Dual Pairs, topology of weak convergence and weak* topology

Question

Edit for Bounty: I decided to put a bounty on this question because I would really like to get it properly. Thus, I would like to get feedbacks on my basic questions, and a detailed answer on my question (3) written in a language as accessible as possible, which is the basic source of confusion.

[Of course, any additional explanation regarding dual pairs is most welcome]

Here there is my question. To ask it, I will disseminate this text with few numbered questions.

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Assume we have:

• a Polish space $X$;
• the set of continuous bounded functionals on $X$, denoted by $C_b (X)$;
• the Borel $\sigma$-algebra on $\Omega$, denoted by $\mathcal{B} ( X)$;
• the set of probability measures on $\Omega$, denoted by $\Delta (X)$.

Immagine that instead of simply endowing $\Delta(X)$ with the topology of weak convergence, we translate this all set up in terms of topological vector spaces, and consequently dual spaces, in order to get the same result by talking about the weak* topology.
Of course, we obtain the same result, but by doing so, I understand how this all thing works... :-)

1. Does what I have written here technically make sense?

If we do that, then we can say that we have a dual pair given by $\langle C_b (X), \Delta (X) \rangle$.

2. Is this literally (!) correct, or the correct expression of dual pair is on of the two:

1. $\langle \mathcal{B} (X), \Delta (X) \rangle$, or
2. $\langle X, \Delta (X) \rangle$?

I think (1) is wrong, because $\mathcal{B}(X)$ is not a vector space, thus it should not make sense to talk about $\langle \mathcal{B} (X), \Delta (X) \rangle$ as a dual pair. However, I am not sure about (2).

3. If one or the two, or both, are wrong and they cannot be dual pairs, why is the case?

Thus, we can say that we have a topological space $( \Delta (X), \sigma (X^*, X))$, where $\sigma (X^*, X)$ denotes the weak* topology, i.e.
$$ \mu_n \overset{w^*}{\longrightarrow} \mu \in \Delta (X) \Longleftrightarrow \forall f \in C_b (X), \ \langle f, \mu_n \rangle \to \langle f, \mu \rangle \in \mathbb{R},$$

where $\sigma (X^*, X) = w^*$.

4. Is this correct?

Thus, we do have that $( \Delta (X), w^*)$ is a topological space, and is nothing more than $\Delta (X)$ endowed with the topology of weak convergence. Hence, this all process also shows why it should be more appropriate to write that $\Delta (X)$ is actually endowed with the topology of weak convergence* (even though the convention is another).

5. Again, is this sound?

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Any feedback or answer is welcome, and it will be enormously appreciated.

Thank you for your time.

Answer 1

Disclaimer

This is just a formal treatment.
(It requires tedious verifications!)

Ground Space

Given a Borel space: $$\mathcal{B}(\Omega)=\sigma(\mathcal{T}(\Omega))$$ (Candidates: Polish Spaces)

Function Space

Consider the continuous functions: $$\mathcal{C}(\Omega):=\left\{f:\Omega\to\mathbb{C}:U\in\mathcal{T}(\mathbb{C}):f^{-1}U\in\mathcal{T}(\Omega)\right\}$$ They form a vector space by: $$(f+g)(\omega):=f(\omega)+g(\omega)\quad(\lambda f)(\omega):=\lambda f(\omega)$$ (Note: No topology yet!)

Measure Space

Consider the complex measures: $$\mathcal{M}(\Omega):=\left\{\mu:\Omega\to\mathbb{C}:A_k\in\mathcal{B}(\Omega):\mu\left(\bigsqcup_kA_k\right)=\sum_k\mu(A_k)\right\}$$

They form a vector space by: $$(\mu+\nu)(A):=\mu(A)+\nu(A)\quad(\lambda\mu)(A):=\lambda\mu(A)$$ (Note: No topology yet!)

Dual Pair

They become a dual pair by: $$\mathcal{C}(\Omega)\times\mathcal{M}(\Omega):\quad\langle f,\mu\rangle:=\int_\Omega f\mathrm{d}\mu$$ This endows the measures with a topology: $$\mathcal{T}(\mathcal{M}(\Omega)):=\sigma\{\mathcal{M}(\Omega);\mathcal{C}(\Omega)\}$$ (Hint: Reversing endows the functions with a topology.)

Modifications

Various function spaces: $$\mathcal{C}^\infty(\Omega)\quad\mathcal{C}_0(\Omega)\quad\mathcal{C}_\infty(\Omega)\quad\mathcal{C}^\infty_0(\Omega)\quad\ldots$$ Various measure spaces: $$\mathcal{M}_\mathbb{R}(\Omega)\quad\mathcal{M}_{\mathbb{R}_+}(\Omega)\quad\mathcal{M}_\mathbb{S}(\Omega)\quad\mathcal{M}_{\mathbb{S}_+}(\Omega)\quad\ldots$$ (Warning: Some don't form a vector space!)