Proof of Theorem 1. We start with a simple observation: for any $j\geq 0$ , $\{\liminf Y_t =j\}$ is a tail event, so it has probability 0 or 1. This implies that $\liminf Y_t$ is a.s. a constant (which may be equal to $+\infty$ ).

We use the following representation of the process (see Figure 1): consider a Poisson process in $\mathbb{R}_+^2$ , with the intensity measure $\lambda\, {\mathrm{d}} t \times {\mathrm{d}} F_S(u)$ , where $F_S(u)=\mathbb{P}[S\leq u]$ is the distribution function of S. Then, a point (t, u) of this Poisson process is interpreted in the following way: a customer arrived at time t and the duration of its service will be u. Now, draw a (dotted) line in the southeast direction from each point, as shown in the picture; as long as this line stays in $\mathbb{R}_+^2$ , the corresponding customer is present in the system. If we draw a vertical line from (t, 0) in the upwards direction, then the number of dotted lines it intersects is equal to $Y_t$ .

Figure 1. A Poisson representation of $\mathrm{M}/\mathrm{G}/\infty$ . In this example, there are exactly three customers at time t.

Next, for $k\in\mathbb{Z}_+$ let

\[ T_k:= \{t\geq 0\,\colon\, Y_t=k\}\]

denote the set of times when the system has exactly k customers, and let

\[ U_{t} = \bigl\{(s,u)\in\mathbb{R}_+^2\,\colon\, s\in[0,t], u\geq t-s \bigr\}.\]

We note that $Y_t$ equals the number of points in $U_{t}$ , which has Poisson distribution with mean

\[ \int_{U_{t}}\lambda \, {\mathrm{d}} t\, {\mathrm{d}} F_S(u) = \lambda \mathbb{E}(S\wedge t).\]

Therefore, by Fubini’s theorem, we have (here $|A|$ stands for the Lebesgue measure of $A\subset \mathbb{R}$ )

(3) \begin{equation}\mathbb{E} |T_k| = \mathbb{E} \int_0^{\infty} {\mathbf{1}\mkern -1.5mu}{\{Y_t=k\}}\, {\mathrm{d}} t= \dfrac{\lambda^k}{k!}\int_0^{\infty} (\mathbb{E}(S\wedge t))^k\text{exp}({-}\lambda\mathbb{E}(S\wedge t))\, {\mathrm{d}} t.\end{equation}

Now, assume that $\mathbb{E}|T_k|<\infty$ for some $k\geq 0$ ; it automatically implies that $\mathbb{E}|T_\ell|<\infty$ for $0\leq\ell\leq k$ . This means that $|T_0|,\ldots,|T_k|$ are a.s. finite, and let us show that $T_0,\ldots,T_k$ have to be a.s. bounded (this is a small technical issue that we have to resolve because we are considering continuous time). Probably the cleanest way to see this is as follows. First notice that, in fact, $T_0$ is a union of intervals of random i.i.d. (with Exp( $\lambda$ ) distribution) lengths, because each time the system becomes empty it will remain so until the arrival of the next customer. Therefore, $|T_0|<\infty$ clearly means that $\sup T_0\leq K_0$ for some (random) $K_0$ . Now, after $K_0$ there are no longer any $1\to 0$ transitions, so the remaining part of $T_1$ again becomes a union of such intervals, meaning that it should be bounded as well; we then repeat this reasoning a suitable number of times to finally obtain that $T_k$ must be a.s. bounded. This implies that $\liminf_{t\to \infty} Y_t \geq k_0$ a.s.

Next, assume that $\{0,\ldots,k\}$ is a transient set, in the sense that $\liminf_{t\to \infty} Y_t \geq k+1$ a.s.; let us show that this implies that $\mathbb{E}|T_k|<\infty$ . Indeed, first we can choose a sufficiently large $h>0$ in such a way that

\[ \mathbb{P}[Y_t\geq k+1 \text{ for all }t\geq h]\geq \dfrac{1}{2}.\]

Define a stopping time $\tau=\inf\{t\geq h \,\colon\, Y_t\leq k\}$ (again, with the convention $\inf\emptyset = +\infty$ ). Then, a crucial observation is that what one sees after $\tau$ is a superposition of two independent systems: one is formed by those customers (with their remaining lifetimes) present at $\tau$ , and the other is a copy of the original system. Then, a simple coin-tossing argument, together with the fact that an initially non-empty system (i.e. with some customers being served, with any assumptions on their remaining service times) dominates an initially empty system, shows that $|T_k|$ (in fact, $|T_0|+\cdots+|T_k|$ ) is dominated by $h\times\mathrm{Geom}_0 ( {\tfrac{1}{2}} )$ random variable and therefore has a finite expectation. It means that we have $\liminf_{t\to \infty} Y_t \leq k_0$ a.s. (because otherwise, in the situation when $k_0<\infty$ , we would have $\mathbb{E}|T_{k_0}|<\infty$ , which, by definition, is not the case). This concludes the proof of Theorem 1.

Proof. Let

\[ H_q = \{t\geq 0\,\colon\, Y_t < q\lambda \mathbb{E}(S\wedge t) \};\]

our goal is to show that $H_q$ is a.s. bounded in the case when (4) holds. We recall a standard (Chernoff) tail bound: if X is Poisson( $\mu$ ) and $q\in(0,1)$ , then

(6) \begin{equation} \mathbb{P}[X\leq q\mu]\leq \text{exp}({-}(q\mu\ln q + \mu-q\mu))= \text{exp}({-}\gamma_q \mu).\end{equation}

Then, analogously to (3), we obtain from (6) that

(7) \begin{equation} \mathbb{E} |H_q| \leq \int_0^\infty \text{exp}({-}\gamma_q\lambda \mathbb{E}(S\wedge t))\, {\mathrm{d}} t;\end{equation}

so, by (4), we have $\mathbb{E}|H_q|<\infty$ , meaning that $|H_q|<\infty$ a.s. To see that this has to imply that $H_q$ is a.s. bounded, analogously to the proof of Theorem 1, one can reason in the following way. If $t\in H_q$ , then $s\in H_q$ for all $s\in (t, A_t)$ , where $A_t$ is the first moment after t when a customer arrives in the system. This implies that the lengths of the intervals that constitute $H_q$ dominate a sequence of i.i.d. random variables with Exp( $\lambda$ ) distribution; in its turn, this clearly implies that if $|H_q|$ is finite then it has to be bounded.