Let $L\subset(0,\infty)$ and $\lambda\in L$.
Let $\mu_\lambda\sim$Poiss$(\lambda)$.
How do we show that $(\mu_\lambda)_{\lambda\in\Lambda}$ is tight if and only if $L$ is bounded?
I was hoping that there exists a way to work only with the definition.
Tightness means that for $\epsilon>0$ there exists an $M>0$ such that for all $m\geq M$ we have
$$1-\inf_{\lambda\in L}\mu_\lambda[-m,m]=1-\inf_{\lambda\in L}\sum_{i=0}^me^{-\lambda}\frac{\lambda^i}{i!}<\epsilon$$
Is there a way of showing this directly or must we use other results?
If $L$ is bounded (say contained in $(0,R)$), then $$ 1-\inf_{\lambda\in L}\sum_{i=0}^me^{-\lambda}\frac{\lambda^i}{i!} =\sup_{\lambda\in L}\sum_{i=m+1}^{+\infty}e^{-\lambda}\frac{\lambda^i}{i!} \leqslant \sum_{i=m+1}^{+\infty} \frac{R^i}{i!} $$ so for a fixed $\varepsilon$, it suffices to choose $M$ such that $ \sum_{i=M+1}^{+\infty} \frac{R^i}{i!}\lt\varepsilon$.
If $L$ is not bounded, then there exists a sequence $\left(\lambda_n\right)_{n\geqslant 1}$of elements of $L$ which goes to infinity. Then for all $m$, $$ \inf_{\lambda\in L}\sum_{i=0}^me^{-\lambda}\frac{\lambda^i}{i!} \leqslant \inf_{n\geqslant 1}\sum_{i=0}^me^{-\lambda_n}\frac{\lambda_n^i}{i!}=0 $$ because $\lim_{t\to +\infty}e^{-t}t^i=0$ for all $i$ hence the definition is contradicted for $\varepsilon\lt 1$.