The problem I pose below is a simplified version of my actual problem: Instead of $C([0,1], \mathbb{R})$ I actually have $C([0,+\infty], H)$, where $H$ is some separable, (infinite-dimensional), real Hilbert space and the norm is the usual one (i.e. $\sum2^{-k}||1_{[0,k]}\cdot||_{\infty}$), but I guess the simpler setting is sufficient to clarify the issue. Here we go:

Let $\mu$, $\nu$ be two probability measures on $C:= C([0,T],\mathbb{R})$, where the space is endowed with the Borel $\sigma$-algebra of the $\sup$-norm (which is also generated by the collection of all projections $\pi_t$). Suppose

  • $\int_{C}f\text{d}\mu = \int_{C}f\text{d}\nu$ for all bounded, measurable $f$, which are continuous w.r.t. the topology of pointwise convergence on $C$.

Does $\mu = \nu$ follow?

A couple of thoughts on this: I am convinced the statement is true when the bold condition above is replaced by

  • $\int_{C}f\text{d}\mu = \int_{C}f\text{d}\nu$ for all bounded, measurable $f$, which are continuous w.r.t the sup-norm topology.

Unfortunately there are at most as many functions, which fulfill bold then which fulfill $\textit{the}$ $\textit{later}$ $\textit{one}$. I guess one would try to show that a functions satisfies one condition if and only it satisfies the other, which should fail, because the generated topologies are definitely not equal (the topology of point wise convergence is strictly smaller than the uniform topology) or one would try to approximate functions, which fulfill $\textit{italic}$ by functions, which fulfill bold. Right now I am not able to solve this.

Does anybody know an answer to this problem? I hope some of you find this interesting as well :)

  • $\begingroup$ I modified the question a little bit so that it reads better. Hope this does not alter the style of the question that you intended. $\endgroup$ – Sangchul Lee Nov 12 '18 at 18:59

As pointed out, it is well-known that $\mathscr{B}(C) = \sigma(\pi_t : t \in [0, T])$. Now for any $\varphi$ in the set $C_b(\mathbb{R})$ of all bounded continuous functions on $\mathbb{R}$ and for any $t \in [0, T]$,

$$ \varphi \circ \pi_t : C \ni \omega \mapsto \varphi(\omega(t)) $$

is bounded and continuous either w.r.t. the pointwise topology on $C$ or w.r.t. the supremum topology on $C$. So, any of two conditions will imply

$$ \int_C \varphi \circ \pi_t \, \mathrm{d}\mu = \int_C \varphi \circ \pi_t \, \mathrm{d}\nu. $$

For each open set $U \subseteq \mathbb{R}$, we can find a sequence $\varphi_n \in C_b(\mathbb{R})$ so that $0 \leq \varphi_n \uparrow \mathbb{1}_U$ pointwise. Then by the monotone convergence theorem,

\begin{align*} \mu ( \pi_t^{-1}(U) ) &= \int_C \mathbf{1}_{U} \circ \pi_t \, \mathrm{d}\mu = \lim_{n\to\infty} \int_C \varphi_n \circ \pi_t \, \mathrm{d}\mu \\ &= \lim_{n\to\infty} \int_C \varphi_n \circ \pi_t \, \mathrm{d}\nu = \int_C \mathbf{1}_{U} \circ \pi_t \, \mathrm{d}\nu = \nu ( \pi_t^{-1}(U) ) \end{align*}

Since the sets of the form $\pi_t^{-1}(U)$ for $U \subseteq \mathbb{R}$ open and $t \in [0, T]$ generates $\mathscr{B}(C)$, it follows that $\mu = \nu$.

  • $\begingroup$ Convincing and true. Thanks a lot! $\endgroup$ – Marco Nov 13 '18 at 10:43

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