A confusion in the proof of-"A convex function over a Martingale is a Submartingale."

Question

Suppose $\left\{ Z_{n},n\geq1\right\} $ is a (discrete) martingale, $f$ is a convex function. Now, we want to prove that $\{f(Z_n),n\geq1\}$ is a submartingale.

As it is stated in A convex function over a Martingale is a Submartingale --- Proof, the proof can be written as \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*}

My question is: why can we check \begin{eqnarray*} E\left[\left.f\left(Z_{n+1}\right)\right|Z_{1},\ldots,Z_{n}\right] & \geq & = & f\left(Z_{n}\right) \end{eqnarray*} instead of \begin{eqnarray*} E\left[\left.f\left(Z_{n+1}\right)\right|f(Z_{1}),\ldots,f(Z_{n})\right] & \geq & = & f\left(Z_{n}\right)? \end{eqnarray*} Since now, $\{f(Z_n),n\geq1\}$ should be seen as a stochastic process.

Answer 1

The result should be stated as follows:

Suppose that $\{Z_n\}$ is a martingale w.r.t. $\{\mathcal{F}_n:=\sigma(Z_1,\ldots,Z_n)\}$ and $f$ is a convex function. Then $\{f(Z_n)\}$ is a submartingale w.r.t. $\{\mathcal{F}_n\}$.