I read the infinitesimal generator for the M/M/1 queue and thought to generalize to the G/G/1 queue. More specifically, though the queue length process is not Markovian anymore, we could consider an extended state $X_t = (Q_t, A^\lambda_t, A^\mu_t)$ in space $\Omega = \mathbb{N}\times \mathbb{R}_+\times \mathbb{R}_+$. Here $Q_t$ is the queue length/customer in the system, $A^\lambda_t$ is the elapsed time since the last arrival comes, and $A^\mu_t$ is elapsed time since the current job started. The subscript $t$ stands for the time and we consider the time interval $t\in[0,\infty)$.
Now consider a function $f: \mathbb{R}^3\rightarrow \mathbb{R}$ with continuous and bounded first partial derivative. By the integration by part (or generalized Ito's formula), we could yield the following (not that straightforward): $$ \begin{align} \mathbb{E}[f(X_t)] - f(x) =&\mathbb{E}[\int_0^t \partial_2 f(X_{s-}) + \mathbb{1}(Q_{s-} > 0)\partial_3 f(X_{s-}) ds]\label{firstroderterms2}\\ +& \mathbb{E}[\int_0^t [f\big((Q_{s-}+1, 0, A^\mu_{s-})\big) - f\big((Q_{s-}, A^\lambda_{s-}, A^\mu_{s-})\big)]r_\lambda(A^\lambda_s) ds]\label{arrivaljump2}\\ +& \mathbb{E}[\int_0^t [f\big((Q_{s-}-1, A^\lambda_{s-}, 0)\big) - f\big((Q_{s-}, A^\lambda_{s-}, A^\mu_{s-})\big)]\mathbb{1}(Q_{s-} >0 ) r_\mu(A^\mu_{s}) ds]\label{servicejump2} \end{align} $$ To give some information about the above formula, the first row is just the continuous part of function change as what we saw in the fundamental theorem of calculus, the second the third part corresponds to two types of jumps: a new arrival comes and the elapsed time $A^\lambda_s$ becomes 0, or a job has finished so the elapsed time $A^\mu_s$ becomes 0. We do not consider the event that two types of jump happen at the same time, as in the next step we will take the expectation and such event has probability zero. The term $r_\lambda(\cdot)$ and $r_\mu(\cdot)$ are the hazard rate for the arrival and service distribution, and those terms appear due to the stochastic intensity.
So now we could calculate the infinitesimal generator: assume we consider the point $x = (q, a^\lambda, a^\mu)$
$$\lim_{t\rightarrow 0} \frac{\mathbb{E}[f(X_t)] - f(x)}{t}$$
My questions is at the first row: could we have $$ \begin{align} &\lim_{t\rightarrow 0} \frac{\mathbb{E}[\int_0^t \partial_2 f(X_{s-}) + \mathbb{1}(Q_{s-} > 0)\partial_3 f(X_{s-}) ds]}{t}\text{ By Fubini to change the integral and expectation}\\ =&\lim_{t\rightarrow 0} \frac{\int_0^t \mathbb{E}[\partial_2 f(X_{s-}) + \mathbb{1}(Q_{s-} > 0)\partial_3 f(X_{s-})] ds}{t}\\ =& \partial_2 f(x) + \mathbb{1}(q > 0)\partial_3 f(x) \end{align} $$ I want to apply a theorem like the integral mean value theorem to take the limit, but since $X_t$ is a jump process, I am not sure about the continuity of the integrand in the above formula. I believe the continuity of this integrand $\mathbb{E}[\partial_2 f(X_{s-}) + \mathbb{1}(Q_{s-} > 0)\partial_3 f(X_{s-})] $ could be proved, but I am not very comfortable with this three dimensional Markovian process ( with two dimensions of uncountable points), so I am not sure how to prove this.
I apologize for my abused notation and not that well description, if something is not clear please feel free to comment below. I would really if someone could offer a hint for this.
