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?



Could I affirm that $f$ is not identically 0?
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.)


