Is $\inf\left\{t\in\left[0,1\right]\vert t+B^2_t=1\right\}$ a stopping time?

Question

Problem

Let $\left(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq 0},\mathbb{P}\right)$ be a filtered probability space such that $(\mathcal{F}_t)_{t\geq 0}$ is a complete and right-continuous filtration and let $B=(B_t)_{t\geq 0}$ be a standard $\mathbb{R}$-valued $(\mathcal{F}_t)_{t\geq 0}$-Brownian motion. Is $S:=\inf\left\{t\in\left[0,1\right]\vert t+B^2_t=1\right\}$ a stopping time?

Similiar Problems

The definition of a stopping time doesn't seem to be immediately useful here:

$T:\Omega\to\mathbb{R}_+\cup\left\{\infty\right\}$ is called a stopping time if $\left\{T\leq t\right\}\in\mathcal{F}_t$ for all $t\geq 0$.

Normally, I make use of the fact that $B$ is (a.s.) continuous and:

If $(\mathcal{F}_t)_{t\geq 0}$ is a right-continuous filtration and $X=(X_t)_{t\geq 0}$ is an (a.s.) continuous $\mathbb{R}$-valued adapted process, then $T_A := \inf\left\{t\geq 0\vert X_t\in A\right\}$ is a stopping time for every open or closed Borel set $A\in\mathcal{B}(\mathbb{R})$.

As an example, consider $S_a := \inf\left\{t\geq 0\vert \left\lvert B_t\right\rvert = a\right\}$ with $a\geq 0$. It is a stopping time since $S_a = \inf\left\{t\geq 0\vert B_t\in A\right\}$ where $A=\left\{-a\right\}\cup\left\{a\right\}$ is the union of two closed Borel sets and therefore a closed Borel set as well.

Intuition

We have $S:=\inf\left\{t\in\left[0,1\right]\vert \left\lvert B_t\right\rvert=\sqrt{1-t}\right\}$. Intuitively, the (a.s.) continuous $B_t$ is caged in by the continuously decreasing upper bound $\sqrt{1-t}$ and the continuously increasing lower bound $-\sqrt{1-t}$. Both bounds reach 0 at the end which means that there must be a point in time $t$ such that $\left\lvert B_t\right\rvert=\sqrt{1-t}$. This intuition may be nice, but it's far from the definition of a stopping time.

Question

1. Are there other useful theorems about stopping times?
2. I am convinced that $S:=\inf\left\{t\in\left[0,1\right]\vert t+B^2_t=1\right\}$ is a stopping time, but how can I show that?

Answer 1

(Edited) Useful theorems

1. Assuming nothing about the filtration:

If $X=(X_t)_{t\geq 0}$ is a continuous $\mathbb{R}^d$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process, then $T_A:=\inf\left\{t\geq 0 \vert X_t ∈ A\right\}$ is a stopping time for every closed Borel set $A\in\mathcal{B}(\mathbb{R}^d)$.

2. Assuming that the filtration is right-continuous:

If $(\mathcal{F}_t)_{t\geq 0}$ is a right-continuous filtration and $X=(X_t)_{t\geq 0}$ is a continuous $\mathbb{R}^d$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process, then $T_A:=\inf\left\{t\geq 0\vert X_t\in A\right\}$ is a stopping time for every open or closed Borel set $A\in\mathcal{B}(\mathbb{R}^d)$.

3. Assuming that the filtration is right-continuous and complete:

If $(\mathcal{F}_t)_{t\geq 0}$ is a right-continuous and complete filtration and $X=(X_t)_{t\geq 0}$ is an (a.s.) continuous $\mathbb{R}^d$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process, then $T_A:=\inf\left\{t\geq 0\vert X_t\in A\right\}$ is a stopping time for every Borel set $A\in\mathcal{B}(\mathbb{R}^d)$.

(Edited) Solution

$(\mathcal{F}_t)_{t\geq 0}$ is a right-continuous and complete filtration. $(t + B_t^2)_{t\geq 0}$ is an (a.s.) continuous $\mathbb{R}$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process because the Brownian motion $(B_t)_{t\geq 0}$ is. $S:=\inf\left\{t\geq 0\vert t+B^2_t\in A\right\}$ is therefore a stopping time because $A=\left\{1\right\}$ is a closed Borel set. Note that $t+B^2_t=1 \Rightarrow t\in\left[0,1\right]$ which means that $S=\inf\left\{t\in\left[0,1\right]\vert t+B^2_t=1\right\}$.

Remark

Note that $\inf\emptyset := \infty$ and that stopping times $S$ with $S=\infty$ exist. Your intuition only shows that $S<\infty$.