TITLE: Exercise from Rogers and Williams's Diffusions, Markov processes and martingales
QUESTION [5 upvotes]: I'm stuck trying to do an exercise (see below) in the first volume of the book by Rogers and Williams and any help would be great (my actual question is right at the end).

Let $E$ be a locally compact Hausdorff space with a countable base, let $\cal{E}$ denote the Borel sigma-algebra on $E$, $(C_0(E),||\cdot||)$ denote the space of continuous real valued functions on $E$ which vanish at infinity (with supremum norm) and $(\rm b \cal{E},||\cdot||)$ that of bounded Borel measurable functions on $E$ (again with supremum norm) . An extension of Riesz representation theorem gives the following

Theorem 1: A bounded linear functional $\phi$ on $C_0(E)$ may be written uniquely in the form 
$$\phi(f)=\mu(f):=\int f(x)\mu(dx)$$
where $\mu$ is a signed measure on $E$ of finite total variation.

Exercise: Derive the following theorem from Theorem 1 using the Monotone Class Theorem.

Theorem 2: Suppose that $V:C_0(E)\to \rm b\cal{E}$ is a bounded linear operator such that $0\leq f\leq 1$ implies $0\leq Vf\leq1$. Then there exists a unique kernel $N:E\times\cal{E}\to\mathbb{R}$ such that
$i)$ for every $x$ in $E$ and $f$ in $C_0(E)$ $$Vf(x)=Nf(x):=\int N(x,dy)f(y),$$
$ii)$ for every $x$ in $E$ $N(x,\cdot)$ is a (non-negative) measure on $(E,\cal{E})$ such that $N(x,E)\leq 1$,
$iii)$ for every $B$ in $\cal{E}$, $N(\cdot,B)$ is $\cal{E}$-measurable.

Since for each $x$ in $E$, the map $f\mapsto Vf(x)$ is a linear function on $C_0(E)$, Theorem gives us signed measure $\mu_x$ of finite total variation $\mu_x$ such that $V f (x) = \mu_x(f)$. It seems to me that the way to start is to set $N(x,\cdot):=\mu_x(\cdot)$ for every $x$ in $E$ so that $N$ satisfies $i)$. But from there I'm unsure how to proceed; my immediate reaction was to come up with a way to approximate (w.r.t. to the supremum norm) $\rm b\cal{E}$ functions (in particular, indicator functions) with $C_0(E)$ functions. Then I realised that since $C_0(E)$ is complete, we can't do this. Furthermore, the exercise explicitly states that one can proof Theorem 2 using the Monotone Class Theorem, so there's something I'm missing.
Could someone please give a hint on how to proceed, maybe what $\pi$-system and what space of functions I should be thinking of when trying to apply the Monotone Class Theorem?

REPLY [2 votes]: Following up on @saz's comment, here's a proof that seems to do the trick (if I've done something silly and someone can point it out, that would be great). I'm leaving the question open a bit in case someone could post a proof that uses the Monotone Class Theorem (I'd really like to see one for pedagogical reasons).

For each $x$ in $E$, $f\mapsto(Vf)(x)$ defines a bounded linear functional on $C_0(E)$ and so by Theorem 1 there exists some finite measure $\mu_x$ such that $\mu_x(f)=Vf(x)$. Define $N(x,\cdot):=\mu_x(\cdot)$. 
We need to show that for every $x$ in $E$, $N(x,\cdot)$ is a non-negative measure and less or equal than $1$. Pick any compact subset $K$ of $\cal{E}$. Since $E$ is LCCB, it is metrisable. Let $d$ be any metric that induces the topology on $E$. Since $K$ is closed, there exists a sequence $(f_n)\subset C_0(E)$ such that $f_n^m\to 1_{K}$ pointwise and $0\leq f_n\leq 1$. For example
$$f_n(x):=\left\{\begin{array}{c l}1 &\text{if }x\in  K\\
1-n d(x,K)&\text{if }d(x, K)\leq 1/n\\
0&\text{otherwise}
\end{array}\right.$$
Since $N(x,\cdot)$ is finite for every $x$ we can apply the bounded convergence theorem to get that
$$N(x,K)=\int N(x,dy)1_{K}=\int N(x,dy)\lim_{n\to\infty}f_n=\lim_{n\to\infty}\int N(x,dy)f_n=\lim_{n\to\infty}(Vf_n)(x).$$
Since $V f_n$ is $\cal{E}$-measurable, the above shows that $N(\cdot,K)$ is the pointwise limit of a sequence of $\cal{E}$-measurable functions and thus $\cal{E}$-measurable itself (that is, $N(x,K)$ satisfies $iii)$). Let $\cal{H}$ be the set of functions $f:E\to\mathbb{R}$ such that $N f$ is $\cal{E}$-measurable. Hence $\cal{H}$ is closed under pointwise limits, linear combinations and we just argued that $1_k\in \cal{H}$ for every compact $K$. Since the compact sets generate $\cal{E}$, the Monotone Class Theorem implies that $\cal{H}$ contains all bounded $\cal{E}$-measurable functions and in particular $1_B$ for every $B$ in $\cal{E}$. Hence $iii)$ holds.
Since $V$ is linear and bounded it is continuous, so the above combined with the assumptions on $V$ implies that $0\leq N(x,K)\leq 1$.