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Sigma-finite ergodic oscillation sets have finite measure

Statement

Let $μ$ be sigma-finite, let $T$ preserve $μ$ , and let $f \in L^{1} (μ)$ be real valued. For rationals $β < α$ , put

$E_{α, β} : = {x : {lim inf}_{n} A_{n} f (x) < β < α < {lim sup}_{n} A_{n} f (x)} .$

Then $E_{α, β}$ is invariant (in particular, invariant modulo null sets) and has finite measure.

Facts & Assumptions

Given: The sigma-finite system, $f$ , and rationals $β < α$ in the Statement.

[F1]

The maximal ergodic theorem applies to every real integrable representative on an arbitrary measure space (Maximal ergodic theorem).

[F2]

Sigma-finiteness supplies a countable finite-measure cover (Finite, sigma-finite, and semifinite measures).

[F3]

For integrable $h$ , $∣ \int h ∣ \leq \int ∣ h ∣$ (The modulus of an integral is bounded by the integral of the modulus).

Proof

technique · cases on the sign of $\alpha$

1.1

The exact identity $A_{n} f (T x) = \frac{n + 1}{n} A_{n + 1} f (x) - \frac{1}{n} f (x)$ shows, by taking lower and upper limits, that both limiting envelopes have the same value at $T x$ as at $x$ ; finite-valuedness of $f (x)$ makes the last term tend to zero. Thus $T^{- 1} E_{α, β} = E_{α, β}$ .

givenalgebra

1.2

Assume first that $α > 0$ , and let $C \subseteq E_{α, β}$ be measurable with $μ (C) < \infty$ . The function $g = f - α 1_{C}$ is integrable. For $x \in E_{α, β}$ , some $n$ satisfies $S_{n} f (x) > n α$ , while $S_{n} 1_{C} (x) \leq n$ ; hence $S_{n} g (x) > 0$ . Therefore $C$ lies in $G : = {\sup_{n} S_{n} g > 0}$ .

assume-case alphapositivegivenalgebra

2.1

By [F1], $\int_{G} g \geq 0$ . Since $C \subseteq G$ , $α μ (C) \leq \int_{G} f d μ \leq ∣ \int_{G} f d μ ∣ \leq \int_{G} ∣ f ∣ d μ \leq ∥ f ∥_{1} .$ Here the middle absolute-value inequality follows because the left side is nonnegative.

F1F3step 1.2

3.1

From a sigma-finite cover form the increasing finite-measure exhaustion $X_{m}$ by finite unions, and take $C_{m} = E_{α, β} \cap X_{m}$ . Step 2.1 gives $μ (C_{m}) \leq ∥ f ∥_{1} / α$ , while $C_{m} ↑ E_{α, β}$ . Continuity from below, which follows from countable additivity of the measure, gives $μ (E_{α, β}) \leq ∥ f ∥_{1} / α < \infty$ .

F2step 2.1

4.1

If $α \leq 0$ , then $- β > 0$ . The same set is the oscillation set for $- f$ with upper threshold $- β$ and lower threshold $- α$ , because $lim sup (- A_{n} f) = - lim inf A_{n} f$ and $lim inf (- A_{n} f) = - lim sup A_{n} f$ . Applying steps 1.2–3.1 to $- f$ proves its measure finite.

assume-case alphanonpositivestep 3.1cases-exhaustive∎

Depends on

Used by

Birkhoff pointwise ergodic theorem Theorem

Dependency tree · two levels

22 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)
Omri Sarig, Lecture Notes on Ergodic Theory (2023) (standard reference, not scraped)