To keep things simple, we assume $X$ to be a polish space (think of $X$ as $\mathbb{R}^n$ for example). Let's denote with $P(X)$ the space of all Borel probability measure on $X$. We say $\{\mu_n\}\subset P(X)$ converges weakly to $\mu\in P(X)$, denoted by $\mu_n\Rightarrow \mu$ if
$$\int fd\mu_n\to\int fd\mu,\forall f\in C_b(X)$$
where $C_b(X)$ is the space of all bounded continuous real valued functions. Let this definition be seen as the purely probabilistic one.
From functional analysis we also have a concept of weak convergence and weak topology. Let $E$ be a Banach space and denote by $E'$ the dual. Then by considering the family $\{\phi_f:f\in E'\}$, where $\phi_f:E\to\mathbb{R}$ is the linear functional $\phi_f(x):=\langle f,x\rangle$, the weak topology on $E$ is the coarsest topology which makes all $\phi_f$ continuous. One can prove $x_n\to x$ weakly (in weak topology) if and only if $\phi_f(x_n)\to\phi(x)$ for all $f\in E'$.
Since $X$ is Polish, we have that $P(X)$ is Polish too. Now I can define a continuous linear functional $\phi_f: P(X)\to\mathbb{R}$ for $f\in C_b(X)$ by
$$\phi_f(\mu):=\int fd\mu$$
Infact $\phi_f\in P(X)'$. Therefore we have $C_b(X)\subset P(X)'$.
My questions are
$\textbf{1. Question}$ Is the dual $P(X)'$ known? Is it isomorphic to a well known space? Is $C_b(X)$ a proper subspace?
$\textbf{2. Question}$ The notation of weak convergence in the probabilistic sense ($\mu_n\Rightarrow \mu$) is weaker than the weak convergence in the functional analytical sense. What I mean is: If $\mu_n\to\mu$ in the weak toplogy, i.e. $\langle f,\mu_n\rangle \to \langle f,\mu\rangle $ for all $f\in P(X)'$ this implies $\mu_n\Rightarrow \mu$ since $C_b(X)\subset P(X)'$. Is this the reason why one calls $\mu_n\Rightarrow \mu$ weak convergence, or is there any other reason?