Generalized Hausdorff Integral and Its Applications: Measurabilityby@hausdorff

# Generalized Hausdorff Integral and Its Applications: Measurability

July 17th, 2024

In this article we will look at the notion of measurability of generalized functions. We will also look at applications to the theory of Borel sets. The paper is written by Attila Losonczi and will be available on the Hackernoon website. For more information about the paper, please visit: http://hackernoon.com/.

Author:

(1) Attila Losonczi.

Abstract and 1 Introduction

1.1 Basic notions and notations

1.2 Basic definitions from [7] and [8]

2 Generalized integral

2.1 Multiplication on [0, +∞) × [−∞, +∞]

2.2 Measurability

2.3 The integral of functions taking values in [0, +∞) × [0, +∞)

3 Applications

4 References

## 2.2 Measurability

First we need to define measurability of generalized functions.

Definition 2.10. Let (K, S) be a measurable space (i.e. S is a σalgebra on P(K)). Let f : K → [0, +∞) × [0, +∞) be a function. We say that f is measurable if for each (d, m) ∈ [0, +∞) × [0, +∞) {x ∈ K : f(x) < (d, m)} ∈ S holds.

Proposition 2.11. Let (K, S) be a measurable space and f : K → [0, +∞) × [0, +∞). Then the following statements are equivalent.

1. f is measurable.

2. For each (d, m) ∈ [0, +∞) × [0, +∞], {x ∈ K : f(x) ≤ (d, m)} ∈ S.

3. For each (d, m) ∈ [0, +∞) × [0, +∞], {x ∈ K : (d, m) < f(x)} ∈ S.

4. For each (d, m) ∈ [0, +∞) × [0, +∞], {x ∈ K : (d, m) ≤ f(x)} ∈ S.

Proof. All equivalences simply follow from the fact that the space [0, +∞) × [0, +∞] is first countable and T2.

Proposition 2.12. Let (K, S) be a measurable space and f : K → [0, +∞) × [0, +∞) be measurable. Then the following statements hold.

Let K = [0, 1], S be the Borel sets, g: [0, 1] → [0, 1] be a non Borel measurable function and f(x) = (x, g(x)) when x ∈ [0, 1]. Clearly {x ∈ K: f(x) < (d, m)} equals to either {x ∈ K: x < d} or {x ∈ K: x ≤ d}, and both sets are Borel, hence f is measurable.

Proposition 2.14. Let (K, S) be a measurable space and f: K → [0, +∞) × [0, +∞) be measurable and (d′, m′) ∈ [0, +∞)×[0, +∞). Then (d′, m′)f is measurable as well.

Proof. Let (d, m) ∈ [0, +∞) × [0, +∞).

If (d ′, m′) = (0, 0), then the statement is trivial.

If m′ = 0, d′ > 0, then {x: (d′, m′)f(x) ≤ (d, m)} = {x: f(x) ≤ (d − d′, +∞)}, where similar argument works

Proposition 2.15. Let (K, S) be a measurable space and f, g : K → [0, +∞) × [0, +∞) be measurable. Then f + g is measurable as well.

This paper is available on arxiv under CC BY-NC-ND 4.0 DEED license.

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