Question on $\mathbb E_{X_{t_{n-2}}}[1_{B_{n-1}}(X_{t_{n-1}-t_{n-2}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]]$ I have a question on these lecture notes:
http://page.math.tu-berlin.de/~scheutzow/WT3main.pdf Page 46-47 Lemma 4.15
In this proof, we use, amongst other things that:
$\mathbb E_{x}[1_{B_{n-1}}(X_{t_{n-1}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]\lvert \mathcal{F}_{t_{n-2}}]=\mathbb E_{X_{t_{n-2}}}[1_{B_{n-1}}(X_{t_{n-1}-t_{n-2}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]](*)$
which comes from the Markov Property which states for any and bounded measurable function $f$ we have:
$\mathbb E_{x}[f((X_{t+h})_{t\geq 0})\lvert \mathcal{F}_{h}]=\mathbb E_{X_{h}}[f((X_{t})_{t\geq 0})]$
But my issue in $(*)$ would be that in our case the function $f$ is indeed: $f(X_{t_{n-1}})=1_{B_{n-1}}(X_{t_{n-1}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]$ and thus
$\mathbb E_{x}[f(X_{t_{n-1}})\lvert \mathcal{F}_{t_{n-2}}]=\mathbb E_{X_{t_{n-2}}}[f(X_{t_{n-1}-t_{n-2}})]=\mathbb E_{X_{t_{n-2}}}[1_{B_{n-1}}(X_{t_{n-1}-t_{n-2}})\mathbb E_{X_{t_{n-1}-t_{n-2}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]]$
Note the difference in this computation compared to $(*)$. My question is rather why would
$\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]$ remain the same when evaluated under $\mathcal{F}_{t_{n-2}}$ even though it is a function of $f(X_{t_{n-1}})$. Is this a mistake or am I simply missing something?
 A: The notation is a bit confusing (although, it is typical to these settings). Here is the same proof presented differently. First, for $g\in \text{b}\mathcal{E}$, a transition semigroup $(P_t)_{t\ge 0}$, and $s<t$, the Markov property is
$$
\mathsf{E}_x[g(X_{t})\mid \mathcal{F}_{s}]=\mathsf{E}_{X_s}[g(X_{t-s})]=P_{t-s}g(X_s) \quad(\mathsf{P}_x\text{-a.s.}),
$$
which is a function of $X_s$. Applying this property recursively, for $f_1,\ldots,f_n\in \text{b}\mathcal{E}$, one has
\begin{align}
&\mathsf{E}_x[f_1(X_{t_1})\cdots f_{n-1}(X_{t_{n-1}})f_n(X_{t_n})] \\
&\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots f_{n-1}(X_{t_{n-1}})\mathsf{E}_x[f_n(X_{t_n})\mid \mathcal{F}_{t_{n-1}}]] \\
&\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots (f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_{n-1}})] \\
&\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots f_{n-2}(X_{t_{n-2}})\mathsf{E}_x[(f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_{n-1}})\mid \mathcal{F}_{t_{n-2}}]] \\
&\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots (f_{n-2}P_{t_{n-1}-t_{n-2}}f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_{n-2}})] \\
&\qquad =\ldots \\
&\qquad=\mathsf{E}_x[(f_1P_{t_2-t_1} f_{2}\ldots P_{t_{n-1}-t_{n-2}}f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_1})] \\
&\qquad= P_{t_1}[f_1P_{t_2-t_1}f_2 \ldots P_{t_{n}-t_{n-1}}f_n](x).
\end{align}
For example, in the fourth line we apply the Markov property to
$$
g(x)=(f_{n-1}P_{t_{n}-t_{n-1}}f_n)(x) \in \text{b}\mathcal{E},
$$
evaluated at $X_{t_{n-1}}.$
