There are many realistic cases where this phenomenon arises, but I will show one main class of cases. Suppose we have a reliability problem with some object that will fail within a finite time. Let $T \geqslant 0$ be the (continuous) random failure time for that object, and let $\lambda$ be the hazard function for the failure time. Since the object is assumed to fail within a finite time, we must have:
$$\int \limits_0^\infty \lambda(t) dt = \infty.$$
Now, suppose we start observing this object at some time $t_0 > 0$, and we are unable to observe what state the object is in, but we can observe the failure event itself. Let $T_*$ be the remaining time until failure, from the time at which we start observing the object. Now, if $0 \leqslant T < t_0$ then the object has already failed before we start observing it, so we will never observe the failure event, and so $T_* = \infty$. This outcome occurs with probability $\mathbb{P}(T_*=\infty) = F_T(t_0)$.
Assuming that this event occurs with non-zero probability (i.e., there is some non-zero probability of failure prior to $t_0$), let's see what this means for the hazard function of $T_*$. For all $0 \leqslant t < \infty$ we have:
$$\begin{equation} \begin{aligned}
F_{T_*}(t) &= F_{T}(t_0+t) - F_{T}(t_0), \\[6pt]
f_{T_*}(t) &= f_{T}(t_0+t). \\[6pt]
\end{aligned} \end{equation}$$
Thus, the hazard rate for the remaining failure time $T_*$ is:
$$\begin{equation} \begin{aligned}
\lambda_*(t)
&= \frac{f_{T_*}(t)}{1-F_{T_*}(t)} \\[6pt]
&= \frac{f_{T}(t_0+t)}{1-F_{T}(t_0+t) + F_{T}(t_0)} \\[6pt]
&= \frac{1-F_{T}(t_0+t)}{1-F_{T}(t_0+t) + F_{T}(t_0)} \cdot \frac{f_{T}(t_0+t)}{1-F_{T}(t+t_0)} \\[6pt]
&= \frac{1-F_{T}(t_0+t)}{1-F_{T}(t_0+t) + F_{T}(t_0)} \cdot \lambda(t+t_0). \\[6pt] \\[6pt]
\end{aligned} \end{equation}$$
We can see that this hazard function is composed of the initial hazard function for $T$, multiplied by an adjustment term. If there is a non-zero probability that the object has already failed prior to observation, then the adjustment term diminishes to zero as $t \rightarrow \infty$. In fact, it can be shown that:
$$\int \limits_0^\infty \lambda_*(t) dt = - \ln \mathbb{P}(T_*=\infty) < \infty.$$
We therefore see that, if there is a non-zero probability that the object has already failed prior to observation, then we have a random variable with a hazard rate that does not integrate to infinity. Note that this latter random variable arises as a simple variation on the initial problem, where we begin observing the object after a period of time, and we only observe the failure event, not the state of the object.
Example: Suppose we have constant hazard rate $\lambda(t) = 1$, so that $F_T(t) = 1-\exp(- t)$ and $\mathbb{P}(T_*=\infty) = 1-\exp(- t_0)$. Then the hazard function for $T_*$ is:
$$\lambda_*(t) = \frac{\exp(-(t_0+t))}{\exp(-(t_0+t)) + 1-\exp(-t_0)}.$$
You can plot this hazard function and see that it reduces quite rapidly towards zero. Integrating this function gives:
$$\begin{equation} \begin{aligned}
\int \limits_0^\infty \lambda_*(t) \ dt
&= \Bigg[ - \ln \big| \exp(-t) + \exp(t_0) - 1 \big| \Bigg]_{t=0}^{t \rightarrow \infty} \\[6pt]
&= \Bigg[ (- \ln \big| \exp(t_0) - 1 \big|) - (- \ln \big| \exp(t_0) \big| ) \Bigg] \\[6pt]
&= - \ln \bigg( \frac{ \exp(t_0)-1}{\exp(t_0)} \bigg) \\[6pt]
&= - \ln ( 1-\exp(-t_0) ) \\[6pt]
&= - \ln \mathbb{P}(T_*=\infty). \\[6pt]
\end{aligned} \end{equation}$$