3
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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 $x<0$.

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$\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\epsilon}$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 $\infty\in C_F$). $\Box$

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  • $\begingroup$ Thanks for the response. Is there any reference for the probability measures on the one-point compactification of $[0,\infty)$? Indeed, according to the definition of the one-point compactification, what is the corresponding metric and what are the corresponding continuous functions on $[0,\infty]$ (to define the weak convergence of 1)? $\endgroup$ – Neymar May 27 at 8:41
  • $\begingroup$ @Neymar : (i) I don't know references concerning precisely this situation. (ii) To define continuous functions on a set $X$, one does not need to have a metric on $X$; to do that, it is sufficient (and necessary) to have a topology on $X$. In your case, of $X=[0,\infty]$, you defined the corresponding standard topology in your post. (iii) Still, if you wish, you can define a metric on $[0,\infty]$ say by the formula $d(x,y):=|g(x)-g(y)|$, where $g(x):=x/(1+x)$ for $x\in[0,\infty)$ and $g(\infty):=1$. $\endgroup$ – Iosif Pinelis May 27 at 12:34
  • $\begingroup$ @Neymar : I have now provided all the details. $\endgroup$ – Iosif Pinelis May 27 at 22:34
  • $\begingroup$ Many thanks for the reply $\endgroup$ – Neymar May 28 at 10:00
  • $\begingroup$ @Neymar : So, are you satisfied with this answer? $\endgroup$ – Iosif Pinelis May 30 at 2:14

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