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Convolution on $L^{1} (R^{n})$ is independent of the chosen Borel representatives

Statement

Let $f, g \in L^{1} (R^{n})$ . If ${\tilde{f}}_{1}, {\tilde{f}}_{2}$ are Borel representatives of $f$ and ${\tilde{g}}_{1}, {\tilde{g}}_{2}$ are Borel representatives of $g$ , then for almost every $x \in R^{n}$ ,

$\int {\tilde{f}}_{1} (x - y) {\tilde{g}}_{1} (y) d y = \int {\tilde{f}}_{2} (x - y) {\tilde{g}}_{2} (y) d y .$

Facts & Assumptions

Given: Two Borel representatives for each of the $L^{1}$ classes $f$ and $g$ .

[L1]

The integrands from Borel representatives are measurable (Borel representatives make the convolution integrand Borel measurable).

[L2]

The Lebesgue integral respects almost-everywhere equality (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

[L3]

Lebesgue measurability and null sets are translation invariant (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

Proof

technique · direct

1.1

Let $N_{f} : = {{\tilde{f}}_{1} \neq {\tilde{f}}_{2}}$ and $N_{g} : = {{\tilde{g}}_{1} \neq {\tilde{g}}_{2}}$ . [L3, given, algebra] These are null sets. For a fixed $x$ , the set ${y : {\tilde{f}}_{1} (x - y) \neq {\tilde{f}}_{2} (x - y)} = x - N_{f}$ is also null by [L3]. Hence the two section integrands agree for almost every $y$ , outside the null set $(x - N_{f}) \cup N_{g}$ .

L3givenalgebra

2.1

By [L1], both section integrands are measurable, and step 1.1 says they are [L1, L2, step 1.1] equal almost everywhere in $y$ . Therefore [L2] gives equality of their integrals whenever either side is defined as an absolutely convergent Lebesgue integral.

L1L2step 1.1

3.1

This holds for every fixed $x$ , so in particular it holds for almost every [step 2.1] $x$ on the domain where the $L^{1}$ convolution is defined. Thus the convolution does not depend on the chosen Borel representatives.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

24 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)

Convolution on L1(Rn)L1(Rn) is independent of the chosen Borel representatives