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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?

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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

1 Answer 1

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 $L_i(\mu) = \int f_i\,d\mu$

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