Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space and $\mathcal G = (\mathcal G_t, t \ge 0)$ a filtration. Let $M$ be a real-valued continuous martingale w.r.t. $\mathcal G$ such that $$ (\star) \quad M_0 = 0 \quad \text{and} \quad (\star \star) \quad \lim_{t \to \infty} \langle M \rangle_t=\infty. $$
Fix $r >0$ and define a stopping time $\tau := \inf \{t \ge 0 : \langle M \rangle_t \ge r \}$. By $(\star \star)$, we have $\tau$ is finite. I would like to prove that
Theorem $M_\tau$ is square-integrable.
Could you have a check on my below attempt?
Proof By OST, $M^\tau = (M_{t \wedge \tau}, t \ge 0)$ is a martingale. We define a process $X$ by $X_t := M_t^2 - \langle M \rangle_t$ for all $t \ge0$. Then $X$ and thus $X^\tau$ are martingales. By $(\star)$, we have $X^\tau_0=0$. Hence $$ \mathbb E [X_{t \wedge \tau}] = \mathbb E [ |M_{t \wedge \tau}|^2 - \langle M \rangle_{t \wedge \tau} ] = 0. $$
We have $\mathbb E [ \langle M \rangle_{t \wedge \tau} ] \le \mathbb E [ \langle M \rangle_{\tau} ]$. Because $M$ and thus $\langle M \rangle$ have continuous sample paths, we get $\mathbb E [ \langle M \rangle_{\tau} ] = r$. Hence $$ \mathbb E [ |M_{t \wedge \tau}|^2 ] = \mathbb E [ \langle M \rangle_{t \wedge \tau} ] \le r. $$
By Doob's maximal inequality, $$ \mathbb E \big [ \sup_{s \in [0, t]} | M_{s \wedge \tau} |^2 \big ] \le 4 \mathbb E [ | M_{t \wedge \tau} |^2 ] \le 4 r. $$
By monotone convergence theorem, $$ \mathbb E \big [ \sup_{s \in [0, \infty)} | M_{s \wedge \tau} |^2 \big ] \le 4r. $$
Clearly, $$ | M_{\tau} |^2 \le \sup_{s \in [0, \infty)} | M_{s \wedge \tau} |^2. $$
Hence $\mathbb E [ | M_{\tau} |^2 ] \le4r$. This completes the proof.
