Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $I\subseteq\mathbb R$, $(\mathcal F_t)_{t\in I}$ be a filtration on $(\Omega,\mathcal A)$, $\mathcal F_{\sup I}:=\sigma(\mathcal F_t:t\in I)$ and $p\ge 1$.
I would like to show that $\bigcup_{t\in I}\mathcal L^p(\mathcal F_t,\operatorname P)$ is a dense subspace of $\mathcal L^p(\mathcal F_{\sup I},\operatorname P)$.
Let $$\mathcal C:=\left\{A\in\mathcal A:1_A\in\overline{\bigcup_{t\in I}\mathcal L^p(\mathcal F_t,\operatorname P)}^{L^p(\operatorname P)}\right\}.$$ If I'm not missing something, $\mathcal C$ should be a $\sigma$-algebra, but I'm not sure whether we need to assume that $I$ is countable to obtain this.
Now, since $\bigcup_{t\in I}\mathcal F_t\subseteq\mathcal C$, we obtain $$\mathcal F_{\sup I}\subseteq\mathcal C.$$ And since $$\mathcal L^p(\mathcal G,\operatorname P)=\overline{\left\{f:\Omega\to\mathbb R:f\text{ is }\mathcal G\text{-measurable and }|f(\Omega)|\in\mathbb N\right\}}^{L^p(\operatorname P)}\tag1,$$ for every $\sigma$-algebra $\mathcal G\subseteq\mathcal A$ on $\Omega$, we should immediately obtain the claim.
So, the only thing I'm unsure about is the necessity of assuming $I$ to be countable.