Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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Finite-measure Lr includes into Lp for p<r

Statement

Let (X,A,μ) be a measure space with μ(X)<.

  1. If 1p<r< and fLr(μ), then fLp(μ) and fpμ(X)1/p1/rfr.
  2. If 1p< and fL(μ), then fLp(μ) and fpμ(X)1/pf.

Facts & Assumptions

Given: A finite measure space (X,A,μ).

[L1]

Holder's inequality for integrals is available (Holder's inequality for integrals, including the endpoint cases).

[L2]
[L3]
[L4]

If hL(μ), then hh almost everywhere (The essential supremum is attained as the least essential bound).

Proof

Proof technique: Write fp as fp1 and apply Holder with exponents r/p and r/(rp). The finite total measure contributes the factor μ(X)1/p1/r.

1.1

Suppose 1p<r< and put [L1, L2, L3, given, algebra] a:=rp,b:=rrp. Then a,b(1,) and 1/a+1/b=1, so [L2] makes them conjugate. Apply [L1] to the functions fp and 1: fpdμ(fpadμ)1/a(1bdμ)1/b=(frdμ)p/rμ(X)1p/r. Taking p-th roots gives the claimed bound.

1.2

If fL(μ), then [L3, L4, given, algebra] fpdμfpμ(X). Indeed, fpfp almost everywhere by [L4]. Thus fLp(μ) and fpμ(X)1/pf.

2.1

Steps 1.1 and 1.2 prove the finite-measure inclusion laws.

step 1.1step 1.2

Depends on

Used by

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Sources