\ 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.

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f is measurable.

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

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

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

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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.

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\ 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.

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\ 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.

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

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This content originally appeared on HackerNoon and was authored by Hausdorff