Let $(\Omega,\Sigma,P)$ be a propability space and $(\mathcal{F}_n)_{n\in\mathbb{N}_0}=:\mathbb{F}$ a filtration. Let $Y_0,Y_1,\dots$ be a adapted process of integrable random variables. Further let $(u_n)_{n\in\mathbb{N}}$ and $(v_n)_{n\in\mathbb{N}}$ be real-valued sequences, such that: $$E[Y_{n+1}|\mathcal{F}_n]=u_nY_n+v_n\forall n\in\mathbb{N}$$ Now try to find two real-valued sequences $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$, such that $M_n=a_nY_n+b_n$ defines a Martingale $(M_n)_{n\in\mathbb{N}_0}$.
I try to define $a_n$ and $b_n$ out of the property of the Martingale, that the conditional expected value of the next observation, given all the past observations, is equal to the most recent observation. According to that: $$\begin{align}E[M_{n+1}|\mathcal{F}_n]&=E[a_{n+1}\cdot Y_{n+1}+b_{n+1}|\mathcal{F}_n]\\&=a_{n+1}\cdot E[Y_{n+1}|\mathcal{F}_n]+b_{n+1}\\&=a_{n+1}(u_nY_n+v_n)+b_{n+1}\\&=\underbrace{(a_{n+1}u_n)}_{\substack{=:a_n}}\cdot Y_n+\underbrace{(a_{n+1}v_n+b_{n+1})}_{\substack{=:b_n}}=M_n \end{align}\\\Rightarrow a_n=\begin{cases} &\frac{a_{n-1}}{u_{n-1}} \text{ if } u_{n-1}\neq0 \\ & 0 \text{ ,else } \end{cases}\wedge b_n=b_{n-1}-a_n\cdot v_{n-1}$$
In such a way the sequence $M_n$ would be proper defined and would fulfill the martingale-property mentioned above. Because $(Y_n)_{n\in\mathbb{N}}$ is a adapted process to $(\mathcal{F}_n)_{n\in\mathbb{N}_0}$ the $M_n=a_nY_n+b_n$ sould also be a adapted process to $(\mathcal{F}_n)_{n\in\mathbb{N}_0}$ (Is this correct? And if yes: how can i proof it?). So it remains to show that the expected value $E(|M_n|)$ is finite for all $n\in\mathbb{N}$ (... or rather that $\{M_n\}\in\mathcal{L}^1$ intgrable). Now it is a requirement that the $(Y_n)$ are integrable, whith that $(Y_n)\in\mathcal{L}^1$ holds for all $n\in\mathbb{N}$. Hence follows from linearity of the integration that $a_n\cdot Y_n+b_n=M_n$ must be integrable, too.
Is this correct so far? Or do I make it too easy for me when it comes to the point that $M_n$ is adapted to $(\mathcal{F}_n)_{n\in\mathbb{N}_0}$?
Any assistance or thoughts would be much appreciated.
EDIT: Claim:
"If $(Y_n)_{n\in\mathbb{N}_0}$ is adapted to $\mathbb{F}$, then $(M_n)_{n\in\mathbb{N}_0}=(a_nY_n+b_n)_{n\in\mathbb{N}_0}$ is also adapted to $\mathbb{F}$.
Because $(Y_n)_{n\in\mathbb{N}_0}$ is adapted to $\mathbb{F}$ we know that for each $n\in\mathbb{N}$: $Y_n:(\Omega,\Sigma)\rightarrow(E\subseteq\mathcal{B}(\mathbb{R}),\mathcal{E})$ is $\mathcal{F_n}$-$\mathcal{E}$-measurable. Meaning: $\forall A\in\mathcal{E}:Y_n^{-1}(A)\in\mathcal{F}_n $. Lets take a closer look at $a_nY_n+b_n=M_n:(\Omega,\Sigma)\rightarrow(E',\mathcal{E}':=\{M_n(B)|B\in\mathcal{E}\})$. Surely $E'=a_n\cdot E+b_n$ applies and $\mathcal{E}'\subseteq\mathcal{B}(\mathbb{R})$ is a $\sigma$-algebra, too. Let $A'\in\mathcal{E}'$. Then there exists a $A\in\mathcal{E}$, such that $A'=M_n(A)$. Therefor $M_n^{-1}(A')=Y_n^{-1}(A)\in\mathcal{F}_n$ (because $Y_n$ is $\mathcal{F_n}$-$\mathcal{E}$-measurable). Thus is $M_n$ $\mathcal{F_n}$-$\mathcal{E'}$-measurable and accordingly is $M_n$ adapted to $\mathbb{F}$.
Is this correct?