Is the product of two independent martingales necessarily a martingale? I'm studying for a probability theory exam and dont know how to show the following:

Let $M_n$ and $N_n$ ($n\in\mathbb{N}$) be martingales for the filtrations $F_n^M = \sigma(M_k, k\leq n)$ and $F_n^N = \sigma(N_k, k\leq n)$ respectively. $M_n$ and $N_n$ are independent, i.e. $\sigma(M_k, k\leq n)$ and $\sigma(N_k, k\leq n)$ are independent. Is the following claim true?
$(M_nN_n)_{n\in\mathbb{N}}$ a martingale for the filtration $F_n^{MN} = \sigma(M_k N_k, k\leq n)$.

My approach is to simply check the definition of a martingale, that is: A martingale is a sequence $M_n$ of RV's satisfying
(i) $M_n$ is measurable wrt $F_n$, $n\in \mathbb{N}_0$
(ii) $M_n \in L^1$ for all $n$
(iii)$E[M_{n+1}|F_n] = M_n $, $n\in \mathbb{N}_0$
The first one (i) holds by definition (using independence of sigma fields). The second one is true since(again using independence):\$ E[|M_n N_n|] = E[|M_n|] E[|N_n|] < \infty$ (which follows from $E[|M_n|] < \infty$ and $E[|N_n|] < \infty$).

My question is: Is my conclusion for (i) and (ii) right? And how can (iii) be shown?

 A: $i)$ $M_nN_n$ is in the generating set for $F_n^{MN}$ so $M_nN_n$ must be $F_n^{MN}$ measurable.
$ii)$ You will need to use Tonneli's Theorem on the unsigned function $|M_nN_n|$. You also need to justify why the functions $|M_n|$ and $|N_n|$ are independent functions. For this you may want to prove that if $f,g$ are Borel measurable function and $X,Y$ are independent function, then $f(X), g(Y)$ are independent functions as well.
$iii) $ This is an idea that I am unsure about but try finding events $A_x, B_x$ such that $P(M_nN_n \leq x) = P(M_n\in A_x)P(N_n\in B_x)$. Then show that $A_x\times B_x$ can be used to generate $\sigma(M_nN_n)$. If you can do this, then for any $A\times B\in \sigma(M_nN_n)$ in the generating set you will have $$ \int_{A\times B}M_{n+1}N_{n+1} d(P\otimes P)= \Big(\int_{A}M_{n+1} dP\Big)\Big(\int_{B}N_{n+1} dP\Big)$$
You can then use the Martingale property and independence again to calculate that we have
$$ \int_{A\times B}M_{n+1}N_{n+1} d(P\otimes P)  =  \int_{A\times B}M_{n}N_{n} d(P\otimes P)$$
for all $A\times B$ that I have described. From here I am stuck, but hopefully something in here is helpful to you!
