
\begin{proof}(\cref{lem2})~

Par définition du conducteur, on a : $\left.\psi\right|_{\varpi^m \OO} \equiv 1, \left.\psi\right|_{\varpi^{m-1} \OO} \not \equiv 1 \text {. }$ \\
Donc, si $n \geqslant m$, on a $\varpi^n \OO \subseteq \varpi^m \OO$ et l'intégrale vaut $\operatorname{Vol}(\varpi^m \OO, dx)$. \\

Supposons que $n < m$, alors $\exists a \in \varpi^{n} \OO$ tel que $\psi(a) \neq 1$. Alors,
$$
\int_{\varpi^n \OO} \psi(x) \,dx = \int_F \1_{\varpi^n \OO}(x) \psi(x) \,dx
$$
$$
\stackrel{dx \text{ de Haar}}{=} \int_F \1_{\varpi^n \OO}(x+a) \psi(x+a) \,dx
$$
$$
\stackrel{\text{car }a \in \varpi^n \OO}{=} \psi(a) \int_F \1_{\varpi^n \OO}(x) \psi(x) \,dx
$$
$$
 = \psi(a) \int_{\varpi^n \OO} \psi(x) \,dx
$$
Ainsi, $\overbrace{(1-\psi(a))}^{\neq 0} \int_{\varpi^n \OO} \psi(x) \,dx = 0$ et donc  $\int_{\varpi^n \OO} \psi(x) \,dx = 0$.
\end{proof}


Preuve: $\left.\psi\right|_{\varpi^m O} \equiv 1, \left.\psi\right|_{\varpi_{\omega-1}^{m-1}} \neq 1.$
    
Donc si $n \geqslant m$, on a $w^n O \leqslant \varpi^m O$ et l'intégrale vaut $\operatorname{Vol}(\varpi^m O, dx)$.

Si $n<m$, $\exists a \in \omega^0 O$ tel que $\psi(a) \neq 1$. Alors,
$$
\begin{aligned}
& \int_{\tan 0} \psi(x) d x=\int_F \int_\pi(x) \psi(x) d x \\
& \stackrel{\rightharpoonup}{=} \int_F \mathbb{1}_{\pi^m 0}(x+a) \psi(x+a) d x \\
& \text{can ac ஐ" } 0 \\
& =\psi(a) \int_F \mathbb{1}_{\text {tां } 0}(x) \psi(x) d x \\
& \neq 0=\psi(a) \int_{\omega^n 0} \psi(x) d x \\
&
\end{aligned}
$$

D'où $\overline{(1-\psi(a)}) \int_{w^* 0} \psi(x) d x=0$ et $\int_{w^* 0} \psi(x) d x=0$. Ainsi,
$$
