Denote by $[0,\infty]\equiv [0,\infty)\cup \{\infty\}$ the one-point compactification of $[0,\infty)$, i.e. all the open sets related to $[0,\infty]$ are either the open sets of $[0,\infty)$ or the sets of the form $G\cup \{\infty\}$, where $G\subset [0,\infty)$ is an open subset s.t. $[0,\infty)\setminus G$ is compact.


Let $\mathcal P$ be the set of probability measures $\mu$ on $[0,\infty]$. If we define the cumulative distribution function $F\_{\mu}:[0,\infty]\to [0,1]$, i.e. $F\_{\mu}(x)=\mu([0,x])$ for all $x\in [0,\infty]$, then we have the equivalence of the following claims :


1. $\mu\_n$ converges weakly to $\mu$;
2. $d(F\_{\mu\_n},F\_{\mu}):=\inf\big\{\epsilon>0 : F\_{\mu}(x-\epsilon)-\epsilon\le F\_{\mu\_n}(x) \le F\_{\mu}(x+\epsilon)+\epsilon \mbox{ for all } x\in [0,\infty]\big\}$ converges to zero;
3. $F\_{\mu\_n}(x)$ converges to $F\_{\mu}(x)$ for all continuity points $t\in [0,\infty]$.


Here we set $F\_{\mu\_n}(x)=F\_{\mu}(x)\equiv 0$ for $x<0$.



Weak convergence of probability measures on the one-point compactification of $[0,\infty)$
$\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\epsilon}$The implications 1$\iff$3 follow because $[0,\infty]$ is homeomorphic to $[0,1]$, with the preservation of the order. Such an order-preserving homeomorphism $g\colon[0,\infty]\to[0,1]$ is given by $g(x):=x/(1+x)$ for $x\in[0,\infty)$, with $g(\infty):=1$. Consider the pushforward probability measures
\begin{equation\*}
\text{$\nu:=\mu g^{-1}$ and $\nu\_n:=\mu\_n g^{-1}$} \tag{1}
\end{equation\*}
on $[0,1]$. Since $g$ is a homeomorphism, a function $f\colon[0,1]\to\R$ is continuous and bounded iff the function $f\circ g\colon[0,\infty]\to\R$ is continuous and bounded. Also, by the change of variables for measures, for any bounded Borel-measurable function $f\colon[0,1]\to\R$,
\begin{equation\*}
 \int\_{[0,1]}f\,d\nu=\int\_{[0,\infty]}(f\circ g)\,d\mu, \quad
 \int\_{[0,1]}f\,d\nu\_n=\int\_{[0,\infty]}(f\circ g)\,d\mu\_n. \tag{2}
\end{equation\*}
It follows that $\mu\_n\to\mu$ (weakly) iff $\nu\_n\to\mu$. So, Claim 1 holds for $\mu,\mu\_n$ iff it holds for $\nu,\nu\_n$.


Also, since $g$ preserves the order, $F\_\mu=F\_\nu\circ g$ and $F\_{\mu\_n}=F\_{\nu\_n}\circ g$, where $F\_\nu$ and $F\_{\nu\_n}$ are the cumulative distribution functions for $\nu$ and $\nu\_n$. So, Claim 3 holds for $\mu,\mu\_n$ iff it holds $\nu,\nu\_n$.


So, the implications 1$\iff$3 for $\mu,\mu\_n$
follow from the usual implications 1$\iff$3 for probability measures $\nu,\nu\_n$ on $[0,1]$, which latter may be identified with probability measures on $\R$.




