Show that $X \in \mathcal{M}^{loc}_c$ implies: $X$ is locally square summable and locally bounded.

Question

If $X$ is a continuous local martingale with $X_0= 0$ a.s., then:

1) the process $X$ is a locally bounded martingale;

2) $X$ is locally square integrable.

Localising sequence definition. There exists a sequence of stopping times $T^n$ s.t. some process belongi ng to class $C^{loc}$ stopped at $T^n$ will belong to $C$.

I know that Lebesgue integrable functions( which cts.loc.mgle is) imply their boundedness.

Could you throw hint what to look at so that I can finish the proofs?

Answer 1

Hints: Since $X$ is a local martingale, there exists a sequence of stopping times $(\tau_k)_{k \in \mathbb{N}}$ such that $\tau_k \uparrow \infty$ and $(X_{t \wedge \tau_k})_{t \geq 0}$ is a martingale.

1. Show that $$\sigma_k := \tau_k \wedge \inf\{t \geq 0; |M_t| \geq k\}$$ is a stopping time for each $k \in \mathbb{N}$ and that $\sigma_k \uparrow \infty$ as $k \to \infty$.
2. Conclude from the optional stopping theorem that $(M_{t \wedge \sigma_k})_{t \geq 0}$ is a martingale.
3. Show that $|M_{t \wedge \sigma_k}| \leq k$. Deduce that $(M_{t \wedge \sigma_k})_{t \geq 0}$ is a bounded martingale.
4. Recall that any bounded random variable $X$ on a probability space is square integrable. Hence, by step 3, $(M_{t \wedge \sigma_k})_{t \geq 0}$ is square integrable.