# Quadratic variation condition and square integrability

Assume $$M$$ is local continuous martingale started from $$0$$. How would one go about showing that if it is a martingale bounded in $$L^2$$, then $$E[\langle M\rangle_{\infty}]\lt \infty$$?

This is not true. The condition $$\mathbb{E}[\langle M,M \rangle_\infty] < \infty$$ implies $$M$$ is a true martingale, but you can have continuous local martingales that are bounded in $$L^2$$ but not true martingales. For example, if $$B_t$$ is a Brownian motion in $$\mathbb{R}^3$$ started from $$(1,0,0)$$ then $$M_t := \frac{1}{|B_t|}$$ is a continuous local martingale bounded in $$L^2$$, but is not a true martingale. We can look at $$M_t - 1$$ instead to satisfy $$M_0 = 0$$.
• The iff in that exercise (19.7) is that $M \in \mathcal M_0^{2,c}$ iff $\mathbb{E}[\langle M,M \rangle_\infty] < \infty$, i.e. we need to assume $M$ is a true martingale rather than just a local martingale to conclude $\mathbb{E}[\langle M,M \rangle_\infty] < \infty$. May 5, 2021 at 0:43
Under your hypotheses the process $$K_t:=M_t^2-\langle M\rangle_t$$ is a continuous local martingale. Let $$(T_n)$$ be an increasing sequence of stopping times with $$\lim_nT_n=\infty$$ (a.s.) and each stopped process $$K^{T_n}$$ a martingale. By Doob's inequality, for each $$t>0$$ and each $$n$$, $$\Bbb E[\langle M\rangle_{T_n\wedge t}]=\Bbb E[M^2_{T_n\wedge t}]\le\Bbb E[\sup_{0\le s\le t}M^2_s]\le 4\Bbb E[M^2_t]\le C<\infty,$$ for some constant $$C$$. By monotonicity, $$\Bbb E[\langle M\rangle_\infty]\le C<\infty$$.