transition kernel I've got some trouble with transition kernels. 
We look at Markov process with statspace $(S,\mathcal{S})$ and initial distribution $\mu^0$. We have a transition kernel $P:S\times \mathcal{S}\to[0,1]$ 
Now I have to show that $P^n(x,B):=\int_{S}P^{n-1}(y,B)P(x,dy)$ for $n\geq 2$ and $P^1:=P$ is also a transition kernel. How do I proof that $P^n$ is measuable for every fixed $x\in S$? 
can I write
$\int_{S}P^{n-1}(y,B)P(x,dy)=\int_{S}P^{n-1}(y,B)dP(x,y)$
and now look at the measurable function $f(x):=1_{A_1\times A_2}$ so I have
$f(x)P^{n-1}(y,B)P(x,dy)$
is measurable. I take a family $f_n$ of elementary functions with $f_n\to f$ for $n\to\infty$. I'm not realy sure, measure theory wasn't my best course. And I hope that anyone is able to understand me, because I'm not realy able to write english in a nice way. 

We have to show that for each $n\geqslant 1$, 


*

*if $x\in S$ is fixed, then the map $S\in\mathcal{S}\mapsto P^n(x,S)$ is a probability measure, and 

*if $B\in\cal S$ is fixed, the map $x\in S\mapsto P^n(x,B)$ is measurable.


We proceed by induction. We assume that $P^{n-1}$ is a transition kernel. If $x\in S$ is fixed, since for all $y$ we have $P^{n-1}(y,S)=1$, we have $P^n(x,S)=1$. We have $P^n(x,\emptyset)=0$ and if $S_i$ are pairwise disjoint measurable sets we get what we want using linearity of the integral and a monotone convergence argument. 
We now show the second bullet. It indeed follows from an approximation argument. Fix $B\in \cal S$. Since $y\mapsto P^{n-1}(y,B)$ is non-negative and measurable, there exists a non-decreasing sequence of measurable functions $(g_k)$ which have the form $\sum_{i=1}^{N_k}c_{k,i}\chi_{A_{k,i}}$, and such that $g_k(y)\uparrow P^{n-1}(y,B)$. We have by a monotone convergence argument that 
$$P^n(x,B)=\lim_{k\to\infty}\sum_{i=1}^{N_k}c_{k,i}\int \chi_{A_{k,i}}(y)P(x,dy)=\lim_{k\to\infty}\sum_{i=1}^{N_k}c_{k,i}P(x,A_{k,i}).$$
We wrote the map $x\mapsto P(x,B)$ as a pointwise limit of a sequence of measurable functions.
