Hi everyone this is an exercise from Cohn's book. I'd appreciate if someone can check part (d) and (e) where I have more problems because this concept is completely new for me.
Let $(X,\mathscr{A})$ and $(Y,\mathscr{B})$ measurable spaces. A function $K:X\times \mathscr{B}\to [0,+\infty]$ is called a Kernel from $(X,\mathscr{A})$ to $(Y,\mathscr{B})$ if
(i) for each $x$ in $X$ the function $B\longmapsto K(x,B)$ is a measure on $(Y,\mathscr{B})$, and
(ii) for each $B$ in $\mathscr{B}$ the function $x\longmapsto K(x,B)$ is $\mathscr{A}$-measurable.
Suppose that $K$ is a Kernel from $(X,\mathscr{A})$ to $(Y,\mathscr{B})$, that $\mu$ is a measure on $(X,\mathscr{A})$, and that $f$ is a $[0,+\infty]$-valued $\mathscr{B}$-measurable function on $Y$. Show that
(a) $B\longmapsto \int K(x,B)\mu(dx)$ is a measure on $(Y,\mathscr{B})$,
(b) $x\longmapsto \int f(y)K(x,dy)$ is an $\mathscr{A}$-measurable function on $X$, and
(c) if $\nu$ is the measure on $(Y,\mathscr{B})$ defined in part (a), then $\int f(y)\nu(dy)=\int\big((\int f(y) K(x,dy)\big)\mu(dx)$
Now suppose that $\mu$ is a finite measure on $X$, that $\sup\{K(x,Y):x\in X\}$ is finite and that the measurable function $f$ is bounded but not necessarily nonnegative. Show also that
(d) $x\longmapsto \int f(y)K(x,dy)$ is a bounded $\mathscr{A}$-measurable function on $X$, and
(e) $ \int f(y)\nu(dy)=\int\big((\int f(y) K(x,dy)\big)\mu(dx)$ where $\nu$ is the measure defined in (a).
Proof: (a) Clearly $\int K(x,\varnothing)\mu(dx)=\int 0 \mu(dx)=0$. Suppose $\{B_n\}$ is a sequence of disjoint subset of $Y$ that belong to $\mathscr{B}$. Then, using that $K(x,\cdot)$ is a measure on $(Y,\mathscr{B})$ and the Beppo-Levi's Theorem the assertion follows:
\begin{align}\int K\bigg(x,\bigcup_n B_n\bigg)\mu(dx)&=\int\sum_n K(x,B_n)\mu(dx)\\&=\sum_n \int K(x,B_n)\mu(dx)\end{align}
(b) Let ${f_n}$ a nondecreasing sequence of simple $[0,+\infty)$-valued $\mathscr{B}$-measurable function such that $\lim_n f_n(y)=f(y)$ for each $y$ in $Y$. For each $n$, define $F_n$ as follows $F_n(x)= \int f_n(y)K(x,dy)$. Suppose $b_1,\ldots b_k$ are the nonzero values of $f_n$ and that $B_1,\ldots B_k$ are the set on which these values occur. Thus $f_n = \sum_{j=1}^k b_j \chi_{B_j}$ and so
$$F_n(x)=\sum_{j=1}^k b_j K(x,B_j)$$
Then $F_n$ is an $\mathscr{A}$-measurable function since each $K(\cdot, B_j)$ is $\mathscr{A}$-measurable. Then $\lim_n F_n$ is $\mathscr{A}$-measurable and by the Monotone Convergence Theorem it follows that
$$\lim_n F_n(x)= \lim_n\int f_n(y)K(x,dy)=\int f(y)K(x,dy)$$ (c) We begin with the case where $f$ is a characteristic function. Let $f(y)=\chi _B(y)$ where $B$ is a subset of $Y$ which belongs to $\mathscr{B}$. Then $\int \big (\int \chi _B(y) K(x,dy) \big)\mu(dx)=\int K(x,B)\mu(dx)=\nu(B)$ and similarly $\int \chi _B(y) \nu(dy)=\nu(B)$, in which case the equality follows.
