Uniform integrability of the following martingale Let $A_{n,k}\in F$ where $1\leq k\leq 2^n$ such that
$A_{n,k}$ are disjoint for fixed $n$ and different $k$,
$A_{n,k}=A_{n+1,2k-1}\cup A_{n+1,2k}$.
Let $F_n=\sigma(\{A_{n,k}:1\leq k\leq 2^n\})$ and let $\mu$ a probability measure on $F$ such that $\mu<<\mathbb{P}$.
Define $$X_n(\omega) = \left\{
     \begin{array}{lr}
       \frac{\mu(A_{n,k})}{\mathbb{P}(A_{n,k})} & : \omega \in (A_{n,k}),\mathbb{P}(A_{n,k})\neq 0\\
       0 & : \text{else}.
     \end{array}
   \right.$$

Why is $X_n$ uniformly integrable and why does this imply that there exists an integrable $X_\infty$ such that $\mu(A)=\int_A X_\infty d\mathbb{P}$?

I have shown that $X_n$ is a martingale and that for all $A\in F_n$ we have
$$\mu(A)=\int_A X_n d\mathbb{P}.$$
By absolute continuity I know that for all $\epsilon$ there is a $\delta$ such that $\mathbb{P}(A)<\delta\implies\mu(A)<\epsilon$.
Then we want to show that
$$\lim\limits_{K\to\infty}\sup_n\int_{\{X_n>K\}}X_nd\mathbb{P}=0.$$
We know that $\{X_n>K\}=\cup_{1\leq k\leq 2^n}\{\mu(A_{n,k})>K\mathbb{P}(A_{n,k})\}$, how do I continue from here?
 A: I think the formulation is unnecessarily complicated than it actually is. First note that by Radon-Nikodym theorem, there exists $X\ge 0$ such that
$$
\mu(A) =\int_A X \mathrm{d}\Bbb P,\quad \forall A\in F.
$$ We can find that for each $k\le 2^n$
$$
X_n(\omega) =\frac{\mu(A_{n,k})}{\Bbb P(A_{n,k})}=\frac{1}{\Bbb P(A_{n,k})}\int_{A_{n,k}} X\mathrm{d}\Bbb P\quad\text{ on }\ \ A_{n,k}.
$$ Since $F_n$ is generated by disjoint family $(A_{n,k})_{k\le 2^n}$, this is equivalent to
$$
X_n =\Bbb E[X|F_n].
$$ (Recall discrete version definition of conditional expectation.) We can also check that $(F_n)_n$ is a filtration, i.e. $F_n\subset F_{n+1}$ by the assumption $A_{n,k}=A_{n+1,2k-1}\cup A_{n+1,2k}$. Since $X$ is integrable, i.e. $\Bbb E[|X|]=\Bbb E[X]=1$, this immediately shows that $(X_n,F_n)$ is a uniformly integrable martingale. To see uniform integrability, observe that
$$
\int_{\{X_n>M\}}X_n\mathrm{d}\Bbb P = \int_{\{X_n>M\}}X\mathrm{d}\Bbb P.
$$ For given $\epsilon>0$ there exists $\delta>0$ such that $\int_A X\mathrm{d}\Bbb P<\epsilon$ for all $\Bbb P(A)<\delta$. Since for all $M>\Bbb E[X]/\delta$,$$
\Bbb P(X_n>M)\le \frac{\Bbb EX_n}{M}\le \frac{\Bbb EX}{M}<\delta
$$  it follows $\int_{\{X_n>M\}}X_n\mathrm{d}\Bbb P=\int_{\{X_n>M\}}X\mathrm{d}\Bbb P <\epsilon$ for all $n\ge 1$.
