Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-09-04
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Step functions on one period are dense in L^1 on the torus

Statement

Assume the Axiom of Countable Choice.

Let f be integrable on one period. For every ε>0 there is a step function s on [0,1) such that

01f(t)s(t)dt<ε.

Equivalently, one-period step functions are dense in L1(T).

Facts & Assumptions

Given: The Axiom of Countable Choice, a one-period integrable function f, and a real ε>0.

[L1]

The torus conventions identify one-period integrable functions with the L1(T) objects used on this page (Period-one Fourier coefficients, partial sums, and convolution on the torus).

[L2]

Assuming the Axiom of Countable Choice, finite linear combinations of interval indicators are dense in L1(R) (Finite linear combinations of box indicators are dense in Lp(Rn) for 1p<).

Proof

technique · direct
1.1

Define g:RR by g(t)=f(t) for 0t<1 and g(t)=0 otherwise. Then gL1(R). By [L2], choose a finite linear combination of interval indicators u with Rg(t)u(t)dt<ε.

givenL2choose
2.1

Restrict u to [0,1) and call the restriction s. Intersecting each interval in u with [0,1) produces only finitely many subintervals, so s is a step function on [0,1). Since g=f on [0,1), 01f(t)s(t)dt=01g(t)u(t)dtRg(t)u(t)dt<ε.

L1step 1.1algebra
3.1

Step 2.1 is exactly the claimed density statement on one period.

step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources