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There's quite a bit of set-up for this, but please bear with it! Let $K$ be a number field, and let $E/K$ be an elliptic curve. Let $v$ be a finite place of $K$, and suppose that at $v$, $E$ has minimal model $$y^2+a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6.$$ Let $\omega$ be the corresponding minimal differential; that is $$\omega = \frac{dx}{2y+a_1x+a_3}.$$ Let $\mu_v$ be the Haar measure on $K_v$ such that $\mu_v(\mathcal{O}_{K_v})=1$. Finally, define a measure $\mu_v(\omega,-)$ on $E(K_v)$ by $$\mu_v(\omega,B)=\int_{B}|\omega|_v=\int_{P\in B}|2y(P)+a_1x(P)+a_3|_v^{-1}d\mu_v(x(P)),$$ where $B$ is some Borel set of $E(K_v)$, and $|\cdot|_v$ is the normalized absolute value corresponding to $v$. It's easy enough to show that this is a Haar measure on $E(K_v)$.

Right, here's the actual question! When $v(2)=0$, I have been able to show that $$\mu_v(\omega,E_1(K_v))=\frac{1}{|k_v|},$$ where $k_v$ is the residue field of $K_v$ and $E_1$ is the kernel of reduction. But I am totally stuck trying to prove it when $v(2)>0$. I'm certain it should hold in general. The thing that makes it easier when $v(2)=0$ is that you can easily find the value of $|2y+a_1x+a_3|_v$ based on the value of $v(x)$. And also it's easier to use Hensel's lemma. But when $v(2)>0$, that pesky factor of $2$ appearing next to the $y$ ruins everything, and Hensel's lemma becomes much harder to use. Any ideas how to prove this?

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