Let $(\Omega, \mathcal F, P)$ be a probability space and $(\Omega_i, \mathcal F_i)$ be measurable spaces for $i=1, \ldots, m$. Let $Z_i:\Omega\to \Omega_i$ be a measurable function for each $i$, and let $Z:\Omega\to \Omega_1\times \cdots\times \Omega_m$ be defined as $Z=(Z_1, \ldots, Z_m)$. Assume that $Z_1, \ldots, Z_m$ are independent.
Let $f:\Omega_1\times \cdots\times \Omega_m\to \mathbf R$ be Borel measurable with the property that $$|f(\omega_1, \ldots, \omega_m)-f(\omega_1', \ldots, \omega_m')|<C$$ whenever $(\omega_1, \ldots, \omega_m)$ and $(\omega_1', \ldots, \omega_m')$ differ in only one coordinate.
For each $i=1, \ldots, m$, define $S_i:\Omega\to \mathbf R$ as $$S_i=E(f\circ Z|\ Z_1, \ldots Z_i)$$ It is clear that $S_1, \ldots, S_m$ forms a martingale on $\Omega$.
I want to show that $|S_{i+i}-S_i|\leq C$ for each $i\geq 1$.
This is easy to show when each $Z_i$ has finite image. For then $S_i$ is constant on each part of the common refinement of the partitions of $\Omega$ formed by the fibres of $Z_1, \ldots, Z_i$. But I fail to see how to do this in general.
What I tried doing is:
Let's try to show $|S_2-S_1|\leq C$. But $$S_1=E(F\circ Z|\ Z_1) = E(E(f\circ Z|\ Z_1)|\ Z_1, Z_2)$$ Thus $$S_2-S_1 = E(f\circ Z|\ Z_1, Z_2) - E(E(f\circ Z|\ Z_1)|\ Z_1, Z_2)$$ which gives $$S_2-S_1 = E(f\circ Z - E(f\circ Z|\ Z_1)|\ Z_1, Z_2)$$ but I am getting nowhere from here.
Becasue the $Z_i$ are independent, $S_1=E(f(Z)|Z_1)=h_1(Z_1)$, where $$ h_1(x):=E(f(x,Z_2,\ldots,Z_m)). $$ Likewise, $S_2=h_2(Z_1,Z_2)$, where $$ h_2(x,y) :=E(f(x,y,Z_3,\ldots,Z_m)). $$ So $S_2-S_1=g(Z_1,Z_2)$, where $$ g(x,y) =E\left[f(x,Z_2,Z_3,\ldots,Z_m)-f(x,y,Z_3,\ldots,Z_m)\right]. $$ The claim follows because $|f(x,Z_2,Z_3,\ldots,Z_m)-f(x,y,Z_3,\ldots,Z_m)|\le C$ by the hypothesis on $f$. The inequalities for $i=2,3,\ldots,m-1$ follow in the same way.