Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, and let $\lambda$ be Lebesgue measure. Consider a square integrable continuous martingale $M$ with quadratic variation $\langle M\rangle$. The Doléans measure $\mu_M$ on $([0,\infty)\times \Omega,\mathcal{B}([0,\infty))\otimes\mathcal F)$ is defined by is defined by $$\mu_M(A)=\int_{\Omega}\int_0^{\infty} \mathbf{1}_A(t,\omega)d\langle M\rangle_t \mathbb P(d\omega).$$ Does this measure have the property that $\mu_M(\mathcal N)=0$ for all $\lambda\times \mathbb P$-null sets $$\mathcal N\in \mathcal B[0,\infty))\otimes\mathcal F?$$
My attempt
First, I thought that for every fixed $\omega$ and $\mathcal B([0,\infty))$-null set $N$, we have $\int_0^{\infty}\mathbf{1}_N(t,\omega)d\langle M\rangle_t=0$. This could lead to an affirmative answer to the question. However, I found a counterexample: if $N$ is the Cantor set and $\langle M\rangle_t$ is the Cantor function, then this integral equals $1$. Since the Cantor set is a Lebesgue null-set, we cannot use this line of reasoning.
