Does the notion “weak convergence” coincide with that using in functional analysis? Let $\mu_k$ be a sequence of probability (Borel) measures on $\mathbb{R}^n$. We say, $\mu_k$ converges to a probability measure $\mu$ weakly if $\int gd\mu_k \to \int gd\mu$ for every continuous bounded real function $g$.
Let $M_b(\mathbb{R}^n)$ be the space of probability measures on $\mathbb{R}^n$. What would be the topology on $M_b(\mathbb{R}^n)$ so that a sequence in this spaces converges with respect to the topology if and only if it converges weakly with respect to the above definition?
And, is this topology the weak topology of some space in the sense of functional analysis?
The only natural topology I know on $M_b(\mathbb{R^n})$ is that induced by the total variation, but I think this topology is not the answer for my question.
Thank you in advance. 
 A: For any probability measure $\mu$ you can define the linear continuos operator   in $A_\mu\in (L^\infty(\mathbb{R}^n)\cap C^0(R^n))^*$ (that is the dual space of the continuos bounded, that are oviously function Borel measurable) in this way:
$A_\mu: L^\infty(\mathbb{R}^n)\cap C^0(\mathbb{R}^n)\to \mathbb{R}$
such that for every $g\in L^\infty(\mathbb{R}^n)\cap C^0(\mathbb{R}^n) $
$A_\mu(g)=\int gd\mu$
Now you can consider the weak$^* $Topology on $(L ^\infty(\mathbb{R}^n)\cap C^0(\mathbb{R}^n) )^*$ for which the succession $\{A_{\mu_k}\}_k\to A_\mu$ if and only if for any $g\in L ^\infty(\mathbb{R}^n)\cap C^0(\mathbb{R}^n) $ you have that $j(g)(A_{\mu_k})\to j(g)(A_{\mu})$ (where $j$ is the canonical injection) that means :
for every $g\in L ^\infty(\mathbb{R}^n)\cap C^0(\mathbb{R}^n) $ 
$\int gd\mu_k\to \int gd\mu$
So you can interpretate the definition of the convergence of probability measure as the converge in the weak$^* $ Topology of the space $(L ^\infty(\mathbb{R}^n)\cap C^0(\mathbb{R}^n) )^*$
A: This is a long comment / supplement to Frederico's answer, and gives a relatively explicit description of the dual. The main result I will use is the following:

Theorem (Riesz representation theorem): Let $X$ be a compact Hausdorff space, $C(X)$ be the space of continuous functions on $X$ and $M(X)$ be the space of regular signed Borel measures on $X.$ Then the map,
  $$ \varphi : M(X) \longrightarrow C(X)^* $$
  defined by,
  $$\varphi(\mu)(f) = \int_X f \,\mathrm{d}\mu$$
  is an isometric isomorphism.

The case $X = \mathbb R^n$ is a bit more subtle; the identification we get is,
$$ C_b(\mathbb R^n)^* \cong M(\beta\, \mathbb R^n),$$
where $C_b(\mathbb R^n)$ is the space of bounded continuous functions on $\mathbb R^n,$ and $\beta\, \mathbb R^n$ is the Stone–Čech compactification of $\mathbb R^n.$ This follows from the universal property, which implies that $C(\beta\,\mathbb R^n) \cong C_b(\mathbb R^n).$ Tracing through the isomorphisms, we get the pairing is,
$$ \langle f, \mu \rangle = \int_{\beta\, \mathbb R^n} \tilde f \,\mathrm{d}\mu,$$
where $f \in C_b(\mathbb R^n),$ $\mu \in M(\beta\,\mathbb R^n)$ and $\tilde f$ is the unique continuous extension of $f$ to $\beta\, \mathbb R^n.$
Note that we can easily embed $M_b(\mathbb R^n)$ into $C_b(X)^*,$ but it's not obvious how a probability measure on $\mathbb R^n$ extends to a signed measure on $\beta\,\mathbb R^n.$ [See the edit below.]
Note: An important point is that $M_b(\mathbb R^n)$ only identifies with a subspace of $C_b(X)^*,$ so it may be the case that a weakly Cauchy sequence does not converge to any probability measure. I'm not actually sure whether this statement is true or false however.
Additional note: The space of signed (regular) Borel measures on $\mathbb R^n$ can be identified with $C_0(\mathbb R^n)$; the space of continuous functions vanishing at infinity. This essentially follows by taking the one-point compactification of $\mathbb R^n.$

Edit: Actually, there is a natural way to embed $M_b(\mathbb R^n)$ into $M(\beta\,\mathbb R^n);$ you extend the measure by saying $\beta\,\mathbb R^n \setminus \mathbb R^n$ is a null set. More precisely, for $\mu \in M_b(\mathbb R^n),$ we define $\tilde \mu \in M(\beta\,\mathbb R^n)$ by,
$$ \tilde \mu(A) = \mu(A \cap \mathbb R^n)$$
for any $A \subset \beta\,\mathbb R^n$ Borel measurable. Regularity of $\tilde \mu$ follows by regularity of $\mu$ (namely inner approximation by compact subsets), and $\tilde \mu$ coincides with $\mu$ with respect to the identifications we are making.
