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?
Best Answer
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).
A commenter asked about an almost sure version. The following holds:
Let $C_c(\mathbb{R})$ be the space of compactly supported functions with the uniform norm. This space is separable even though $C_b(\mathbb{R})$ is not, so let $f_1, f_2, \dots$ be a countable dense subset of $C_0(\mathbb{R})$. Then we can find a null set $E$ so that $\langle L_N(\omega), f_k \rangle \to \langle \sigma(\omega), f_k \rangle$ for all $\omega \notin E$ and all $k$. Using the triangle inequality, we can show that for such $\omega$, we actually have $\langle L_N(\omega), f \rangle \to \langle \sigma(\omega), f \rangle$ for all $f \in C_c(\mathbb{R})$. Indeed, fix $f \in C_c(\mathbb{R})$, let $\epsilon > 0$ be arbitrary, and choose an $f_k$ with $\|f-f_k\|_\infty < \epsilon$. Now $$\begin{align*}|\langle L_N(\omega), f \rangle - \langle \sigma(\omega), f \rangle| &\le |\langle L_N(\omega), f-f_k\rangle |+ |\langle L_n(\omega), f_k \rangle - \langle \sigma(\omega), f_k \rangle| + |\langle \sigma(\omega), f_k - f \rangle| \\ &\le |\langle L_n(\omega), f_k \rangle - \langle \sigma(\omega), f_k \rangle| + 2\epsilon\end{align*}$$ since $L_N(\omega), \sigma(\omega)$ are probability measures. Taking limsup of both sides, we have $$\limsup_{N \to \infty} |\langle L_N(\omega), f \rangle - \langle \sigma(\omega), f \rangle| \le 2 \epsilon.$$ Since $\epsilon$ was arbitrary, we conclude $\limsup_{N \to \infty} |\langle L_N(\omega), f \rangle - \langle \sigma(\omega), f \rangle| = 0$, which is to say $\langle L_N(\omega), f \rangle \to \langle \sigma(\omega), f \rangle$. But $f \in C_c(\mathbb{R})$ was arbitrary as well.
But this implies that $L_N(\omega) \to \sigma(\omega)$ weakly; see Defining weak* convergence of measures using compactly supported continuous functions. As this holds for all $\omega \notin E$ we are done.