# Is individual boundedness of the pre-visible process sufficient for the martingale transform of a martingale to be a martingale?

From Probability With Martingales by David Williams:

A fundamental principle: you can't beat the system!

The proof is a straightforward application of the 'taking known stuff out' property of conditional expectations:

'Taking out what is known'

where it's implied that X is integrable.

My question is, in (i) and (ii), do we actually need to require $C_n$ to be a bounded process in the sense that there is a uniform bound $K$ such that $|C_n|<K$ for all $n$? Couldn't we still apply (j) under the weaker condition that the $C_n$ are individually bounded random variables (i.e. for all $n$ there is a $K_n>0$ such that $|C_n(\omega)| < K_n$ for all $\omega$)?

The only other reason for the requirement I can think of is maybe $C_n$ not being a bounded process spoils the integrablility of $Y_n,$ but I would think each $C_n$ being a bounded RV would be sufficient there too.

You are right. This would be a variation of (iii), which gives a generalization of the condition on $C$. Anyway, since we check the (in)equality $\mathbb E\left[Y_n-Y_{n-1}\mid\mathcal F_{n-1}\right]= (\geqslant )0$ for a fixed $n$, the fact that the bound doesn't depend on $n$ do not bring restriction.