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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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Every nonnegative measurable function is the increasing limit of simple measurable functions

Statement

Let f:X→[0,+∞] be measurable. Then there is an increasing sequence of nonnegative simple measurable functions (sn) such that sn(x)↑f(x) for every x∈X.

Facts & Assumptions

Given: A measurable function f:X→[0,+∞].

[L1]

For a measurable f, the sets f−1(B) are measurable for every Borel B⊆R‾ (Extended-real-valued measurable functions).

[L2]

A nonnegative measurable function with finite range is a nonnegative simple measurable function (Nonnegative simple measurable functions).

[L3]

Increasing pointwise suprema of measurable functions are measurable, and measurable functions remain measurable under the elementary truncations used below (Closure properties of measurable functions used by the integral).

Proof

technique · direct
1.1L1L2construct

Set s0:=0. For n≥1 and 0≤k<n2n, put En,k:={x:k2−n≤f(x)<(k+1)2−n} and set sn:=∑k=0n2n−1k2−nχEn,k+nχ{f≥n}. Each En,k and {f≥n} is measurable by [L1], the range of sn is finite, and therefore each sn is simple by [L2]; s0 is also simple.

2.1step 1.1algebra

For each x, one has 0≤sn(x)≤f(x). If f(x)<+∞ and n>f(x), then f(x)−2−n<sn(x)≤f(x); if f(x)=+∞, then sn(x)=n. Hence sn(x)→f(x).

3.1step 1.1step 2.1algebra∎

The functions are increasing. Indeed, sn(x) is the largest multiple of 2−n at most f(x)∧n, hence is also a multiple of 2−(n+1) at most f(x)∧(n+1). The maximality of the latter dyadic truncation gives sn(x)≤sn+1(x). Together with step 2.1, this proves sn↑f.

Depends on

Used by

Dependency tree · two levels

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Sources