If quadratic variation of a local martingale is zero then it is itself zero

Question

Let $M$ be a local martingale, if we need it, we can assume that $M$ is continuous. We know that $\langle M\rangle =0$. This implies that $M$ and $M^2$ are local martingale. Can we conclude that $M=0$? If so, how?

Answer 1

Hint: Suppose $M$ is a martingale such that $\langle M \rangle = 0$. Then $M^2$ is a martingale, thus

$$\mathbb{E}(M_t^2-M_0^2) = 0$$

Now use that $$\mathbb{E}(M_t^2 - M_s^2) = \mathbb{E}((M_t-M_s)^2)$$ for all $s \leq t$.