Exercise in Probability/Measure Theory Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let also $A_{n,j}\in\mathcal{F},n\in\mathbb{N}_0,j\in\{1,2,3,...,2^n\}$, be such that for all $n\in\mathbb{N}_0:\cup_{j=1}^{2^n}A_{n,j}=\Omega$ and: $\forall i,j\in\{1,2,3,...,2^n\}i\neq j:A_{n,i}\cap A_{n,j}=\emptyset$ and $A_{n,i}=A_{n+1,2i-1}\cup A_{n+1,2i}$.
Now define: $\mathcal{F}_n=\sigma(\{A_{n,j}:j\in\{1,2,3,...,2^n\}\})$.
i.) Prove that $\mathcal{F}$ is a filtration.
ii.) Define $\mu:\mathcal{F}\rightarrow[0,\infty)$ probability measure on $(\Omega,\mathcal{F})$. Assume that if $\mathbb{P}(A)=0$, for $A\in\mathcal{F}$ then $\mu(A)=0$.
Define: $M_n(\omega)=\frac{\mu(A_{n,j})}{\mathbb{P}(A_{n,j})},\omega\in A_{n,j},\mathbb{P}(A_{n,j})\neq 0$ and $M_n(\omega)=0$ if $\mathbb{P}(A_{n,j})=0$.
Prove that for all $n\in\mathbb{N}_0$ and all $j\in\{1,2,3,...,2^n\}$: 
$\mathbb{E}(M_{n+1}1_{A_{n,j}})=\mathbb{E}(M_n1_{A{_n,j}})$.
For question i.) we need to show that $\mathcal{F}_n\subset\mathcal{F}_{n+1}\forall n$ and also that all $\mathcal{F}_n$ are $\sigma-$algebras. 
I have already shown that $\Omega\in\mathcal{F}_n\forall n$ and that if $E_n\in\mathcal{F}_n$ then also $\cup_{n=1}^{\infty}E_n\in\mathcal{F}_n$. However I have some hard time showing the condition for the complements.
Now for question ii.), first of all if $\mathbb{P}(A_{n,j})=0$ then the equality is obvious. Hence we can assume that $\mathbb{P}(A_{n,j})\neq 0$.
Now we have that: $\mathbb{E}(M_{n+1}1_{A{n,j}})=\int M_{n+1}1_{A_{n,j}}=\int \frac{\mu(A_{n+1,j})}{\mathbb{P}(A_{n+1,j})}1_{A_{n,j}}=\frac{\mathbb{P}(A_{n,j})\mu(A_{n,j})}{\mathbb{P}(A_{n+1,j})}$.
On the other hand, we have that: 
$\mathbb{E}(M_{n}1_{A{n,j}})=\int M_{n}1_{A_{n,j}}=\int\frac{\mu(A_{n,j})}{\mathbb{P}(A_{n,j}}1_{A_{n,j}}=\mu(A_{n,j})$.
I am pretty sure there is something wrong in what I have done, but can not figure out exactly what. Any help would be appreciated.
 A: *

*By assumption, $$A_{n,i} = A_{n+1,2i-1} \cup A_{n+1,2i}.$$ Since $A_{n+1,2i-1}, A_{n+1,2i} \in \mathcal{F}_{n+1}$, this shows $A_{n,i} \in \mathcal{F}_{n+1}$ for all $i=1,\ldots,2^n$. Consequently, $$\mathcal{F}_n = \sigma(A_{n,i}; i \in \{1,\ldots,2^n\}) \subseteq \mathcal{F}_{n+1}.$$ Mind that $\mathcal{F}_n$ is a $\sigma$-algebra; it is by definition the smallest $\sigma$-algebra which contains $A_{n,i}$ for all $i \in \{1,\ldots,2^n\}$.

*Your first calculation is not correct because $M_{n+1}$ does not equal $\frac{\mu(A_{n+1,j})}{\mathbb{P}(A_{n+1,j})}$ on $A_{n,j}$. By definition, $$M_{n+1} = \sum_{i=1}^{2^{n+1}} \frac{\mu(A_{n+1,i})}{\mathbb{P}(A_{n+1,i})} 1_{A_{n+1,i}} \tag{1}$$ (in abuse, we set $\frac{0}{0} := 0$). Since the sets $A_{n+1,i}$, $i =1,\ldots,2^{n+1}$, are disjoint and $A_{n,j} = A_{n+1,2j-1} \cup A_{n+1,2j}$, we have $$1_{A_{n,j}} \cdot 1_{A_{n+1,i}} = \begin{cases} 0, & i \notin \{2j-1,2j\}, \\ 1_{A_{n+1,i}}, & i \in \{2j-1,2j\}. \end{cases} \tag{2}$$ Consequently, by $(1)$ and $(2)$, $$\mathbb{E}(M_{n+1} 1_{A_{n,j}}) = \sum_{i=2j-1}^{2j} \mathbb{E}\left( \frac{\mu(A_{n+1,i})}{\mathbb{P}(A_{n+1,i})} 1_{A_{n+1,i}} \right).$$ We can calculate the right-hand side explicitly: $$\begin{align*}\sum_{i=2j-1}^{2j} \mathbb{E}\left( \frac{\mu(A_{n+1,i})}{\mathbb{P}(A_{n+1,i})} 1_{A_{n+1,i}} \right) &= \frac{\mu(A_{n+1,2j-1})}{\mathbb{P}(A_{n+1,2j-1})} \mathbb{P}(A_{n+1,2j-1}) + \frac{\mu(A_{n+1,2j})}{\mathbb{P}(A_{n+1,2j})} \mathbb{P}(A_{n+1,2j}) \\ &= \mu(A_{n+1,2j-1}) + \mu(A_{n+1,2j}) \\ &= \mu(A_{n,j}) \end{align*}$$ where we have used in the last step that $A_{n,j}$ is the disjoint union of $A_{n+1,2j-.1}$ and $A_{n+1,2j}$. On the other hand (that's the calculation you already did), it follows straight from the definiton that $$\mathbb{E}(M_n 1_{A_{n,j}}) = \mu(A_{n,j}).$$ Combining both results proves $$\mathbb{E}(M_{n+1} 1_{A_{n,j}}) = \mu(A_{n,j}) = \mathbb{E}(M_n 1_{A_{n,j}}).$$

