I am looking at p-adic distributions, and in this case p-adic measures. To say that $\mu$ is a distribution means that the arguments of $\mu$ are compact open subsets of $\mathbb{Z}\_p$, $\mu$ is finitely additive, and the values $\mu$ takes are in $\mathbb{C}\_p$. To say that $\mu$ is a measure means that $\mu$ is a distribution and the values $\mu$ takes are bounded. Let $\mu$ be a measure. 


Suppose that $lim\_{n\to\infty}\mu(a+p^n\mathbb{Z}\_p)=0$ for all $a\in\mathbb{Z}\_p$. Does this imply that $\mu\equiv 0$?



Trivial p-adic measures
Yes. Suppose that $\mu$ of some set is nonzero. Then $\mu$ of some interval of the form $a+p^n \mathbb Z\_p$ is zero. Then it must be nonzero modulo $p^k$ for some $k$. By finite additivity:


$\sum\_{t=0}^{p-1} \mu(a+p^{n}t+ p^{n+1} \mathbb Z\_p) = \mu(a+p^n \mathbb Z\_p) \not\equiv 0$ modulo $p^k$.


So for some $t$, $\mu(a+p^{n}t+ p^{n+1} \mathbb Z\_p) \not \equiv 0$ modulo $p^k$.


Let $a\_1 = a+p^n t$ then repeat this process using $a\_1$ and $n+1$ to produce $a\_2$ , and so on. Then $a\_i$ is a sequence that converges $p$-adically, so


$\mu( \lim\_{i\to \infty} a\_i + p^{n+j} \mathbb Z\_p) = \mu( a\_j + p^{n+j} \mathbb Z\_p) \not \equiv 0$ modulo $p^k$.


and a sequence which is never $0$ modulo $p^k$ cannot converge to $0$ $p$-adically.


