My probability textbook gives the following problem:
For the family $(\pi_{\alpha})_{\alpha>0}$ of Poisson distributions on $\mathbb{R}$ show that
$$\lim_{\alpha\to 0, \alpha>0} \pi_{\alpha}=\varepsilon_0 $$
in the sense of weak convergence. Is there a probability measure $\mu$ on $\mathcal{B}(\mathbb{R})$ such that $\pi_{\alpha}\to\mu$ as $\alpha\to \infty$?
For the first part I used the definition of weak convergence in my book: Let $f\in C_b(\mathbb{R})$ be a continuous real-valued bounded function on $\mathbb{R}$. Then
$$ \int f d\pi_{\alpha}=\sum_{k=0}^{\infty} e^{-\alpha}\frac{\alpha^k}{k!}f(k)=e^{-\alpha}f(0)+\sum_{k=1}^{\infty} e^{-\alpha}\frac{\alpha^k}{k!}f(k)$$
and $\bigg \lvert \sum_{k=1}^{\infty} e^{-\alpha}\frac{\alpha^k}{k!}f(k)\bigg\rvert\leq M e^{-\alpha} (e^{\alpha}-1)\to 0$ as $\alpha\to 0$ so we indeed have
$$\int f d\pi_{\alpha} \to \int f d\varepsilon_0 \hspace{1cm} \text{as } \hspace{0.2cm} \alpha\to 0$$
But am stuck on second part. I believe the answer is $\mu$ normal distribution but I wasn't able to show it. Bear in mind we haven't covered characteristic functions yet.
Any help is greatly appreciated.
$\pi_{\alpha} ([0,N])= \sum\limits_{k=0}^{N} e^{-\alpha}\frac {\alpha^{k}} {k!} \to 0$ as $\alpha \to \infty$ for each $n$. Hence if $\pi_{\alpha}$ does convergse weakly to some $\mu$ then we get $\mu [0,N)=0$ for each $N$ which is clearly impossible.