Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck 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:X→R‾ be measurable. Then there is a sequence of simple functions sk:X→R such that

∣sk∣≤∣f∣for every k,

and sk(x)→f(x) for every x∈X.

Facts & Assumptions

Given: A measurable function f:X→R‾.

[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
1.1L2L1L3

By [L2], the functions f+ and f− are measurable and nonnegative. Applying [L1, L2, L3] [L3] to them gives simple functions uk↑f+ and vk↑f−. [L1, L2, L3].

2.1step 1.1L1∎

Put sk:=uk−vk. 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+vk≤f++f−=∣f∣. Also sk→f+−f−=f pointwise because uk→f+ and vk→f−. [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