This blog is devoted to the proof (given in P.A. Meyer, Probability and Potentials) of the following theorem (The idea is to "untersify" the proof given there and make it more explicit and also as a reference point for me later on)
First, a few supporting results are given.
The algebra $A_0$ is closed under uniform convergence. This is because if $g \in \bar{A_0}$, then as $\mathcal{H}$ is closed under uniform convergence, we have $g \in \mathcal{H}$. Then the algebra generated by $A_0$ and $g$ is strictly "greater" than $A_0$ and is contained in $\mathcal{H}$ (if $f\in A_0, \|fg_n-fg\| \leq \|f\|\|g_n-g\| \to 0$ uniformly and hence $fg\in \mathcal{H}$) which contradicts the maximality of $A_0$. It can also be easily seen that the algebra $A_0$ contains all the constants as it contains $1$. So, we have $A_0 \subset \mathcal{H}$ satisfying all the properties $\mathcal{H}$ satisfies.
Now, it is shown that if $f \in A_0$ then $|f| \in A_0$. The map $x \to |x|$ can be approximated uniformly by polynomials on each compact interval of $\mathbb{R}$ using Stone-Weierstrass theorem (i.e., there exist a sequence of polynomials $p_n$ such that for every compact subset $K$ of $\mathbb{R}$ and $\epsilon > 0$ there exists an $N(\epsilon,K)$, so that $\sup_{x\in K}|p_n(x) -|x|| < \epsilon$ for all $n \geq N$). Therfore, the composition $|\cdot| \circ f : \Omega \to \mathbb{R}$ can be uniformly approximated by $p_n\circ f : \Omega \to \mathbb{R}$ where $p_n$ are some polynomials. To show this, let $|f(\omega)| \leq M$ for all $\omega\in \Omega$. Then there exists an $n$ such that for all $n \geq N(\epsilon)$, $\sup_{x\in [-M,M]}|p_n(x) -|x|| < \epsilon$. Hence, we also have $\sup_{\omega \in \Omega} ||f(\omega)|-p_n(f(\omega))| < \epsilon$. This proves the assertion. Now, $p_n \circ f$ is a polynomial in $f$ and hence belongs to $A_0$. As $f \wedge g = ((f+g) + |f-g|)/2, f \vee g = -((-f) \wedge (-g))$, we have that $f\wedge g , f \vee g$ are both contained in $A_0$ if $f,g \in A_0$.
Similarly, let $g$ be the limit of an increasing uniformly bounded sequence of positive elements of $A_0$. Then $g \in \mathcal{H}$. The algebra generated by $A_0$ and $g$ is contained in $\mathcal{H}$ (if $f\in A_0$, write $f=f^+-f^-$ and then there exists a sequence of uniformly bounded non-negative functions $f_n^+ \uparrow f$ pointwise. Then $0 \leq f_n^+g_n \uparrow f^+g$ pointwise. Hence, $f^+g \in \mathcal{H}$ etc.) and therefore, it is equal to $A_0$.
Now, let $\mathcal{G}= \{ E \subset \Omega | 1_E \in A_0\}$. Then, $\mathcal{G}$ is a field. Let $E_i \in \mathcal{G}$ so that $E_i \uparrow$. Then, as $A_0$ is closed w.r.t. limit of non-negative uniformly bounded increasing functions, we have that $\cup_i E_i \in \mathcal{G}$. By De-Movire's formula, $\cap_i E_i \in \mathcal{G}$ if $E_i \downarrow$. Therefore, $\mathcal{G}$ is a $\sigma$-algebra. If $f : \Omega \to \mathbb{R}$ is a bounded $\mathcal{G}$-measurable function, then there exist a sequence of elementary $\mathcal{G}$-measurable functions $f_n$ (see the previous proposition) which uniformly converge to $f$. As $f$ is bounded $f_n \in A_0$ and hence $f \in A_0$.
