Integral identity with Markov kernel Let $K:E\times\mathcal{G}\rightarrow\left[0,+\infty\right]$ be a positive Markov kernel. The $n$-th power of the $K$ is defined iteratively as
$$K^0(x,A)=\delta_{x}(A)$$
where $\delta_x(A)$ is the Dirac measure and
$$
K^{n} = K^{n-1}\circ K
$$
where $\circ$ denotes the kernel composition
$$
(K^{n-1}\circ K)(x,A) = \int_{E}K^{n-1}(x,dx^{\prime})\,K(x^{\prime},A)
$$
In a book by Revuz it is said that the following chain of identity holds
\begin{eqnarray}
K^{n}(x,f) &=& \int_{E}K^n(x,dx^{\prime})f(x^{\prime}) \\
&=& K(x,K^{n-1} f) \\
&=& \int_{E}K(x,dx^{\prime})K^{n-1}f(x^{\prime})\\ 
&=& \int_{E}K(x,dx^{\prime})\int_{E}K^{n-1}(x^{\prime},dx^{\prime\prime})f(x^{\prime\prime}) \\
&=&K^{n-1}(x,Kf) \\
&=&\int_{E}K^{n-1}(x,dx^{\prime})Kf(x^{\prime}) \\
&=& \int_{E}K^{n-1}(x,dx^{\prime})\int_E K(x^{\prime},dx^{\prime\prime})f(x^{\prime\prime})
\end{eqnarray}
From the first to the fourth equation there are no problems, just definitional equations, but how from the fourth are we going to the fifth? I suspect that behind there is an application of the Fubini theorem that I cannot see.
 A: It's not Fubini's Theorem: the key is "$K$ commutes with $K^{n - 1}$".  (Basically, this is analogous to the identity $a \cdot a^{n - 1} = a^{n - 1} \cdot a$.)
Claim: for all $n \in \mathbb{N} \cup \{0\}$ and all test functions $f$,
\begin{align*}
K^{n}(x,Kf) = K(x,K^{n}f) = K^{n + 1}(x,f).
\end{align*}
Proof by induction: when $n = 0$, this follows from the fact that $K^{0}(x,g) = g(x)$.
Assume the identity holds when $n = m$.  In the case $n = m + 1$, we write
\begin{align*}
K^{m + 1}(x,Kf) &= \int_{E} K(x,dx') K^{m}(x',Kf)
\end{align*}
Here we used the definition of $K^{m + 1}$ in terms of $K$ and $K^{m}$, nothing more.  Now we are in a position to use the inductive hypothesis:
\begin{align*}
K^{m + 1}(x,Kf) &= \int_{E} K(x,dx') K^{m}(x',Kf) \\
&= \int_{E} K(x,dx') K^{m + 1}(x',f) \\
&= K(x,K^{m+1}f).
\end{align*}
Finally, observe that, by definition,
\begin{align*}
\int_{E} K(x,dx') K^{m + 1}(x',f) = K^{m + 2}(x,f).
\end{align*}
Putting it all together, we get $K^{m+1}(x,Kf) = K(x,K^{m+1}f) = K^{(m+1)+1}(x,f)$.
Finally, let me say this is an argument against this kernel notation in favor of transition matrices.  Denote by $\mathcal{F}$ the space of test functions and let us define $P : \mathcal{F} \to \mathcal{F}$ as follows:
\begin{equation*}
(Pf)(x) = K(x,f).
\end{equation*}
Notice this is a linear operator --- a matrix, if you will.  (In the context of Markov chains, $P$ is called the transition matrix.)  With this notation, the above identities reduce to the familiar one: $P \circ P^{n} = P^{n} \circ P = P^{n + 1}$, where $\{P^{n}\}$ are the "matrix powers" defined through composition of operators.
