Martingale converges to the boundary I asked an almost same question before and it is solved by considering adjacent $Z_n$ can not be far away and obtain a contradiction. However, if the setting is altered a bit, I wonder whether it is still true.
Suppose $D$ is a bounded, connected, open subset of $\mathbb{R}^2$ with boundary $\partial D$. 
Consider a Markov chain $\{Z_n\}_{n\geq 0}$ on $D$ which evolves as follows: for each $n\geq 0$, conditional on $\sigma(Z_k)_{k\leq n}$, the random variable $Z_{n+1}$ is uniformly distributed on the disk of radius $R_n$ centered at $Z_n$, where $2R_n$ is the distance from $Z_n$ to $\partial D$. 
Prove that $$Z_n  \rightarrow  Z_{\infty} \;a.s., Z_{\infty}\in \partial D. $$
It is not too hard to see that coordinates of $Z_n$ are bounded (as $D$ is). It is also easy to see that the coordinates of $Z_n$ are martingale. Then by the boundedness of the coordinates and the martingale convergence theorem, we know that $Z_n$ indeed converges to some $Z_{\infty}$. 
But how to show $Z_\infty\in \partial D$? I am not able to produce a contradiction to $Z_\infty \in D$ with positive probability.
I did some simulation on $\mathbb{R}^2$ with $D$ being a disk and it seems true that $Z_n$ will hit the boundary finally. 
 A: Let's model your Markov chain as follows. Let 
$(U_n)_{n\geq 0}$ be i.i.d. random variables 
uniformly distributed on the unit disk in $\mathbb{R}^2$.
Set $Z_0$ take values in $\bar D$, and for $n\geq 0$ let
$$Z_{n+1}=Z_n+{1\over 2} \varphi(Z_n)\, U_n,$$
where $\varphi(z)=\inf\{\|x-z\|: x\in D^c\}$.
Since 
\begin{eqnarray*}\mathbb{E}(Z_{n+1}\mid {\cal F}_n)
&=&\mathbb{E}(Z_n+{1\over 2} \varphi(Z_n)\, U_n\mid {\cal F}_n)\\[5pt]
&=&Z_n+{1\over 2} \varphi(Z_n)\,\mathbb{E}(U_n\mid {\cal F}_n)\\[5pt]
&=&Z_n+{1\over 2}\varphi(Z_n)\,\mathbb{E}(U_n)\\[5pt]
&=&Z_n+0,\end{eqnarray*}
we see that $(Z_n)$ is a martingale. 
Since $Z_n\in \bar D$, it is a bounded martingale
 and so $Z_n\to Z_\infty \in\bar{D}$ almost surely. 
Now, on the one hand, $\mathbb{E}\|Z_{n+1}-Z_n\|\to 0$
by bounded convergence. On the other hand, noting that 
$\varphi$ is continuous, and using Fatou's lemma 
we get 
$$ \mathbb{E}(\varphi(Z_\infty))\leq \lim_n\mathbb{E}(\varphi(Z_n)) 
 =2 \lim_n \mathbb{E}(\|Z_{n+1}-Z_n\|)/ \mathbb{E}\|U\|=0.$$
Therefore $\mathbb{E}(\varphi(Z_\infty))=0$ so $Z_\infty\in D^c$
almost surely. Combined with the fact that $Z_\infty \in \bar{D}$
almost surely, we get $Z_\infty\in \partial D$ almost surely.   