So, it remains to prove that $C^\prime \subset \mathcal{G}$. This can be shown if every set $B=\{\omega: f(\omega) \geq 1\}$ belongs to $\mathcal{G}$ for $f \in C^\prime$. Now, $g=(f \wedge 1)^+$ belongs to $A_0$ and $1_B = \lim_{n \to \infty} g^n$ (If $\omega \in B$ then $g^n(\omega) = 1$ for every $n$ and if $\omega \in B^c$ then $g^n(\omega) \downarrow 0$ as $n \to \infty$).- The map $t \to \dot{Y}_t$ takes values in a seprable subspace of $\mathcal{M}$
- The inverse image of every open ball in $\mathcal{M}$ under this map is a Borel subset of $\mathbb{R}_+$
- The map $t \to \dot{Y}_t$ is the uniform limit of a sequence of measurable elementary functions with values in $\mathcal{M}$
Now, the first two properties imply the third property and vice versa. These are shown next. Firstly, if the map $t \to \dot{Y}_t$ satisfies the first two properties, then by proposition above, there exist $\mathcal{M}$ valued elementary measurable functions which converge uniformly to the given map. For the converse, let the map $g: \mathbb{R}_+ \to \mathcal{M}$ be the uniform limit of the maps $f_n : \mathbb{R}_+ \to \mathcal{M}$. Let $A_n = \text{range}(f_n)$ and $S=\cup_n A_n$. Clearly, $S$ is countable. As $g$ is the uniform limit of $f_n$, it is also the pointwise limit and so $g(t) = \lim_{n \to \infty} f_n(t)$. As each $f_n(t) \in S$ and $S$ is countable, we have that $g$ takes values in a separable subspace of $\mathcal{M}$ (namely the closure of $S$ in $\mathcal{M}$). As each $f_n$ is measurable, we immediatley have that $g$ is measurable.
Every process $(X_t)$ is a map $(t,\omega) \to X_t(\omega)$. Addition of two processes is defined as pointwise addition and scalar multiplication is defined in the obvious manner. Then, $\mathcal{C}$ is a vector space. This is because, if $f_n \to (t \to \dot{Y}_t)$ and $g_n \to (t \to \dot{Z}_t)$ uniformly where $f_n,g_n$ are elementary, then $f_n + g_n \to (t \to \dot{Y}_t + \dot{Z}_t ) =(t \to \dot{(Y+Z)}_t)$ uniformly. Hence, $\mathcal{C}$ is closed w.r.t. addition. It will be shown next that $\mathcal{C}$ is also closed w.r.t. sequential pointwise convergence. The processes $Y^n$ converge to $Y$ pointwise if $Y^n_t \to Y_t$ for each $t$ almost surely.
Let $Y^n \in \mathcal{C}$ for each $n \in \mathbb{N}$. Then, $g_n(t) = \dot{Y}^n_t$ takes values in a separable subspace $E_n$ of $\mathcal{M}$ and $S_n$ be a countable dense subset of $E_n$. As before, take $S=\cup_n S_n, E=\cup_n E_n$. Then, clearly $E$ is separable and $S$ is a countable dense subset of $E$. Assume now that $Y^n_t \to Y_t$ for each $t$ almost surely. Then, $\dot{Y}_t^n \to \dot{Y}_t$ for each $t$ as $E(|Y^n_t-Y_t|\wedge 1) \to 0$. Let $g(t) = \lim_n g_n(t) = \lim_n \dot{Y}^n_t = \dot{Y}_t$. Then, as $g_n(t) \in E$ for every $n,t$, $g$ takes values in the separable subspace $\bar{E}$. Measurability of $g$ follows from the measurability of $g_n$.
Now, note that $\mathcal{C}$ contains all processes of the form $Y_t(\omega) = I(t)Y(\omega)$, (where $I$ is the indicator function of an interval of $\mathbb{R}_+$ and $Y$ is a bounded r.v.) as the range of the map $t \to \dot{Y}_t$ is $\{0,\dot{Y}\}$. Properties 1) and 2) above are then obviously satisfied by such processes. Now, as $\mathcal{C}$ is closed under sequential pointwise convergence, it is closed under uniform convergence. Let $\mathcal{H} \subset \mathcal{C}$ such that $\mathcal{H}$ contains all the bounded processes in $\mathcal{C}$. Then, clearly, $\mathcal{H}$ is closed under pointwise convergence. By applying the previous theorem, we get that $\mathcal{C}$ contains all bounded measurable processes. Any measurable process is a pointwise limit bounded measurable processes. Hence $\mathcal{C}$ contains all measurable processes.
Consider now a measurable process $(X_t)$ adapted to $\mathcal{F}_t$. From what has just been proven, the map $t \to \dot{X}_t$ is the uniform limit of a sequence of measurable elementary functions $f_n : \mathbb{R}_+ \to \mathcal{M}$. This sequence of functions satisfy the following:- As the $f_n$ are elementary, there exists a partition of $\mathbb{R}_+$ into a sequence of Borel sets $A_k^n$ and a sequence of elements of $\mathcal{M}$ denoted by $M_k^n$ such that $f_n = M_k^n$ on $A_k^n$.
- As $f_n$ converges uniformly, $\pi(\dot{X}_t, f_n(t)) \leq \frac{1}{2^{n+1}}$ for every $t$.