---


Let us now prove implication 2$\implies$3. Assume indeed that Claim 2 holds.
Let $F:=F\_\mu$ and $F\_n:=F\_{\mu\_n}$.
Let $C\_F$ denote the set of all points of continuity of $F$ in $[0,\infty]$. Take any $x\in C\_F$. Take any real $\de>0$. Then for some $\ep=\ep\_\de\in(0,\de)$ we have
\begin{equation\*}
 0\le\max(|F(x-\ep)-F(x)|,|F(x+\ep)-F(x)|)\le\de. 
\end{equation\*}
Since Claim 2 holds, there is some natural $n\_\ep=n\_{\ep\_\de}$ such that
\begin{equation\*}
 F(x-\ep)-\ep\le F\_n(x)\le F(x+\ep)+\ep
\end{equation\*}
for $n\ge n\_\ep$. So,
\begin{align\*}
 |F\_n(x)-F(x)|&\le\max(|F(x-\ep)-\ep-F(x)|,|F(x+\ep)+\ep-F(x)|) \\ 
 &\le\ep+\max(|F(x-\ep)-F(x)|,|F(x+\ep)-F(x)|) \\ 
 &\le\ep+\de\le2\de. 
\end{align\*}
This completes the proof of implication 2$\implies$3.




---


Implication 3$\implies$2 does not hold in general. E.g., let here $\mu$ and $\mu\_n$ be the Dirac measures $\de\_\infty$ and $\de\_n$, respectively. Then $F(x)=1(x=\infty)$ and $F\_n(x)=1(x\ge n)$ for $x\in[0,\infty]$, so that $C\_F=[0,\infty)$ and hence Claim 3 holds. However, for all $\ep\in[0,1)$ and all natural $n$ we have $F\_n(n)=1>\ep=F(n+\ep)+\ep$, so that $d(F\_n,F)\ge1$ and hence Claim 2 does not hold.


However, implication 3$\implies$2 will hold if we additionally assume $\infty\in C\_F$, that is, $\mu(\{\infty\})=0$. Indeed, assume that Claim 3 holds and take any real $\ep>0$. Since $\infty\in C\_F$ and $F(\infty)=1$, and because $C\_F$ is dense in $[0,\infty)$, there is some real $c\_\*\in[0,\infty)\cap C\_F$ such that
\begin{equation\*}
 1-F(c\_\*)\le\ep/2. \tag{3}
\end{equation\*}
Moreover, there exist some natural $k$ and some $c\_0,\dots,c\_k$ in $C\_F$ such that
\begin{equation\*}
 (-\infty,0]\ni c\_0<\cdots<c\_k=c\_\*
\end{equation\*}
and
\begin{equation\*}
 c\_{j+1}-c\_j\le\ep
\end{equation\*}
for all $j=0,\dots,k-1$.
Next, for some natural $n\_\ep$, all natural $n\ge n\_\ep$, and all $j=0,\dots,k$,  

\begin{equation\*}
 |F\_n(c\_j)-F(c\_j)|\le\ep/2.
\end{equation\*}


Take now any $x\in[0,\infty]$ and any natural $n\ge n\_\ep$. If $x\ge c\_k[=c\_\*]$, then, in view of (3),
\begin{equation\*}
 F\_n(x)\le1\le F(c\_k)+\ep/2\le F(x)+\ep/2\le F(x+\ep)+\ep, 
\end{equation\*}
\begin{equation\*}
 F\_n(x)\ge F\_n(c\_k)\ge F(c\_k)-\ep/2\ge1-\ep\ge F(x-\ep)-\ep, 
\end{equation\*}
so that
\begin{equation}
 F(x-\ep)-\ep\le F\_n(x)\le F(x+\ep)+\ep. \tag{4}
\end{equation}
If $x<c\_k$, then for some $j\in\{0,\dots,k-1\}$ we have $c\_j\le x\le c\_{j+1}$ and hence
\begin{equation\*}
 F\_n(x)\le F\_n(c\_{j+1})\le F(c\_{j+1})+\ep/2\le F(x+\ep)+\ep, 
\end{equation\*}
\begin{equation\*}
 F\_n(x)\ge F\_n(c\_j)\ge F(c\_j)-\ep/2\ge F(x-\ep)-\ep. 
\end{equation\*}
So, (4) holds for all $x\in[0,\infty]$ and all natural $n\ge n\_\ep$. So, Claim 2 holds.
This completes the proof of implication 3$\implies$2 (assuming $\infty\in C\_F$).
$\Box$


