Clarification about weak topology in the space of probability measure

general-topologymeasure-theorypolish-spacesweak-convergence

In Jacod and Shiryaev book, page 347, we find the definition of weak convergence of probability measures.

Definition. Let $E$ be a Polish space (completely metrizable space which is also separable) and let $\mathcal{E}$ be its Borel $\sigma$-algebra (the $\sigma$-algebra generated by the collection of all open subsets of $E$). We denote by $\mathcal{P}(E)$ the space of all probability measures on $(E,\mathcal{E})$.
The weak topology on $\mathcal{P}(E)$ is defined as the coarsest topology for which the mappings $\mu\rightarrow \mu(f)$ are continuous for all bounded continuous functions $f$ on $E$.

I am not very familiar with the concept of "coarsest topology for which a mapping" is continuous, but this wikipedia page helps a bit. The idea is, among all those topologies that make the mapping $\mu\rightarrow \mu(f)$ continuous, to consider the smallest one. The problem is: does this topology exist?

Furthermore, I see form this post that the weak topology is sometimes defined in this way. We have $\mu_n\longrightarrow\mu$ whenever
$$
\int f d\mu_n \to \int f d\mu.
$$

for all bounded continuous functions $f$ on $E$.

Do the two definitions coincide? Any reference would help.

Best Answer

Yes, such a topology always exists. The intersection of any families of topologies is, again, a topology. So just take the intersection of all topologies (which is nonempty, it includes the discrete one) under which all these functions are continuous.

The two topologies coincide. Indeed, if $\mu\mapsto\int f\mathrm d\mu=\mu(f)$ is continuous, then $\langle\mu_n\rangle$ converging to $\mu$ implies that $\langle \mu_n(f)\rangle$ converges to $\mu(f)$; continuity implies sequential continuity. The other direction is a bit more difficult; sequential continuity does not always imply continuity. You would get immediate equivalence if you replace sequences with nets. But in a metrizable space, sequential continuity implies continuity. So it is enough to show that the second topology is metrizable, which can be done in various ways (such as by using the Prokhorov metric).

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