I am writting the following proof and would like to know if you guys think it is good enough, or should I maybe explain it further. My concern is that I maybe should prove why the last claim I make (about the integral being different) is true.
Consider the generalization of the Poisson process with a time-dependant intensity of arrivals $\lambda_t$ such that $\lim_{T \to \infty}\int_{0}^T\lambda_tdt=\infty$. The time $t\geq 0$ at which the event occurs has c.d.f: \begin{equation*} \mathbb{P}(t\leq\tau|t>s)=1-e^{-\int_0^{\tau}\lambda_tdt} \end{equation*}
Claim: in general, $\mathbb{P}(t\leq\tau|t>s)\neq\mathbb{P}(t\leq\tau-s)$
Proof. Using the definition of conditional probability: \begin{align*} \mathbb{P}(t\leq\tau|t>s)&=\frac{\mathbb{P}(s<t\leq\tau)}{\mathbb{P}(s<t)}\\ &=\frac{\left(1-e^{-\int_0^{\tau}\lambda_tdt}\right)-\left(1-e^{-\int_0^{s}\lambda_tdt}\right)}{1-\left(1-e^{-\int_0^{s}\lambda_tdt}\right)}\\ &=1-e^{-\left(\int_0^{\tau}\lambda_tdt-\int_0^{s}\lambda_tdt\right)}\\ &\neq 1-e^{-\int_0^{\tau-s}\lambda_tdt}\\ &=\mathbb{P}(t\leq\tau-s)\\ \end{align*} since, in general: \begin{equation*} {\int_0^{\tau}\lambda_tdt-\int_0^{s}\lambda_tdt} \neq {\int_0^{\tau-s}\lambda_tdt} \end{equation*}
if $\lambda_t$ is not a constant for all $t$.