Filtration's Lp Spaces are Dense in Limit Filtration's Lp Space

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\sequence {\FF_n}_{n \mathop \in \N}$ be a filtration of $\Sigma$.

Let $\FF_\infty$ be the limit of $\sequence {\FF_n}_{n \mathop \in \N}$.

Let $p \ge 1$.

Let $\map {L^p} {\cdot}$ denote the $L^p$ spaces.

Then $\ds \bigcup_{n \mathop \ge 0} \map {L^p} {X, \FF_n, \mu}$ is a dense subset of $\map {L^p} {X, \FF_\infty, \mu}$.

Proof
First, it is a subset, since
 * $\forall n \ge 0 : \map {L^p} {X, \FF_n, \mu} \subseteq \map {L^p} {X, \FF_\infty, \mu}$

in view of $\FF_n \subseteq \FF_\infty$.

In the following, we show its density.

Let:
 * $\ds \AA_0 := \bigcup_{n \mathop \ge 0} \FF_n$

By Union of Algebras of Sets is Algebra, $\AA_0$ is an algebra.

By Sigma-Algebra extended from Algebra by Measure:
 * $\ds \BB := \bigcap_{\epsilon \mathop > 0} \bigcup_{A \mathop \in \AA_0} \set {B \in \Sigma : \map \mu {A \symdif B} \le \epsilon}$

is a $\sigma$-algebra.

In particular:
 * $(1):\quad \FF_\infty = \map \sigma {\AA_0} \subseteq \BB$

Let $f \in \map {L^p} {X, \FF_\infty, \mu}$.

Let $\epsilon > 0$.

By Space of Simple P-Integrable Functions is Everywhere Dense in Lebesgue Space, there exist:
 * $n \in \Z_{>0}$
 * $\alpha_1, \ldots, \alpha_n \in \R_{\ne 0}$
 * $B_1, \ldots, B_n \in \FF_\infty$

such that:
 * $\ds (2): \quad \norm {f - \sum_{k \mathop = 1}^n \alpha_k \chi_{B_k} }_p \le \epsilon$

By $(1)$, there exist:
 * $A_1, \ldots, A_n \in \AA_0$

such that:
 * $\ds (3): \quad \forall k \in \set {1, \ldots, n} : \map \mu {A_k \symdif B_k} \le \paren {\frac \epsilon {\size {n \alpha _k} } }^p$

By definition of $\AA_0$, for each $k$ there exists an $n_k \ge 0$ such that:
 * $A_k \in \FF_{n_k}$

Let $N := \max \set {n_1, \ldots, n_k}$.

Then:
 * $A_1, \ldots, A_n \in \FF_N$

In particular:
 * $\ds \sum_{k \mathop = 1}^n \alpha_k \chi_{A_k} \in \map {L^p} {X, \FF_N, \mu}$

Moreover: