I was reading some lecture notes for Poisson processes, in which after proving the following,
$\lim_{t\rightarrow \infty}P(X(t)=j) = e^{-\rho}\frac{\rho^j}{j!}$
it states that this implies convergence in a time average sense as well:
$\lim_{t\rightarrow \infty}\frac{1}{t}\int_0^tP(X(s)=j)ds = e^{-\rho}\frac{\rho^j}{j!}$
Intuitively, it seems fine to me. Only what happens at large values of $t$ matters to us when computing long run time average. However, how should I go about proving this formally?
Your question isn't really about probability since we can just define $f(s) = \mathbb{P}(X(s) = j)$. Then we are given that $\displaystyle{\lim_{s \to \infty} f(s) = c}$ for some constant $c$.
We are asked to prove that $\displaystyle{\lim_{t \to \infty} \frac{1}{t} \int_0^t f(s)ds = c}$. This is true for any bounded $f$. (Clearly $f$ is a bounded function since it is a probability). As you say, only large values of $t$ matter. For any $\varepsilon > 0$, there is some $u$ such that $|f(s) - c| < \varepsilon$ for $s > u$. Then just divide the domain of integration into the intervals $[0,u]$ and $[u,t]$.