TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Maximal ergodic theorem

Statement

Let $(X, A, μ, T)$ be a measure-preserving system for an arbitrary measure $μ$ , and let $f : X \to R$ be a finite-valued measurable representative in $L^{1} (μ)$ . With the unnormalised sums $S_{n} f$ , put

$E : = {x : \sup_{n \geq 1} S_{n} f (x) > 0} .$

Then

$\int_{E} f d μ \geq 0.$

Neither finiteness of $μ$ , invertibility of $T$ , nor ergodicity is assumed.

Facts & Assumptions

Given: The system and real integrable representative in the Statement.

[F1]

Composition by $T$ preserves measurability and the integral of every nonnegative measurable or integrable function (Integral invariance under measure-preserving maps).

[F2]

Integrable functions form a vector space and their integral is linear (The Lebesgue integral is linear on $L^{1} (μ)$ ).

[F3]

Increasing nonnegative measurable functions satisfy monotone convergence (Monotone convergence for the integral).

Proof

technique · direct finite-maximum argument

1.1

For $N \geq 1$ , set $F_{N} : = \max (0, S_{1} f, \dots, S_{N} f), E_{N} : = {F_{N} > 0} .$ Every $S_{j} f$ is integrable, so $F_{N}$ is measurable and integrable because a finite maximum of real functions is obtained from addition and absolute value. Also $F_{N} \geq 0$ and $F_{N} = 0$ on $X ∖ E_{N}$ .

F1F2

2.1

Since $F_{N} \geq S_{j} f$ for $0 \leq j \leq N - 1$ , composition and addition give $F_{N} \circ T + f \geq S_{j + 1} f$ . Hence $F_{N} \circ T + f \geq \max_{1 \leq j \leq N} S_{j} f = F_{N} on E_{N},$ where strict positivity is what permits insertion of the zeroth sum $S_{0} f = 0$ .

step 1.1

3.1

Integrating the preceding inequality over $E_{N}$ , using $F_{N} = 0$ off $E_{N}$ , nonnegativity of $F_{N} \circ T$ , and invariance of its integral, yields $\int_{E_{N}} f d μ \geq \int_{X} F_{N} d μ - \int_{E_{N}} F_{N} \circ T d μ \geq \int_{X} F_{N} d μ - \int_{X} F_{N} \circ T d μ = 0.$ All displayed integrals are finite because $F_{N}$ is integrable.

F1F2step 1.1step 2.1

4.1

The sets $E_{N}$ increase and their union is $E$ . Applying monotone convergence separately to $f^{+} 1_{E_{N}}$ and $f^{-} 1_{E_{N}}$ gives $\int_{E_{N}} f d μ ⟶ \int_{E} f d μ .$ Passing to the limit in the nonnegative inequalities of step 3.1 proves the claim.

F2F3step 3.1∎

Depends on

Used by

Norm convergence alone does not imply pointwise convergence False statement
Sigma-finite ergodic oscillation sets have finite measure Lemma
Birkhoff pointwise ergodic theorem Theorem
Finite-measure identification of the Birkhoff limit Theorem

Dependency tree · two levels

24 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

Alessio Del Vigna, The Birkhoff Ergodic Theorem (standard reference, not scraped)
Charles Walkden, Ergodic Theory lecture notes (standard reference, not scraped)