ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedaudited 2026-09-22

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Indicator of a stopping interval

Example

Assume the Axiom of Choice and the standing hypothesis (H) of Elementary predictable Brownian integrands, and assume the usual conditions required by the localized-integral interface. For every stopping time $τ$ Continuous-time stopping times and stopped sigma-algebras and every $t \geq 0$ , $\int_{0}^{t} 1_{[0, τ]} (s) d B_{s} = B_{t \land τ} - B_{0} almost surely,$ and the two sides are continuous processes over $t$ , hence indistinguishable. In particular for the deterministic stopping time $τ \equiv t_{0}$ the integral is $B_{t \land t_{0}} - B_{0}$ , the Brownian path stopped at $t_{0}$ .

Facts & Assumptions

Given: AC, the standing hypothesis (H), the usual conditions on the filtration, a stopping time $τ$ and $t \geq 0$ .

[F1]

On each finite horizon $[0, T]$ , the process $1_{(0, T]}$ is elementary: with the partition $0 < T$ and coefficient $1 \in F_{0}$ , its defining sum is $I_{t} (1_{(0, T]}) = B_{t} - B_{0}$ . It represents the same $(d t \otimes P)$ -class as the constant process $1$ , because they differ only at time $0$ . Elementary predictable Brownian integrands Ito integral of an elementary predictable process

[F2]

The process $H \equiv 1$ is predictable and locally square-integrable, with energy $\int_{0}^{t} 1^{2} d s = t < \infty$ ; for such an $H$ and any stopping time $τ$ the stopping identity $(H \cdot B)_{t \land τ} = \int_{0}^{t} 1_{[0, τ]} H d B$ holds up to indistinguishability, and both sides are continuous. Locally square-integrable predictable Brownian integrands Stopping an Ito integral

[F3]

AC is declared for the ambient interfaces. The Axiom of Choice

Verification

technique · direct

1.1

The predictable finite-energy process $H \equiv 1$ is represented in $L^{2} (d t \otimes P)$ on each finite horizon by the elementary process $1_{(0, T]}$ of [F1]. Hence its integral is the elementary sum $B_{t} - B_{0}$ , and the stopped quantity is $(1 \cdot B)_{t \land τ} = B_{t \land τ} - B_{0}$ .

F1F2given

2.1

By [F2] the stopping identity applies with $H \equiv 1$ : $(1 \cdot B)_{t \land τ} = \int_{0}^{t} 1_{[0, τ]} (s) d B_{s}$ up to indistinguishability, and substituting step 1.1 for the left-hand side gives the displayed identity; both sides are continuous in $t$ because $B$ is and $t \mapsto t \land τ$ is.

F2step 1.1

3.1

The cases $τ \equiv \infty$ (integral equals $B_{t} - B_{0}$ ), $τ \equiv t_{0}$ (integral equals $B_{t \land t_{0}} - B_{0}$ ) and $τ \equiv 0$ (integral vanishes) are all instances; the identity is a statement about the localized integral of a bounded integrand, so no integrability of $τ$ is required. AC enters only through [F3].

F1F3step 2.1given∎

Source notes

Lawler, Section 3.2.3, records the stopped-integral identity for the constant integrand; the version here is the constant- $H$ case of the general stopping theorem of item 17.

Depends on

Used by

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Sources

Gregory F. Lawler, Stochastic Calculus: An Introduction with Applications, Section 3.2.3 (standard reference, not scraped)