$\left\{ Z_{n},n\geq1\right\} $ is a martingale, which means that $\forall n$, it satisfies the below two conditions,
1) This is the uniformly bounded condition. $$ E\left[\left|Z_{n}\right|\right]<\infty $$
2) Standard martingalge condition. $$ E\left[\left.Z_{n+1}\right|Z_{1},\ldots,Z_{n}\right]=Z_{n} $$
$f$ is a convex function, then the proof that, $\left\{f\left(Z_{n}\right),n\geq1\right\} $ is a submartingale goes as below. This is from the book Stochastic Processes, Second Edition, Sheldon Ross, Lemma 6.4.3, page 314.
\begin{eqnarray*} E\left[\left.f\left(Z_{n+1}\right)\right|Z_{1},\ldots,Z_{n}\right] & \geq & f\left(E\left[\left.Z_{n+1}\right|Z_{1},\ldots,Z_{n}\right]\right)\\ & = & f\left(Z_{n}\right) \end{eqnarray*}
Question:
How do we know that, $f\left(Z_{n}\right)$, is uniformly bounded for it to be a submartingale? That is, what extra assumption do we need to make or is there another result that gives the below? Please give full details and steps. $$ E\left[\left| f\left(Z_{n}\right)\right|\right]<\infty $$
If I am overlooking something basic, please point out.