Let $X$ be a compact metric space and $\mathcal X$ be its Borel $\sigma$-algebra. Let $\mathscr P(X)$ be the set of all the Borel probability measures on $X$. A Markov chain on $X$ is a measurable map $P:X\to \mathscr P(X)$. We write the image of $x$ under $P$ as $P_x$. (Here $\mathscr P(X)$ is quipped with the Borel $\sigma$-algebra coming from the weak* topology).
Intuitively, for $E\in \mathcal X$, we think of $P_x(E)$ as the probability of landing inside $E$ in the next step given that we are sitting at $x$ at the current instant.
Now let $\mu$ be a probability measure on $X$. We define a new measure $\nu$ on $X$ as follows:
$$\nu(E) = \int_X P_x(E)\ d\mu(x)$$
Intuitively, suppose in the firt step we land in $X$ according to $\mu$. The probability of landing on $x\in X$ according to the measure $\mu$ is $d\mu(x)$. Now let the Markov chain $P$ drive us. Then the probability of landing in $E$ in the next step given that we are at $x$ is $P_x(E)$. So the probability of landing in $E$ in two steps is $\int_X P_x(E)\ d\mu(x)$.
So we have a map $P_\sharp:\mathscr P(X)\to \mathscr P(X)$ which takes a probability measure $\mu$ and produces a new measure $\nu$ as defined above.
Composing $P:X\to \mathscr P(X)$ with $P_\sharp:\mathscr P(X)\to \mathscr P(X)$ we again get a map $X\to \mathscr P(X)$, which I will denote by $P^2$.
Question. Is $P^2$ aslo a Markov chain, that is, is $P^2$ aslo Borel measurable?
My guess is that the map $P_\sharp$ is actually continuous, which would answer the question in the affirmative. But I am not sure.
EDIT II: I tried to argue the measurability of $P_\sharp\circ P$ as follows:
Let $f\in C(X)$ be fixed. We need to figure out what is $\int_X f\ d(P_\sharp\mu)$: If $g:X\to \mathbb R$ is the characteristic function of $E\in \mathcal X$, then $\int_X g\ d(P_\sharp\mu)$ is nothing but $\int_XP_x(E)\ d\mu(x)$. So a reasonable guess for $\int_X f\ d(P_\sharp\mu)$ is $$\int_Xf\ d(P_\sharp\mu) = \int_X \left(\int_X f\ dP_x\right)\ d\mu(x)$$ Write $x\mapsto \int_Xf\ dP_x:X\to \mathbf R$ as $\tilde Pf$ (Note that this is a measuable function and thus the integral above is well-defined).
Now since $C(X)$ is separable, there is a countable dense set $D$ in $C(X)$. This shows that the map $i:\mathscr P(X)\to \prod_{f\in D}\mathbf R$ defined as $\mu\mapsto (\int_X f\ d\mu)_{f\in D}$ is an embedding. So to check the measurability of $P_\sharp \circ P$ we just need to check the measurability of $\pi_f\circ i\circ P_\sharp\circ P:X\to \mathbf R$ for each $f\in D$. But this map takes $x$ to $$\int_X f\ d(P_\sharp P_x) = \int_X \tilde P f\ dP_x = \tilde P(\tilde Pf)(x)$$ Hence this map is measurable and we are done.
EDIT I: (Please ignore)
I tried to argue the continuity of $P_\sharp$ as follows:
Let $f\in C(X)$ be fixed. We want to show that the map $\mathscr P(X)\to \mathbb R$ defined as $\mu\mapsto \int_X f\ d(P_\sharp\mu)$ is continuous (This is beacuse $\mathscr P(X)$ has the weak* topology).
So we need to figure out what is $\int_X f\ d(P_\sharp\mu)$. If $g:X\to \mathbb R$ is the characteristic function of $E\in \mathcal X$, then $\int_X g\ d(P_\sharp\mu)$ is nothing but $\int_XP_x(E)\ d\mu(x)$. So a reasonable guess for $\int_X f\ d(P_\sharp\mu)$ is $$\int_Xf\ d(P_\sharp\mu) = \int_X \left(\int_X f\ dP_x\right)\ d\mu(x)$$ Write $x\mapsto \int_Xf\ dP_x:X\to \mathbf R$ as $\tilde Pf$ (Note that this is a measuable function and thus the integral above is well-defined).
So we want to show that the map $\mathscr P(X)\to \mathbb R$ taking $\mu$ to $\int_X\tilde Pf\ d\mu$ is continuous.
This will be true if $\tilde Pf$ is continuous. $\tilde Pf$ is continuous certainly if $P$ is continuous. But we just have mesurability of $P$ to work with.