measurable by hypothesis.

[L2]

The Borel product on Rn×Rn is the Euclidean Borel sigma-algebra on R2n (The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}).

[L3]

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).

[L4]

Sections of product-measurable functions are measurable (Every section of a product-measurable function is measurable).

Proof

technique · direct
1.1

The map T:R2nR2n given by [L2, L3, given, construct] T(x,y):=(xy,y) is continuous, hence Borel measurable. Since (u,v)f~(u) and (u,v)g~(v) are Borel measurable on R2n by [L2] and [L3], the functions (x,y)f~(xy) and (x,y)g~(y) are Borel measurable.

L2L3givenconstruct
2.1

Multiplication on C is continuous, so [L3] makes [L3, L4, step 1.1]

H(x,y)=f~(xy)g~(y)

Borel measurable on R2n. Then [L4] gives measurability of each section yH(x,y).

L3L4step 1.1

Depends on

Used by

Dependency tree · two levels

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Sources