Infinite dimensional Borel-measurable function. I am not quite sure how this statement about infinite dimensional borel-measurable functions is true, but the author says it is an easy observation:

Let $D([0,\infty))$ denote the space of all cádlág functions over
  $[0,\infty)$, and let $\mathcal{H}$ be the $\sigma$-algebra over
  $D([0,\infty))$ generated by the coordinate functionals. If $f$ is an
  $\mathcal{H}$-measurable functional, it is easy to see that $f$ is of
  the form $f(X) =g(X(t_1),X(t_2),\ldots)$ where $g:\mathbb{R}^\infty
 \rightarrow \mathbb{R}$ is a borel-measurable function and $(t_k)$ is
  a countable sequence in $\mathbb{R}_+$.

But why is this the fact? Is it true that $\mathcal{B}(\mathbb{R}^\infty)=\sigma(\{\prod_{i=0}^\infty O_i| O_i \subset \mathbb{R}, O_i \text{ open, and only a finite number of $O_i\ne\mathbb{R}$}\})$. There are two ways to have open sets in $\mathbb{R^\infty}$, the box topology and the product topology, but the product topology is chosen when generating the Borel-sets?
Anyway, if what I wrote above is correct. I need to show both ways that it holds. First if $f(X) =g(X(t_1),X(t_2),\ldots)$, then $f^{-1}(B)=\{X \in D([0,\infty))| (X(t_1),X(t_2),\ldots )\in g^{-1}(B)\}$.  But why is this set $\mathcal{H}$-measurable? It says that $\mathcal{H}$ is generated by the coordinate functionals, does that mean that it is generated by the functionals of the form, $r\in\mathbb{R}, f_r(X)=X(r)$, so $f_r^{-1}(B)=\{X \in D([0,\infty))| X(r)\in B\}$? If this is so, how do I see that I have the required measurability?
And what about the converse? If $f$ is $\mathcal{H}$-measurable. how do I see that it is of the form $f(X)=g((X(t_1),X(t_2),\ldots))$?
Are there any introductory books that deals with these things? Which books deals with infinite dimensional measurable functions, and infinite dimensional borel sets etc.?
 A: "Easy" is in the eye of the beholder.
Fix a sequence $(t_n)$ of elements of $[0,\infty)$. Let $Y:D([0,\infty))\to\Bbb R^\infty$ denote the map $X\mapsto (X(t_1), X(t_2),\ldots)$. Then $Y$ is $\mathcal H/\mathcal B(\Bbb R)^\infty$-measurable. (Here $B(\Bbb R)^\infty$ is the product $\sigma$-algebra on $\Bbb R^\infty$. It coincides with the Borel $\sigma$-algebra when $\Bbb R^\infty$ is given the product topology.) Thus, if $f$ has the indicated form $g\circ Y$ and $B\in\mathcal B(\Bbb R)$ then $f^{-1}(B)=Y^{-1}(g^{-1}(B))\in\mathcal H$ by standard facts about the composition of measurable maps.
The converse is less easy. The first step is to show that if $f: D([0,\infty))\to\Bbb R$ is $\mathcal H$-measurable then there is a countable  subset of $[0,\infty)$, enumerated as $\{t_n\}_{n\in\Bbb N}$ such that $f$ is measurable with respect to the smaller $\sigma$-algebra $\mathcal G:=\sigma\{X(t_1),X(t_2),\ldots)$. This is a consequence of the Monotone Class Theorem. (Note that $\{t_n\}$ may depend on $f$.) With this in hand you can appeal to the Factorization Lemma: Employing the notation for $Y$ as in the previous paragraph, if $f:D\to\Bbb R$ is $\sigma(Y)$-measurable, then there is a $\mathcal B(\Bbb R)^\infty$-measurable function $g$ such that $f=g\circ Y$. ($g$ is not necessarily unique.) For this see, for example, Corollary 1.97 on page 38 of Klenke's Probability Theory.
