Reading lemma 3 in this post, I tried to prove the equality $$ {\mathbb E}[f(X_{t_0},\ldots,X_{t_n})\mid\mathcal{F}_{t_{n-1}}] = \int f(X_{t_0},\ldots,X_{t_{n-1}},y)\,P_{t_n-t_{n-1}}(X_{t_{n-1}},dy). $$ for arbitrary bounded measurable $f\colon E^{n+1}\rightarrow{\mathbb R}$, where $X$ is a Markov process with state space $(E,\mathcal{E})$ and $P$ its transition function.
The monotone class theorem (mct) for functions is:
Let $\mathcal{C}$ be a $\pi$-class on $\Omega$ and $\mathcal{H}$ be a linear space formed by some real functions on $\Omega$. If the following conditions are satisfied
- $1\in\mathcal{H}$
- $f_n\in\mathcal{H}$,$0\leq f_n\uparrow f$, f is finite $\implies f\in\mathcal{H}$
- $A\in\mathcal{C}\implies 1_A\in\mathcal{H}$
I don't see why the proof of the equality in the case where $f(x_0,\ldots,x_n)$ is just a product $u(x_0,\ldots,x_{n-1})v(x_n)$ helps me applying the mct.
For fixed $n \in \mathbb{N}$ and $t_1,\ldots,t_n \geq 0$ denote by $\mathcal{H}$ the set of all functions $f: E^{n+1} \to \mathbb{R}$ which are bounded, measurable and satisfy
$$\mathbb{E}(f(X_{t_0},\ldots,X_{t_n})\mid\mathcal{F}_{t_{n-1}}) = \int f(X_{t_0},\ldots,X_{t_n-1},y)\,P_{t_n-t_{n-1}}(X_{t_{n-1}},dy). \tag{1}$$
for fixed $n \in \mathbb{N}$ and $t_1,\ldots,t_n \geq 0$. Clearly, $1 \in \mathcal{H}$. Moreover, applying twice the monotone convergence theorem, we find that
$$f_k \in \mathcal{H}, 0 \leq f_k \uparrow f \implies f \in \mathcal{H}.$$
If we set $\mathcal{C} := \mathcal{E}^{n+1}$, then $\mathcal{C}$ is $\cap$-stable. Moreover, we have for any set of the form $A = A_1 \times \ldots \times A_{n+1}$, $A_j \in \mathcal{E}$,
$$1_A(x) = \prod_{j=1}^{n+1} 1_{A_j}(x_j), \qquad x=(x_1,\ldots,x_{n+1}) \in E^{n+1}$$
and therefore we can use the statement on products of functions to conclude that $1_A \in \mathcal{H}$; hence property 3 is satisfied. Since $\mathcal{E}^{n+1}$ is a generator of the product $\sigma$-algebra on $E^{n+1}$, we can apply the monotone class theorem to conclude that $(1)$ holds for any measurable bounded function $f$.