Product of independent martingales is adapted Let $(\Omega,F,P)$ be a probability space and suppose $(F_{n})_{n\geq1}$ and $(G_{n})_{n\geq1}$ are independent filtrations. ($F_{n}$ independent of $G_{n}, \forall n\geq1$). Suppose $(M_{n})_{n\geq1}$ is a martingale with respect to $(F_{n})_{n\geq1}$, and  $(L_{n})_{n\geq1}$ a martingale with respect to $(G_{n})_{n\geq1}$. I want to show their product $Z_{n}=M_{n}L_{n}$ is a martingale with respect to $(H_{n})_{n\geq1}$ where $H_{n}=\sigma(F_{n}\cup G_{n})$. 
I'm trying to be very clear when showing that this martingale is adapted with respect to $(H_{n})_{n\geq1}$. I understand that $E(|Z_{n}|)=E(|M_{n}L_{n}|)=E(|M_{n}||L_{n}|)$. It makes sense to me that this should give $E(|M_{n}||L_{n}|)=E(|M_{n}|)E(|L_{n}|)$ since the filtrations are independent, but I'm struggling to work out the details of this. The definition of independence I am using is $\forall A\in F_{n}, \forall B\in G_{n}, P(A\cap B)=P(A)P(B)$. 
What is the correct approach here?
 A: $M_n L_n \in L^1$ if both $M_n$ and $L_n$ are $L^1$ part:
$$ \mathbb E[|M_nL_n|] = \mathbb E[\mathbb E[|M_n||L_n|\ \big|\ F_n]] = \mathbb E[|M_n| \mathbb E[|L_n|\big|\ F_n]] = \mathbb E[|M_n| \mathbb E[|L_n|]] = \mathbb E[|M_n|]\mathbb E[|L_n|] $$
due to the fact that $M_n$ is $F_n$-measurable and $L_n$ is independent of $F_n$. 
The martingale part: It is to show that for any $k<n$
$$ \mathbb E[M_nL_n\big| H_k] = M_k L_k. $$
Claim: $H_n$ is generated by the sets $A\cap B$, where $A\in F_n$ and $B\in G_n$.
Proof: Since $\Omega \in F_n,G_n$ we can see that
$$ F_n, G_n \subset \{A\cap B, A\in F_n,B\in G_n\}. $$
Futhermore for $A\in F_n, B \in G_n$ we have that $A,B\in H_n $ and so, since $H_n$ is a $\sigma$-algebra, $A\cap B \in H_n$, hence
$$ H_n  = \sigma (F_n \cup G_n)\subset \sigma(\{A\cap B, A\in F_n, B\in G_n\})\subset H_n \quad\quad \Box. $$ 
Since $H_n$ is generated by the sets $\{A\cap B, A\in F_n,B\in G_n\}$ it is enough to show that
$$  \mathbb E[\mathbb E[M_nL_n\big| H_k] {\bf 1}(A\cap B)] = \mathbb E[M_k L_k\cdot {\bf 1}(A\cap B)]. $$
for any $A\in F_k, B\in G_k$. It is easy to check that ${\bf 1}(A\cap B) = {\bf 1}(A)\cdot{\bf 1}(B)$, and that $\mathbb E[M_n|H_k]=M_k$ and $\mathbb E[L_n|H_k] = L_k$, hence
$$ \mathbb E[\mathbb E[M_nL_n|H_k]{\bf 1}(A\cap B)] = \mathbb E [M_nL_n {\bf 1}(A) {\bf 1}(B)]= \mathbb E[\mathbb E[M_nL_n{\bf 1}(A){\bf 1}(B)|F_n]] = \mathbb E[M_n{\bf 1}(A)\mathbb E[L_n{\bf 1}(B)|F_n]] = \mathbb E[M_n{\bf 1}(A)]\mathbb E[L_n {\bf 1}(B)] = \mathbb E[M_k {\bf 1}(A)]\mathbb E[L_k{\bf 1}(B)] $$
where the last equality is due to the fact that $M_n$ and $L_n$ are martingales. Now
$$ \mathbb E[M_k{\bf 1}(A)]\mathbb E[L_k{\bf 1}(B)] = \mathbb E[M_kL_k{\bf 1}(A\cap B)] $$
since $M_k$ and $L_k$ are independent, and so are $A\in F_k $ and $B\in G_k$. Thus we showed that 
$$ \mathbb E[M_nL_n | H_k ] = M_kL_k. $$
