# Weak convergence of probability measures on the one-point compactification of $[0,\infty)$

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 $$x<0$$.

$$\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\epsilon}$$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 $$\begin{equation*} (-\infty,0]\ni c_0<\cdots and $$\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 $$x, then for some $$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$$

• 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)? – Neymar May 27 at 8:41
• @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 $g(\infty):=1$. – Iosif Pinelis May 27 at 12:34
• @Neymar : I have now provided all the details. – Iosif Pinelis May 27 at 22:34
• Many thanks for the reply – Neymar May 28 at 10:00
• @Neymar : So, are you satisfied with this answer? – Iosif Pinelis May 30 at 2:14