I'm reading a remark at page 4 of these lecture notes.
Let $\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$be a filtration and $M=\left(M_t, t \in \mathbb{R}_{+}\right)$be a continuous square-integrable martingale with respect to $\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$.
Reminder. The quadratic variation of $M$ is the unique process $\left(\langle M\rangle_t, t \in \mathbb{R}_{+}\right)$which is increasing, continuous and adapted to $\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$, such that $\langle M\rangle_0=0$ a.s. and $\left(M_t^2-\langle M\rangle_t, t \in \mathbb{R}_{+}\right)$is a martingale with respect to $\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$.
Lemma 1.1. For all $t>s \geq 0$, $$ \mathbb{E}\left(\left(M_t-M_s\right)^2 | \mathcal{F}_s\right)=\mathbb{E}\left(\langle M\rangle_t-\langle M\rangle_s | \mathcal{F}_s\right). $$
Remark. In general, $\langle M\rangle_t$ is not deterministic, but when $M$ has independent increments, then $\langle M\rangle_t=\mathbb{E}\left(M_t^2\right)-\mathbb{E}\left(M_0^2\right)$ (and is therefore deterministic).
My understanding In other threads (1, 2, 3, 4, 5), the remark holds if in addition $M$ is a Gaussian process, i.e., if $M$ is Gaussian then $t \mapsto \mathbb{E}\left(M_t^2\right)$ is continuous.
Could you confirm that the remark is not necessarily correct?