= Sample-path area identity for Little's law
{title2=$\int_0^T N(t)\,dt=\sum_k\bigl(\min(d_k,T)-\max(a_k,0)\bigr)_+$}
Let customer $k$ enter a <queue> at $a_k$ and leave at $d_k$, and count it in $N(t)$ for $a_k\leq t<d_k$. On a finite observation interval $[0,T]$, the <integral> of the <queue population> is exactly the sum of the lengths of the overlaps of customer visits with that interval, as displayed above; $x_+=\max(x,0)$ is the <positive part>. This deterministic identity includes visits already in progress at time zero and visits still unfinished at time $T$. It needs no <probability distribution> for arrivals or <service times>. The area argument is also developed in https://web.mit.edu/urban_or_book/www/book/chapter4/4.4.html[Richard C. Larson and Amedeo R. Odoni, Urban Operations Research, section 4.4].
For a fully computed example, start empty and observe through $T=6$ minutes. Three customers arrive at $a=(0,1,4)$ minutes, begin service at $b=(0,2,4)$ minutes, and depart at $d=(2,3,5)$ minutes. These times describe a single <first come first served> server with <service times> $(2,1,1)$ minutes. Their <customer sojourn times> are $(2,2,1)$ minutes, and their <queueing delays> are $(0,1,0)$ minutes. The <queue population> is respectively $1,2,1,0,1,0$ on the consecutive intervals $[0,1),[1,2),[2,3),[3,4),[4,5),[5,6]$, so
$$
\int_0^6 N(t)\,dt=1\cdot1+2\cdot1+1\cdot1+0\cdot1+1\cdot1+0\cdot1=5\ \text{customer-minutes}.
$$
The same area is $2+2+1=5$ when counted by customer. Hence the finite-window averages satisfy
$$
L=\frac56,\qquad \lambda_{\mathrm{eff}}=\frac36=\frac12\ \text{customer/minute},\qquad W=\frac53\ \text{minutes},\qquad \lambda_{\mathrm{eff}}W=\frac56.
$$
The waiting area is one customer-minute, giving $L_q=1/6$ and $W_q=1/3$ minute; $(1/2)(1/3)=1/6$. The server works for four minutes, so its <server utilization> is $4/6=2/3$ and $L-L_q=5/6-1/6=2/3$. Repeating this six-minute pattern indefinitely makes all these values long-run averages, without a <Poisson process> of arrivals.
\Image[/little-law-sample-path.png]
{title=Little's law counts the same five customer-minutes by time slices and by customer visits}
An observation window that cuts visits short requires the overlap terms, rather than full <customer sojourn times>. Starting empty, let $A(T)$ count admissions before $T$ and write $w_k=d_k-a_k$. Then
$$
\int_0^T N(t)\,dt=\sum_{k:a_k<T}w_k-R(T),\qquad R(T)=\sum_{k:a_k<T<d_k}(d_k-T).
$$
For the example cut at $T=2.5$ minutes, the area is $1+2+0.5=3.5$ customer-minutes, whereas the full durations of the two admitted customers sum to $4$; the remaining half-minute of the second visit is $R(2.5)=0.5$. A precise sufficient <sample path> condition for the limiting <Little's law> is $A(T)/T\to\lambda_{\mathrm{eff}}\in(0,\infty)$, $A(T)^{-1}\sum_{a_k<T}w_k\to W<\infty$, and $R(T)/T\to0$. Dividing the identity by $T$ then gives $L=\lambda_{\mathrm{eff}}W$. Equating admission and departure rates additionally uses $N(T)/T\to0$, by the balance $N(T)=A(T)-D(T)$ for an initially empty <queue>, where $D(T)$ counts all departures.
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