Definition of weak convergence of random measures in probability In the book Anderson, Greg W., Alice Guionnet, and Ofer Zeitouni. An introduction to random matrices. (Vol. 118. Cambridge university press, 2010.), the following theorem is stated:

For a Wigner matrix, the empirical measure $L_N$ converges weakly, in probability, to the semicircle distribution.
  In greater detail, Theorem 2.1.1 asserts that for any $f \in \mathcal{C}_b(\mathbb{R})$, and any $\varepsilon > 0$,
  $$\begin{equation}\tag{1}\lim_{N \to \infty} \mathbb{P}(|\left<L_N, f\right> - \left<\sigma, f\right>| > \varepsilon) = 0\end{equation}$$

(The definition of Wigner matrix is standard and the empirical measure is defined as $L_N = N^{-1}\sum_{j = 1}^{N} \delta_{\lambda_j}$ where $\lambda_j$ are the (random) eigenvalues of the Wigner matrix. Also, in this case, $\sigma$ is non-random and is the standard semicircle distribution.)
My question is the following:
Convergence in probability of $M$-valued random variables $L_N$ to $\sigma$ (where $M$ is a Polish space) is defined as $$\begin{equation}\tag{2}\forall \varepsilon > 0, \lim_{N \to \infty} \mathbb{P}(d_M(L_N, \sigma) > \varepsilon) = 0\end{equation}$$ where $d_M$ is the distance on $M$. So, in the aforementioned case of the Polish space of measures on $\mathbb{R}$, we should use any of the standard equivalent metrics which metrize that space.
How do we prove that showing $(1)$ is equivalent to showing $(2)$ in this case?
 A: Let $d$ be your favorite metric which metrizes the weak topology on $\mathcal{P}(\mathbb{R})$.  
Suppose (1) holds.  Let $B(\sigma, \varepsilon)$ denote the open $d$-ball centered at $\sigma$ of radius $\varepsilon$.  This ball is open in the weak topology, which means that there exist $f_1, \dots, f_n \in \mathcal{C}_b(\mathbb{R})$ and $\delta > 0$ such that for any measure $\mu$ satisfying $|\langle \mu, f_i\rangle - \langle \sigma, f_i \rangle| \le \delta$ for $i = 1, \dots, n$, we have $\mu \in B(\sigma, \varepsilon)$.   In other words, if $d(\mu, \sigma) > \epsilon$, then $|\langle \mu, f_i\rangle - \langle \sigma, f_i \rangle| > \delta$ for some $i$.  This implies
$$\mathbb{P}(d(L_N, \sigma) > \varepsilon) \le \mathbb{P}\left(\bigcup_{i=1}^n \{|\langle L_N, f_i \rangle - \langle \sigma, f_i \rangle| > \delta\}\right) 
\le \sum_{i=1}^n \mathbb{P}(|\langle L_N, f_i \rangle - \langle \sigma, f_i \rangle| > \delta) $$
by union bound.  But the right side is a sum of $n$ terms, each of which converges to $0$ as $N \to \infty$ (according to (1)), therefore the left side also converges to 0, which gives (2).
Now suppose (2) holds, and fix any $f \in \mathcal{C}_b(\mathbb{R})$.  The map $\mu \mapsto \langle \mu, f \rangle$ is continuous with respect to $d$, so by the continuous mapping theorem (whose proof is the same in any metric space), we have $\langle L_N, f \rangle \to \langle \sigma, f \rangle$ in probability, which is precisely the statement (1).
