Possible example of stopped martingale not being in $L^1$

Question

I am looking at Lawler's stochastic process book, and for the optional stopping theorem, under the assumption for the stopping time that $\mathbb{P}(T<\infty) = 1$, he had

1. $\mathbb{E}[|M_T|]< \infty$, and
2. $\mathbb{E}[|M_n|{\bf 1}_{\{T>n\}}]$ goes to zero as $n\to \infty$.

I am confused by the first point, if we assume our martingale satisfies $M_n\in L^1$, could you help me with a simple example that $$\mathbb{E}[|M_T|]= \infty?$$

Answer 1

Let $B_t$ be a Brownian motion and $Y$ a $(0,\infty)$-valued random variable such that $Y \not\in L^1$ and is independent from Brownian motion. Let $\mathcal{F}_t$ be the Brownian filtration augmented by $\sigma (Y)$. Define the stopping time $$T(\omega ) = \inf\{t\geq 0 \, : \, |B_t|(\omega ) \geq Y(\omega) \}$$ Since $B_t$ a.s. takes on any value in finite time, we have that $P(T < \infty) = 1$. We also have: $$E(|B_T|) = E(Y) = \infty$$ Despite $B_t \in L^1$ for all $t$.