Lebesgue density theorem: a martingale proof

Question

I'm trying to prove the Lebesgue density theorem using Mantingale convergence theorem. In this post How to prove the Lebesgue density theorem using martingales? I see that someone proved it using a trick due to Morayne and Solecki, but I can't understand it and I can't get access to the referenced article neither. Could anyone kindly guide me or provide any reference so I can solve the following issues?

1. Why to take $A \subseteq \left[\frac{1}{3},1\right)$? Can we do that without loss of generality? I tried to use the fact that $1/3=\sum_{k\geq 1}1/4^k$ to write $$ \left[\frac{1}{3},1\right) = \bigcap_{n=1}^{\infty} \left[\sum_{k = 1}^{n} 1/4^k, 1\right), $$ but it didn't help.
2. If $F_n'(x)\in \mathcal{F}_n'$, Why is $F_n'(x)$ just $F_n(x)$ shifted by one of the four numbers $\pm 2^{-n}\left(\frac{1}{3}\right), \pm 2^{-n}\left(\frac{2}{3}\right)$? I see that $F_n'(x)$ is an element of $\mathcal{F}_n'$, which is the $\sigma$-algebra $\mathcal{F}_n$ shifted by $\frac{1}{3}$, so $F_n'(x)$ does not necessarily has to be related to $F_n(x)$, because it may be a union of elements of the partition generating $\mathcal{F}_n$ but shifted by $\frac{1}{3}$.

I'd really apreciate any help.

Answer 1

I read the paper due to Morayne and Solecki, and the idea is the following:

- Step 1: Define \begin{align} A_m &= \{\ (k2^{-m},(k+1)2^{-m}]:\ k\in \mathbb{Z}\}, \\ A'_m &= \{\ 1/3+ (k2^{-m},(k+1)2^{-m}]:\ k\in \mathbb{Z}\} \end{align} and notice that $A_m$ and $A_m'$ are partitions of $\mathbb{R}.$

- Step 2: Define $A_m(x)$ as the unique element of $A_m$ such that $x\in A_m(x)$. Do the same for $A_m'(x)$.

- Step 3: Show that if $x\in \mathbb{R}$ and $0<\delta<3^{-1}2^{-m}$ then

$$x+(-\delta,+\delta)\subset A_m(x)\cup A_m'(x).$$

- Step 4: Suppose WLOG that $x\in (0,1)$. Show that $$ \begin{align} P(A|\sigma A_m)(x) &\rightarrow_m 1_A(x),\\ P(A|\sigma A'_m)(x) &\rightarrow_m 1_A(x) \end{align} $$ where $\sigma A_m$ is the $\sigma$-algebra generated by $A_m$ restricted to the interval $(0,1)$ (the same for $\sigma A_m'$). Hence $$\frac{1}{P(A_m(x))}\int_{A_m} |1_A(t) - 1_A(x)|dt\rightarrow_m0 $$

- Step 5: Given $\delta>0$, let $m$ be the unique integer such that $ 2^{-m}3^{-1} > \delta \geq 2^{-m-1}3^{-1} $. Then, by Step 3 you'll be able to finish the theorem by proving that $$ \begin{align} 0&\leq \frac{1}{2\delta} \int_{x+(-\delta,\delta)}|1_A(t)-1_A(x)|dt \\ &\leq \frac{1}{2\delta} \int_{A_m(x)\cup A'_m(x)}|1_A(t)-1_A(x)|dt \rightarrow 0 . \end{align} $$