Convergence of iid random variables to a poisson process Suppose $Y_i$ are iid random variables and $\frac{ \max_{i \leq n} Y_i - a_n}{b_n} \rightarrow X$ weakly, where $X$ is nonconstant and $a_n, b_n > 0$. Let $f$ be continuous of compact support. I wish to show that $ \sum_{i=1}^n f(\frac{Y_i-a_n}{b_n}) \rightarrow \int f dN$ weakly where $N$ is a Poisson point process.
I've managed to show using independence that $nP(\frac{Y_n - a_n}{b_n} \geq \alpha) \rightarrow -\log(F_X(\alpha))$ where $F_X$ is the cdf of $X$. But I'm unsure of how to apply the Poisson convergence theorem, and bring in the $f$. I am sure that whatever $N$ turns out to be will be obtained by the Poisson convergence theorem.
 A: We give a direct proof.
As in the comment above, we define the intensity measure as $\mu(a,b] := \log F_X(b)-\log F_X(a)$. To prove the claim, it suffices to prove convergence of the associated Laplace functionals, i.e., for all $f \in C_c^{\infty}(\Bbb R)$ one has
$$E\bigg[\exp\bigg({f\big(\frac{Y_i-a_n}{b_n}\big)\bigg)}\bigg]^n \;\;\;\stackrel{ n \to \infty}{\longrightarrow} \;\;\;\exp \bigg( \int_{\Bbb R} (e^{f(x)}-1)\mu(dx) \bigg).$$ If $g$ is a $C^1$ function supported on some interval $[a,b]$, then as a consequence of Fubini's theorem, we have $E[g(X)] = \int_a^b g'(x) P(X \geq x) dx$. Applying to $g:=e^f-1$, we find that $E[e^{f(X)}] = 1+\int_a^bf'(x)e^{f(x)}P(X \geq x)dx.$ So we find that \begin{align*}
E\bigg[\exp\bigg({f\big(\frac{Y_i-a_n}{b_n}\big)\bigg)}\bigg] = 1+\int_a^b f'(x)e^{f(x)} \big( 1- F_Y(a_n x+b_n)) \big)dy
\end{align*}
where $F_Y$ denotes the cdf of the $Y_i$. On the other hand, the given convergence in distribution of $\max Y_i$ implies that $F_Y(a_nx+b_n)$ behaves asymptotically like $F_X(x)^{1/n}$, which in turn behaves like $1+\frac{1}{n}\log F_X(x)+O(\frac{1}{n^2})$. Consequently, we find that \begin{align}
\lim_{n \to \infty} E\bigg[\exp\bigg({f\big(\frac{Y_i-a_n}{b_n}\big)\bigg)}\bigg]^n &= \lim_{n \to \infty} \bigg[ 1-\frac{1}{n} \int_a^b f'(x)e^{f(x)} \log F_X(x)dx \bigg]^n \\ &= \exp \bigg[ -\int_a^bf'(x) e^{f(x)} \log F_X(x) dx \bigg] \\ &= \exp \bigg[ \int_{\Bbb R} (e^{f(x)}-1) \mu(dx) \bigg]
\end{align} where we again used fubini's theorem in the final line.
