It is known that the stochastic integral satisfies the following property
$$ \mathbb{E}\left[\left\langle \int_0^{\cdot}X(s)\,dM(s) \right\rangle_t\right]= \mathbb{E}\left[ \int_0^t X^2(s) \, d\left\langle M \right\rangle_s\right], $$
where $M$ is a square integrable martingale, and $\langle\cdot \rangle$ denotes the quadratic variation which can be defined as $$ \langle M \rangle_t=_{\mathbb{P}}\lim_{n\to+\infty}\sum_{i\geq 0} \left(M\left(\frac{i+1}{2^n}\wedge t\right)-M\left(\frac{i}{2^n}\wedge t\right)\right)^2. $$
I want to know if the above property holds ponit-wise a.s., i.e. whether
$$ \left\langle \int_0^\cdot X(s) \, dM(s) \right\rangle_t= \int_0^t X^2(s) \, d\left\langle M \right\rangle_s,\text{ a.s.?} $$