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?

  • $\begingroup$ The kind of the typos here, and the kind of the question, makes a funny impression on me. $\endgroup$ Jun 25, 2016 at 16:30
  • $\begingroup$ @WłodzimierzHolsztyński If you are not part of the solution, you must be part of the problem, thanks for nothing. $\endgroup$
    – user11178
    Jun 25, 2016 at 17:47
  • $\begingroup$ @JuanValdez: posts that are inappropriate for some reason or other show up quite regularly on mathoverflow, for instance, one often sees homework problems posed as if they were research question. Sometimes it is hard to tell if a question is really legitimate or not ... try not to take it personally. $\endgroup$
    – Nik Weaver
    Jun 25, 2016 at 20:54
  • $\begingroup$ @JuanValdez, you're a part of a trouble, I am a part of the solution. You still didn't correct your typos. How come? (Or are these simply errors on your part?). $\endgroup$ Jun 26, 2016 at 6:32
  • $\begingroup$ @JuanValdez, let me give you your first hint. You claim that $\bigcup_k[-k\ k]^\mathbf N = l^\infty(\mathbf R)$. This is false. Do you see why? Can you point out to one of the sides being a separable space, and the other one being non-separable? That's only a start of your troubles. $\endgroup$ Jun 26, 2016 at 6:50

1 Answer 1


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.)


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