Proof of Convergence in Distribution to a Constant Implies Convergence in Probability

Claim 1. Let $B\in\mathcal{B}(X)$. If $c\in B$ then $Y_c^{-1}(B)=\Omega$, and if $c\notin B$ then $Y_c^{-1}(B)=\varnothing$; both sets belong to $\mathcal{F}$ by Sigma-Algebra and Measurable Space. Hence $Y_c$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(X)$, that is, a random element of $(X,d)$ on $(\Omega,\mathcal{F},P)$. By Random Element of a Metric Space and Its Law its law satisfies $\lambda_c(B)=P(\Omega)=1$ when $c\in B$ and $\lambda_c(B)=P(\varnothing)=0$ when $c\notin B$.

Claim 2. The set $\{c\}$ is a nonempty subset of $X$, and for every $x\in X$ the set $\{d(x,a):a\in\{c\}\}$ is $\{d(x,c)\}$, whose greatest lower bound is $d(x,c)$; hence $\operatorname{dist}_d(x,\{c\})=d(x,c)$ in the sense of Distance from a Point to a Nonempty Subset of a Metric Space. By claim 3 of Borel Measurability and Bounded Integration on a Metric Space the map $x\mapsto d(x,c)$ is measurable with respect to $\mathcal{B}(X)$ and the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ of the real line, and by claim 4 of the same lemma the map $\omega\mapsto d\bigl(Y_n(\omega),c\bigr)$ is measurable with respect to $\mathcal{F}_n$ and $\mathcal{B}(\mathbb{R})$. The set $[\varepsilon,\infty)$ is a closed subset of $\mathbb{R}$ and hence a Borel set, and $E_{n,\varepsilon}$ is its preimage under that map; therefore $E_{n,\varepsilon}\in\mathcal{F}_n$.

Claim 3. Let $\varepsilon>0$ be real and let $U=B_d(c,\varepsilon)$ be the open ball, which is open in $(X,d)$ by Open Ball in a Metric Space is Open and hence lies in $\mathcal{B}(X)$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space. Since $d(c,c)=0<\varepsilon$ by condition 2 of Metric Space, we have $c\in U$, so $\lambda_c(U)=1$ by claim 1.

By claim 1 of Image Measures, Measures with Densities, and Change of Variables, each law $\lambda_n$ and the law $\lambda_c$ is a probability measure on $(X,\mathcal{B}(X))$, and by hypothesis together with Convergence in Distribution of Random Elements of a Metric Space the sequence $(\lambda_n)_{n\in\mathbb{N}}$ converges weakly to $\lambda_c$. Claim 3 of Portmanteau Theorem on a Metric Space therefore gives

$$1=\lambda_c(U)\le\liminf_{n}\lambda_n(U).$$

The sequence $(\lambda_n(U))_{n\in\mathbb{N}}$ takes values in $[0,1]$ by claim 2 of Basic Properties of a Measure, so it is bounded, and claim 2 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence gives $\limsup_n\lambda_n(U)\le1$. With claim 1 of that lemma we obtain

$$1\le\liminf_n\lambda_n(U)\le\limsup_n\lambda_n(U)\le1 ,$$

so both are equal to $1$, and claim 5 of that lemma shows that $(\lambda_n(U))_{n\in\mathbb{N}}$ converges to $1$.

For every $\omega\in\Omega_n$ we have $d\bigl(c,Y_n(\omega)\bigr)=d\bigl(Y_n(\omega),c\bigr)$ by the symmetry of $d$ (condition 3 of Metric Space), and exactly one of $d(Y_n(\omega),c)<\varepsilon$ and $\varepsilon\le d(Y_n(\omega),c)$ holds by the totality of the order of $\mathbb{R}$. Hence $Y_n^{-1}(U)=\Omega_n\setminus E_{n,\varepsilon}$, and so $\lambda_n(U)=P_n\bigl(\Omega_n\setminus E_{n,\varepsilon}\bigr)$. Applying claim 3 of Basic Properties of a Measure to the finite measure $P_n$ gives

$$P_n(E_{n,\varepsilon})=P_n(\Omega_n)-\lambda_n(U)=1-\lambda_n(U).$$

Let $\eta>0$ be real. Since $(\lambda_n(U))$ converges to $1$, there is $N\in\mathbb{N}$ with $|\lambda_n(U)-1|<\eta$ for every $n\ge N$, and then

$$\bigl|P_n(E_{n,\varepsilon})-0\bigr|=|1-\lambda_n(U)|=|\lambda_n(U)-1|<\eta ,$$

using claim 2 of Properties of the Absolute Value in an Ordered Field. Hence $\bigl(P_n(E_{n,\varepsilon})\bigr)_{n\in\mathbb{N}}$ converges to $0$.