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Assume that $f_n$ is a sequence of functionals on the space of cádlág functions, and that for every cádlág function $\mathbb{X}$: $f_n(\mathbb{X})=g_n(X(0),X(q_1),X(q_2),\ldots)$, where $q_i$ is a sequence of the rational numbers. And each $g_n$ is borel measurable, with respect to $\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R^\mathbb{N}})$.

We also have that $f_{n+1}(\mathbb{X})\ge f_n(\mathbb{X})$, and we have that $f_{n}(\mathbb{X})\rightarrow f(\mathbb{X})$. And lastly we have that f is also bounded.

My question is then. Do we have that $f(\mathbb{X})=g(X(0),X(q_1),X(q_2),\ldots)$ for a borel-measurable function g?

my difficulty:

My difficulty is that we have that offcourse borel-measurable functions are bounded under pointwise limits. However, we do not have convergence under every $\mathbb{N}$-tuple of real numbers, we only have convergence under $(X(0),X(q_1),\ldots)$, where X is a cádlag function, and this excludes some tuples, because if $q_i'$ is a sequence of rational numbers converging to 0, then we can't have oscilliations between $1$ and $-1$ on $X(q_i')$ because that would violate the right continuity at 0.

Any ideas on how to find the answer?

UPDATE:

One way I think I can solve this is if I can show that the set of $\mathbb{N}$-tuples $A=\{(r_1,r_2,\ldots )\}$, that are such that $\{(0,r_1),(q_2,r_2)\}$ is cádlag, is such that $A \in \mathcal{B}(\mathbb{R^{\mathbb{N}})}$. Here I have used that a cádlág function is uniquely determined by its value on rational points. However, do we know that the set A is borel-measurable?

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    $\begingroup$ Suggestion: Define $g(x_0,x_1,\ldots):=\limsup_n g_n(x_0,x_1,\ldots)$ on all of $\Bbb R^{\Bbb N}$. Then $g$ is Borel measurable, and $f(\Bbb X) =g(X(0),X(q_1),\ldots)$. $\endgroup$ – John Dawkins Oct 3 '16 at 18:55
  • $\begingroup$ @JohnDawkins Thank you very much! That seems to work. $\endgroup$ – user119615 Oct 3 '16 at 19:07

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