Author:
(1) Attila Losonczi.
Table of Links
1.1 Basic notions and notations
1.2 Basic definitions from [7] and [8]
2 Generalized integral
2.1 Multiplication on [0, +β) Γ [ββ, +β]
2.3 The integral of functions taking values in [0, +β) Γ [0, +β)
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 Οalgebra on P(K)). Let f : K β [0, +β) Γ [0, +β) be a function. We say that f is measurable if for each (d, m) β [0, +β) Γ [0, +β) {x β K : f(x) < (d, m)} β S holds.
Proposition 2.11. Let (K, S) be a measurable space and f : K β [0, +β) Γ [0, +β). Then the following statements are equivalent.

f is measurable.

For each (d, m) β [0, +β) Γ [0, +β], {x β K : f(x) β€ (d, m)} β S.

For each (d, m) β [0, +β) Γ [0, +β], {x β K : (d, m) < f(x)} β S.

For each (d, m) β [0, +β) Γ [0, +β], {x β K : (d, m) β€ f(x)} β S.
Proof. All equivalences simply follow from the fact that the space [0, +β) Γ [0, +β] is first countable and T2.
Proposition 2.12. Let (K, S) be a measurable space and f : K β [0, +β) Γ [0, +β) be measurable. Then the following statements hold.
Let K = [0, 1], S be the Borel sets, g: [0, 1] β [0, 1] be a non Borel measurable function and f(x) = (x, g(x)) when x β [0, 1]. Clearly {x β K: f(x) < (d, m)} equals to either {x β K: x < d} or {x β K: x β€ d}, and both sets are Borel, hence f is measurable.
Proposition 2.14. Let (K, S) be a measurable space and f: K β [0, +β) Γ [0, +β) be measurable and (dβ², mβ²) β [0, +β)Γ[0, +β). Then (dβ², mβ²)f is measurable as well.
Proof. Let (d, m) β [0, +β) Γ [0, +β).
If (d β², mβ²) = (0, 0), then the statement is trivial.
If mβ² = 0, dβ² > 0, then {x: (dβ², mβ²)f(x) β€ (d, m)} = {x: f(x) β€ (d β dβ², +β)}, where similar argument works
Proposition 2.15. Let (K, S) be a measurable space and f, g : K β [0, +β) Γ [0, +β) be measurable. Then f + g is measurable as well.
This paper is available on arxiv under CC BYNCND 4.0 DEED license.