Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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Every measurable function admits simple approximations dominated by its absolute value

Statement

Let (X,A) be a measurable space and let f:XR be measurable. Then there is a sequence of simple functions sk:XR such that

skffor every k,

and sk(x)f(x) for every xX.

Facts & Assumptions

Given: A measurable function f:XR.

[L1]

The positive and negative parts satisfy f=f+f, f=f++f, and at each point at least one of f+,f is zero. (The positive and negative parts of a function)

[L2]

The arithmetic-and-lattice theorem makes f+ and f measurable. (Arithmetic and lattice operations preserve measurability whenever they are defined)

[L3]

Every nonnegative measurable function admits an increasing sequence of simple approximations. (Every nonnegative measurable function admits an explicit increasing sequence of simple approximations)

Proof

technique · direct
1.1

By [L2], the functions f+ and f are measurable and nonnegative. Applying [L1, L2, L3] [L3] to them gives simple functions ukf+ and vkf.

L1L2L3
2.1

Put sk:=ukvk. Because uk and vk are simple, sk is a [step 1.1, L1] simple real-valued function. At each point, [L1] makes at least one of uk and vk equal to 0, so

sk=uk+vkf++f=f.

Also skf+f=f pointwise because ukf+ and vkf. [step 1.1, L1] ∎

Depends on

Used by

Dependency tree · two levels

7 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