Time change of an inhomogeneous Poisson process

If $\lambda >0$ and $g={\Lambda }^{-1}$, then $M_t=N_{g(t)}$ is a rate-one Poisson process, becauseConversely, $N_t=M_{\Lambda (t)}$, so $N_t$ has a Poisson distribution of mean $\Lambda (t)$.