Could I affirm that $f$ is not identically 0? Consider the following  situation: Let $\Omega =l^{\infty}(\mathbb{R})$ be the 
space of all bounded sequences of real numbers. We will consider in $\Omega$ the metric:
$
d(x,y)=\sum_{i\geq 1}\frac{|x_i-y_i|}{2^i}.
$
Set $\Omega_k=[-k,k]^{\mathbb{N}},$ is clearly that $\bigcup \Omega_k=\Omega$.
Denote by $B(\Omega,\mathbb{R})$ (or $B(\Omega)$ for short) the space of all bounded real functions, endowed with the sup norm (which makes $(B(\Omega)$ a Banach space). We also consider the subspaces $C_b(\Omega)$ of the bounded continuous functions ($C_b(\Omega)$ is a  Banach space with the induced norm).
My problem: Let $(f_k)$ be a sequence of functions 
in $B(\Omega)$  satisfying $f_k|\Omega_k \in C_b(\Omega_k)$ and
$f_k=0$ outside $\Omega_k$. Assume that $f_n\to f\in C_b(\Omega)$ 
where the above convergence is uniform in compact sets.
Now suppose that $(\nu_k)$ it  is a sequence of probability measures
on the Borel sigma algebra of $\Omega$ satisfying
 $\nu_k(f_k)=\int_{\Omega}f_kd\nu_k=1$
My Question: Suppose that $\nu_k\to \nu$ in the weak topology to another
probability measure $\nu$.  Could I affirm that $f$ is not identically 0?
 A: No, $f$ could be identically zero. First let me describe a counterexample to a simpler situation. Let $\Omega = \mathbb{R}$ and let $\nu$ and each $\nu_k$ be the measure whose restriction to $[n-1,n]$ is $\frac{1}{2^n}$ times Lebesgue measure, for $n = 1, 2, \ldots$. So $\nu_k \to \nu$ because each $\nu_k$ equals $\nu$. Let $f_k$ be a positive continuous function supported on $[k-1,k]$ and normalized so that $\int f_k\, d\nu = 1$. Then $f_k \to 0$ uniformly on compact sets.
To transport this to your situation, identify $\mathbb{R}$ with the set of sequences $(x_i)$ satisfying $x_i = 0$ for $i \geq 2$, define $\nu$ and $\nu_k$ as above, and extend each $f_k$ to a continuous function supported on $\Omega_k$.
(Maybe there is a slight issue about extending arbitrary continuous functions on a non locally compact space, but that could be resolved by writing an explicit formula for $f_k$. Or you could take each $f_k$ to be Lipschitz, that would work because Lipschitz implies continuous and Lipschitz functions on subsets of any metric space can always be extended.)
