Convergence in Distribution to a Constant Implies Convergence in Probability

Let $(X,d)$ be a metric space with Borel $\sigma$-algebra $\mathcal{B}(X)$, let $c\in X$, let $(\Omega,\mathcal{F},P)$ be a probability space, and let $Y_c:\Omega\to X$ be the map with $Y_c(\omega)=c$ for every $\omega\in\Omega$. For each $n\in\mathbb{N}$ let $Y_n$ be a random element of $(X,d)$ on a probability space $(\Omega_n,\mathcal{F}_n,P_n)$, and write $\lambda_n$ for the law of $Y_n$.

1. $Y_c$ is a random element of $(X,d)$ on $(\Omega,\mathcal{F},P)$, and its law $\lambda_c$ satisfies $\lambda_c(B)=1$ for every $B\in\mathcal{B}(X)$ with $c\in B$, and $\lambda_c(B)=0$ for every $B\in\mathcal{B}(X)$ with $c\notin B$.

2. For every $n\in\mathbb{N}$ and every real $\varepsilon>0$, the set

$$E_{n,\varepsilon}=\bigl\{\omega\in\Omega_n\;:\;\varepsilon\le d\bigl(Y_n(\omega),c\bigr)\bigr\}$$

belongs to $\mathcal{F}_n$.

3. If $(Y_n)_{n\in\mathbb{N}}$ converges in distribution to $Y_c$, then for every real $\varepsilon>0$ the sequence $\bigl(P_n(E_{n,\varepsilon})\bigr)_{n\in\mathbb{N}}$ converges to $0$.