This is a follow-up to my former question:
Consider a dice with $f$ faces and let $(X_n)_{1 \le n \le N}$ be the outcomes of the tosses. For $1 \le n \le N$ we set $\mathcal{F}_n = \sigma(X_1,\ldots,X_n)$ and additionally $\mathcal{F}_0 = \{\emptyset, \Omega \}$. Consider the RVs $Z_0 := 0$ and $Z_n := \max_{1 \le k \le n} X_k \quad (1 \le n \le N)$.
Show that for any real-valued function $g$ it holds $\mathbb{E}(g(Z_n) \mid \mathcal{F}_{n-1}) = \mathbb{E}(g(Z_n) \mid Z_{n-1})$
I understand that intuitively $Z_n$ describes precisely the maximum number yielded by the $n$th dice toss when we know the results of the previous $n-1$ tosses. I furthermore suppose that we can model this setting as a measure space $(\Omega, \mathcal{A}, \mathbb{P})$ via $\Omega := \{1,\ldots,f\}$ and consider $X_1,\ldots,X_N$ as uniformly distributed independent canonical RVs, i.e. $X_i(\omega) := \omega$. For the $\sigma$-Algebra $\mathcal{A}$ I suppose it is best to set $\mathcal{A} = \mathcal{P}(\Omega)$. My definition of conditional expectation is as follows:
Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space and let $X$ be a real RV with $E(\lvert X \rvert) < \infty$. $\mathbb{E}(X \mid \mathcal{F})$ is uniquely defined as a RV via the conditions
$\mathbb{E}(X \mid \mathcal{F})$ is $\mathcal{F}$ measurable,
$\mathbb{E}(X \mid \mathcal{F}) \in L^1(\mathbb{P})$
$\int_A \mathbb{E}(X \mid \mathcal{F}) d \mathbb{P} = \int_A X d \mathbb{P}$ for all $A \in \mathcal{F}$.
My idea would have been to show that $\mathcal{F}_{n-1} = \sigma({Z}_{n-1})$, however, I doubt that this is true. Could you please give me a hint?
Note $Z_n=\max(Z_{n-1},X_n)$ and $X_n$ is independent of $\mathscr{F}_{n-1}$ and $Z_{n-1}$ is $\mathscr{F}_{n-1}$-measurable. So $E[f(Z_n)|\mathscr{F}_{n-1}]=E[f(\max(X_n,z))]|_{z=Z_{n-1}}=g(Z_{n-1})$. But, equivalently, $E[f(Z_n)|Z_{n-1}]=g(Z_{n-1})$.