Let $(Y_i)$ be i.i.d. random variables with $\mathbb{P}(Y\geq 0)=1$ and $\mathbb{E}(Y)\leq 1$.
The sequence of partial products $M_n:=\prod_{i=1}^n Y_i$ is a non-negative supermartingale
and so converges almost surely to a limit random variable $M_\infty$.
Martingale Lemma. Either $\mathbb{P}(Y=1)=1$ or $\mathbb{P}(M_\infty=0)=1$.
Proof. We have
$$\lim_{\varepsilon\downarrow 0} \mathbb{P}(|Y-1|>\varepsilon)=\mathbb{P}(|Y-1|>0)=\mathbb{P}(Y\neq 1).$$
If this probability is zero, we are in case 1.
Otherwise, we can choose $\varepsilon>0$ so that $\mathbb{P}(|Y-1|>\varepsilon)>0$. Then the Borel-Cantelli
lemma shows that $\mathbb{P}(|Y_i-1|>\varepsilon \mbox{ infinitely often})=1$. This implies that $Y_i$ fails to
converge to 1 pointwise, so $\mathbb{P}(Y_i\not\to 1)=1$.
If an infinite product converges to a non-zero limit, the factors must converge to 1.
Therefore we finally get
$$1=\mathbb{P}(M_n\to M_\infty)=\mathbb{P}(M_n\to M_\infty, Y_i\not\to 1)=\mathbb{P}(M_\infty=0),$$
which is case 2.
A function $f:\mathbb{R}^d\rightarrow \mathbb{R}$ is differentiable at the point $X$ iff there exists a function $\displaystyle J:\mathbb{R}^d\rightarrow \mathbb{R}$ such that $\lim_{h\rightarrow 0}\frac{f(X+h)-f(X)-J(h)}{||h||}=0$.
This is what I guess might happen in the $d=2$ case:
Given $X=(x,y)\in [0,1]^2$, define $X_{n}=(x_n,y_n)=2^{2n}(\lfloor2^{-n}x\rfloor,\lfloor 2^{-n}y\rfloor)$.
We want to find a function $J=(J_1,J_2)$ such that, at almost all points $X=(x,y)$, $J=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$.
Define
$\displaystyle Y_{1,n}=2^{2n}\int_{y_n}^{y_n+2^{-n}} \left(f(x_n+2^{-n},s)-f(x_n,s) \right)\mathrm{d}s$
and $\displaystyle Y_{2,n}=2^{2n}\int_{x_n}^{x_n+2^{-n}}\left( f(t,y_n+2^{-n})-f(t,y_n)\right)\mathrm{d}t$.
The main motivation for this definition of $Y_{1,n}$ is the fact that, if $f$ could be assumed to be differentiable, then we would have $ \displaystyle 2^{2n}\int_{x_n}^{x_n+2^{-n}}\int_{y_n}^{y_n+2^{-n}}\frac{\partial f}{\partial x} \mathrm{d}s\mathrm{d}t=2^{2n}\int_{y_n}^{y_n+2^{-n}}\left(f(x_n+2^{-n})-f(x_n)\right)\mathrm{d}t$
As before, $(Y_{1,n},Y_{2,n})$ converges to a vector $Y=(Y_1,Y_2)$ and there exists a bounded measurable $J$ such that $Y=J(X)$ a.s.
We have
$\displaystyle Y_{1,n}(X)=\mathbb{E}(Y_1|X_{n})=2^{2n}\int_{x_n}^{x_n+2^{-n}}\int_{y_n}^{y_n+2^{-n}} J_1(s,t)\mathrm{d}s \mathrm{d}t$
$\displaystyle Y_{2,n}(X)=\mathbb{E}(Y_2|X_{n})=2^{2n}\int_{x_n}^{x_n+2^{-n}}\int_{y_n}^{y_n+2^{-n}} J_2(s,t) \mathrm{d}s \mathrm{d}t$
If $L$ is a smooth curve going from $0$ to $X$, then $\displaystyle f(X)=f(0)+\int_L J\cdot\mathrm{d}L$.
Best Answer
A set $A\in\mathcal F_n$ can be written as $(X_1,\dots,X_n)\in B$ for some $B\in\mathcal B(\mathbb R)$. Therefore, one has $$ \mathbb E\left[f\left(Y_{n+1}\right)\mathbf{1}_A\right]= \mathbb E\left[f\left(Y_n+X_{n+1}\right)\mathbf{1}_B(X_1,\dots,X_n)\right] = \mathbb E\left[\int f\left(Y_n+y\right)\mathbf{1}_B(X_1,\dots,X_n)d\sigma(y)\right], $$ where we used Fubini's theorem and independence between $X_{n+1}$ and $\left(X_1,\dots,X_n\right)$. From the assumption on $f$, we derive that $$ \mathbb E\left[f\left(Y_{n+1}\right)\mathbf{1}_A\right]= \mathbb E\left[ f\left(Y_n \right)\mathbf{1}_A \right] $$ which proves the martingale property.
For the question (b), note that if you exchange the first say $10$ indexes, it will not affect $f\left(Y_n\right)$ for $n$ larger than $10$.