Now suppose that $f$ is a simple $[0,\infty)$-valued $\mathscr{B}$-measurable function. Suppose that $b_1,\ldots b_k$ are the nonzero values of $f$ and $B_1,\ldots B_k$ are the set in which these values occur. Then
\begin{align}\int \sum_{j=1}^k b_j \chi_{B_j} \nu(dy)=\sum_{j=1}^k b_j\int \chi_{B_j} \nu(dy)&=\sum_{j=1}^k b_j\int \bigg (\int \chi_{B_j} K(x,dy)\bigg) \mu(dx)\\ &=\int \bigg (\int \sum_{j=1}^k b_j\chi_{B_j} K(x,dy)\bigg) \mu(dx)\end{align}
To conclude suppose that $f$ is a $[0,+\infty]$-valued $\mathscr{B}$-measurable function. Choose a nondecreasing sequence $\{f_n\}$ of simple $[0,+\infty)$-valued measurable functions such that $f=\lim_n f_n$. By the Monotone Convergence Theorem it follows that
\begin{align} \int f\, \nu(dy)=\lim_n \int f_n\, \nu(dy)&=\lim_n \int \bigg (\int f_n K(x,dy)\bigg)\mu(dx)\\ &= \int \lim_n \bigg (\int f_n K(x,dy)\bigg)\mu(dx)\\ &=\int \bigg ( \int \lim_n f_n K(x,dy)\bigg)\mu(dx)\\ &= \int \bigg ( \int f\, K(x,dy)\bigg)\mu(dx)\end{align}
(d) Let $c$ a positive real number such that $\lvert f\rvert \le c$. We choose a sequence $\{f_n\}$ of simple real valued $\mathscr{B}$-measurable functions, and without loss of generality we may assume that $\lvert f_n\rvert\le c$ hold for each positive integer $n$. Then for $x$ in $X$
\begin{align}\int c K(x,dy)=c K(x,Y)\le c\sup\{K(x,Y):x\in X\}<+\infty \end{align}
Thus $f$ and $f_1,\ldots$ are dominated by the $K(x,\cdot)$-integrable function $c$. Define $F(x)=\int f(y)K(x,dy)$ and $F_n(x)=\int f_n(y)K(x,dy)$ for each $n$, then the Dominated Convergence Theorem implies that $F(x)=\lim_n F_n(x)$ so it is sufficient to show that each $F_n$ is $\mathscr{A}$-measurable. Now for a fixed and arbitrary $n$, let $b_1, \ldots b_k$ a finite sequence of nonzero real numbers and $B_1,\ldots B_k$ the sets where these values occur respectively. Thus $f_n =\sum _{j=1}^k b_j \chi _{B_j}$ and so
$$F_n(x) =\sum _{j=1}^k b_j K(x,B_j)$$
Then $F_n$ is $\mathscr{A}$-measurable function since $K(\cdot,B_j)$ is measurable. Hence $F$ is also $\mathscr{A}$-measurable. Now as we have seen $\lvert F_n(x)\rvert \le \int \lvert f(y)\rvert K(x,dy)\le c\sup\{K(x,Y):x\in X\}<+\infty $ is bounded.
(e) Clearly the assertion is true for simple real valued $\mathscr{B}$-measurable functions. Suppose that $f$ is a bounded real valued function and $\{f_n\}$ a sequence of simple real valued $\mathscr{B}$-measurable functions such that $f=\lim_n f_n$. Define $F_n(x)=\int f_n(y)K(x,dy)$ and $F(x) =\int f(y)K(x,dy)$. To simplify notation we set $F_0 \equiv F$ and $f_0 \equiv f$, and let $c$ a bound for $f_j $ for each nonnegative integer $j$. Thus
\begin{align}\int c\, \nu(dy)= c\nu (Y)&= c\int K(x,Y)\mu(dx)\\ &=c\sup \{K(x,Y):x\in X\}\ \mu(X)<+\infty \end{align}
Similarly $\lvert F_n(x)\rvert \le c K(x,Y)$. Using the same argument than above we can derive that $ c K(x,Y)$ is a $\mu$-integrable function and by the Dominated Convergence Theorem $\int f(y)\nu(dy)=\lim_n \int f_n(y) \nu(dy)$ and $\int F(x) \mu (dx) =\lim _n \int F_n(X)\mu(dx)$. Hence
\begin{align}\int f \nu(dy)=\lim_n \int f_n \nu(dy)&=\lim_n \int\bigg(\int f_n(y) K(x,dy) \bigg)\mu(dx)\\ &=\lim_n \int F_n(y)\mu(dx)\\ &= \int F(y)\mu(dx)\\ &= \int\bigg(\int f(y) K(x,dy) \bigg)\mu(dx) \end{align}