Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathbb{X}$ be a complete metric space, $U(\mathbb{X})$ the space of bounded and continuous functions in $\mathbb{X}$ and $\mathcal{L}\big(U(\mathbb{X})\big)$ the space of all linear functionals $L:U(\mathbb{X})\to\mathbb{R}$.

By definition, the weak topology of $U(\mathbb{X})$ is the smallest topology of $U(\mathbb{X})$ $\Big($"smallest" with respect to lower order of inclusion "$\subset$" in $\{ \tau : \tau \mbox{ is topology of } U(\mathbb{X})\}$ $\Big)$ that makes continuous all linear functionals $L:U(\mathbb{X})\to\mathbb{R} $ of $\mathcal{L}\big(U(\mathbb{X})\big)$.

A linear functional $L\in\mathcal{L}\big(U(\mathbb{X})\big) $ is called positive if $f\geq 0$ implies $L(f)\geq 0$, $\forall f\in U(\mathbb{X})$. Let $\mathcal{L}_{\geq 0}\big( U(\mathbb{X})\big)$ the subspace of all linear functionals positives of $\mathcal{L}\big( U(\mathbb{X}) \big)$.

The Riesz Markov Theorem tells us that the space of positive linear functional $\mathcal{L}_{\geq 0}\big( U(\mathbb{X})\big)$ and $\mathcal{M}(\mathbb{X})$ the space of measures with sign $\mu$ on Borel subsets of $\mathbb{X}$ are isomorphic. So it makes sense to speak of the weak topology of $\mathcal{M}(\mathbb{X})$ which is the topology induced by the isomorphism.

But several authors of books on probability in metric spaces ( see for exemple Parthasarathy p. 40 ) define the weak topology in $\mathcal{M}(\mathbb{X})$ as that generated by the following system of neighborhoods:

$$ V_\mu \big( f_1,\dots,f_n,\epsilon_1,\dots\epsilon_n\big)=\bigg\{ \nu\in \mathcal{M}(\mathbb{X}) : \bigg| \int_{\mathbb{X}} f_i d\mu -\int_{\mathbb{X}}f_i d\nu \;\bigg|<\epsilon_i \bigg\} $$ whit $ f_1,\dots,f_n\in U(\mathbb{X})$.

Question: as we prove that these two topologies in space $\mathcal{M}(\mathbb{X})$ are really equals?

share|cite|improve this question
vizinhaças ????? – Byron Schmuland Sep 27 '12 at 18:24
@ByronSchmuland, sorry. "neighborhoods". – MathOverview Sep 27 '12 at 18:28
Aha, cognate with "vincinities" I guess. I should study other languages more. While we are at it, what do you mean by "signal"? – Byron Schmuland Sep 27 '12 at 18:30
@ByronSchmuland, Double sorry, "sign". – MathOverview Sep 27 '12 at 18:34
So: how would you write a base for the first topology? – GEdgar Sep 27 '12 at 19:28
up vote 0 down vote accepted

You write $V_f(L_1,\dots,L_n,\epsilon_1,\dots\epsilon_n)=\{g\in \mathcal{L}(U(\mathbb{X})) : |L_i(f)-L_i(g)|<\epsilon_i \}$. That notation works where the elements of the space are called $f,g$ and the functionals are called $L_i$. For the case of your question: elements of the space $\mathcal M(X)$ are called $\mu, \nu$ and functionals are written as $L_i(\mu) = \int f_i\,d\mu$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.