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The Brownian zero set has Lebesgue measure zero

Statement

Assume the Axiom of Choice. Let $B$ be a standard Brownian motion and let $Z$ be its zero set as in The Brownian zero set, understood through the all-path continuous jointly measurable version. Then for every $0 \leq T < \infty$ the Lebesgue measure of $Z_{T}$ is zero almost surely: $P (λ (Z_{T}) = 0) = 1.$ Here set $Z_{0} = Z \cap {0}$ , extending the positive-horizon notation. There is one measurable full event on which $λ (Z) = 0$ and all finite horizons have zero measure. On its intersection with the supplied event of all-time agreement, the same pathwise assertion holds for the original B.

Facts & Assumptions

Given: AC, a standard Brownian motion $B$ , its normalized version $\hat{B}$ and zero set $Z$ , and a horizon $T \in (0, \infty)$ .

[F1]

The zero set is the closed, nonempty set $Z = {t \geq 0 : {\hat{B}}_{t} = 0}$ of the all-path continuous version $\hat{B}$ , and $Z_{T} = Z \cap [0, T]$ . The Brownian zero set

[F2]

The map $(t, ω) \mapsto {\hat{B}}_{t} (ω)$ is $B ([0, \infty)) \otimes F$ -measurable, so $(t, ω) \mapsto 1_{{{\hat{B}}_{t} = 0}}$ is product measurable. Brownian motion has a jointly measurable continuous version

[F3]

For t>0 the Brownian definition gives $B_{t} - B_{0}$ law N(0,t), and B_0=0 almost surely, hence B_t has that law. By the definition of the normal law, it is the law of sqrt(t) times a standard normal variable. Brownian motion Standard normal and normal laws

[F4]

Tonelli: for a product-measurable $f \geq 0$ on a product of sigma-finite spaces, the section integrals are measurable and the two iterated integrals agree. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

[F5]

AC is the ambient assumption of the Brownian interfaces. The Axiom of Choice

[F6]

A nonnegative measurable function has integral zero exactly when it vanishes almost everywhere. Countable unions of probability-zero events are null. A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere Basic identities for a probability measure

Proof

technique · direct

1.1

For t>0, [F3] gives $P (B_{t} = 0) = γ ({0}) = 0$ , because sqrt(t)>0 and the standard normal density integrates to zero over the Lebesgue-null singleton {0}. The latter singleton convention is in [F1]. The normalized version agrees with B on one measurable full event by [F2], so $P ({\hat{B}}_{t} = 0) = 0$ as well. At t=0 the probability is one, but that time singleton has Lebesgue measure zero. The zero indicator is product measurable by [F2].

F1F2F3

2.1

Applying [F4] to that indicator over the sigma-finite product $[0, T] \times Ω$ with the product measure $d t \otimes P$ gives $E λ (Z_{T}) = \int_{0}^{T} P ({\hat{B}}_{t} = 0) d t = 0$ , and it also shows that $ω \mapsto λ (Z_{T} (ω))$ is measurable.

step 1.1F4

3.1

Since $0 \leq λ (Z_{T}) \leq T$ , its measurability and zero expectation in step 2.1 permit [F6], giving $λ (Z_{T}) = 0$ almost surely.

F6step 2.1

4.1

Intersect the measurable probability-one events from step 3.1 over the explicitly listed horizons N>=1. By [F6] their intersection A is measurable with probability one. On A, every finite nonnegative T has $Z \cap [0, T] \subseteq Z \cap [0, N]$ for some integer N>=T, so its measure is zero by monotonicity. Also $Z = ⋃_{N \geq 1} (Z \cap [0, N])$ , so countable subadditivity of Lebesgue measure gives $λ (Z) = 0$ on A. Conversely a zero-measure whole zero set has zero-measure intersections, which explains the global formulation.

F1F6step 3.1

5.1

At T=0 the zero set is the singleton {0}, so its measure is zero pathwise; the time endpoint t=0 does not affect step 2.1. Let A_* be the measurable full event on which the supplied normalized process agrees with B at all times. On A intersect A_* from step 4.1 the original B path has exactly the same zero set, proving its pathwise nullness there. No claim is needed that arbitrary exceptional paths of B have measurable zero sets or that the entire all-time equality event is measurable. The countable horizon list is fixed; full AC is inherited from [F5], with no further selection of paths or exceptional events.

F1F2F5step 2.1step 4.1∎

Source notes

Durrett, Section 7.4.1, computes $E ∣ Z \cap [0, T] ∣ = \int_{0}^{T} P (B_{t} = 0) d t = 0$ and concludes that the zero set has measure zero; the argument above makes the Tonelli step explicit through the all-path continuous jointly measurable version, so the section integrals are measurable without any auxiliary regularity assumption.

Depends on

Used by

The Brownian zero set is uncountable Corollary
A null uncountable random closed set Example

Dependency tree · two levels

36 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

Rick Durrett, Probability: Theory and Examples, fifth edition, Section 7.4.1 (the zero set has measure zero) (standard reference, not scraped)
Perla Sousi, Advanced Probability, Theorem 6.39 (standard reference, not scraped)