Are all stopped Brownian motions integrable? Let $(W_t)_{t\in[0,\infty)}$ be a standard Brownian motion and let $\tau$ be a stopping time that is finite almost everywhere, but otherwise arbitrary—it need not be a hitting time, in particular.

I am wondering whether the random variable $\omega\mapsto W_{\tau(\omega)}({\omega})$ is integrable for all such stopping times or whether a counterexample can be constructed.


Intuitively, the expected value of $|W_t(\omega)|$ for a fixed $t\geq0$ is proportional to $\sqrt{t}$, which tends to make it less likely that $\omega\mapsto W_{\tau(\omega)}({\omega})$ is integrable if one can make $\tau$ take on large values with sufficiently high probabilities. On the other hand, the probability of $\tau$ being “large enough” is small (given that $\tau$ is finite almost everywhere), which tends to favor integrability.
I cannot see whether a counterexample can be constructed making the first of these two effects “win” against the second, yielding $$\int_{\omega\in\Omega}|W_{\tau(\omega)}(\omega)|\,\mathrm d\mathbb P(\omega)=+\infty.$$

Any hints and comments would be much appreciated.
 A: Using the idea in your answer we can prove something stronger.

For any cumulative distribution function $F$ on $\mathbb{R}$, there is an a.s. finite stopping time such that $W_\tau$ has the distribution $F$.

In particular we can choose an $F$ with infinite first moment.
Recall the following lemma which is fairly standard and not hard to prove:

For any cdf $F$, define $F^{\leftarrow} : (0,1) \to \mathbb{R}$ by $F^{\leftarrow}(y) = \sup\{x : F(x) < y\}$.  If $U$ is a random variable with a uniform(0,1) distribution, then $F^{\leftarrow}(U)$ has the distribution $F$.

Note that if $F$ is continuous and strictly increasing then $F^{\leftarrow} = F^{-1}$.
Next let $\Phi$ be the standard normal cdf.  If $Z$ has a standard normal distribution, then $\Phi(Z)$ has a uniform(0,1) distribution.  In particular this is true for $\Phi(W_1)$.  So let $g = F^{\leftarrow} \circ \Phi$; then $g(W_1)$ has the distribution $F$.
Now let
$$\tau = \inf\{t \ge 1 : W_t = g(W_1)\}.$$
As in your answer, by the strong Markov property, $X_s = W_{s+1} - W_1$ is a Brownian motion independent of $W_1$.  Brownian motion is recurrent, so for any $x$, $X_s$ hits the value $x$ almost surely.  Since $\{X_s\}$ is independent of $W_1$, this implies that $X_s$ hits the value $g(W_1) - W_1$ almost surely.  This happens when $W_t$ hits the value $g(W_1)$, so $\tau < \infty$ almost surely.  And it is clear that $W_\tau = g(W_1)$, so $W_\tau$ has the distribution $F$.
It's worth mentioning here the Skorohod embedding theorem, which says that if $F$ has zero mean and finite variance, then we can find an integrable stopping time $\tau$ such that $W_\tau \sim F$.  You can find a proof in Brownian Motion by Mörters and  Peres.  The assumption of zero mean and finite variance is necessary here, since it follows from the Wald lemmas that if $\tau$ is integrable then $E[W_\tau] = 0$ and $E[W_\tau^2] = E[\tau]$.
A: Here is a (somewhat contrived) counterexample. The idea is this: wait until the period $t=1$. Then, decide when to stop based solely on the value of $W_1$. In particular, choose for each realization of $W_1(\omega)$ such an extremely late time of eventual stopping that makes the first effect mentioned above “win” against the second, rendering an infinite expected absolute value of the stopped process.

For each $n\in\mathbb N$, define
\begin{align*}
p_n\equiv&\;\mathbb P\big(\{\omega\in\Omega\,|\,n-1\leq|W_1(\omega)|<n\}\big),\\
t_n\equiv&\;\frac{\pi}{2p_n^2}+1.
\end{align*}
Since $W_1$ has the standard normal distribution, the sequence $(p_n)_{n\in\mathbb N}$ of probabilities is strictly positive, strictly decreasing, and converges to $0$. Correspondingly, the sequence $(t_n)_{n\in\mathbb N}$ is well-defined, strictly increasing, always greater than $1$, and diverges to $+\infty$.
Now, define $$\tau(\omega)\equiv t_n\quad\text{if $n-1\leq|W_1(\omega)|<n$.}$$
It is not difficult to verify that $\tau$ is a stopping time that is everywhere finite-valued.
The following argument [which exploits the facts that for any $t>1$, (i) $W_1$ and $W_t-W_1$ are independent; and (ii) $W_t-W_1$ has the normal distribution of mean $0$ and variance $t-1$] establishes that $W_{\tau}-W_1$ is not integrable, which implies that $W_{\tau}$ is not integrable, either:
\begin{align*}
&\;\int_{\omega\in\Omega}|W_{\tau(\omega)}(\omega)-W_1(\omega)|\,\mathrm d\mathbb P(\omega)=\sum_{n=1}^{\infty}\int_{\{\omega\in\Omega\,|\,\tau(\omega)=t_n\}}|W_{t_n}(\omega)-W_1(\omega)|\,\mathrm d\mathbb P(\omega)\\
=&\;\sum_{n=1}^{\infty}\int_{\{\omega\in\Omega\,|\,n-1\leq|W_1(\omega)|<n\}}|W_{t_n}(\omega)-W_1(\omega)|\,\mathrm d\mathbb P(\omega)\\
=&\;\sum_{n=1}^{\infty}\mathbb P\big(\{\omega\in\Omega\,|\,n-1\leq|W_1(\omega)|<n\}\big)\times\mathbb E(|W_{t_n}-W_1|)=\sum_{n=1}^{\infty}p_n\times\sqrt{\frac{2}{\pi}\times(t_n-1)}\\
=&\;\sum_{n=1}^{\infty}1=+\infty.
\end{align*}
