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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 $\mu\equiv 0$?

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    $\begingroup$ @Steve: You need to give more details, otherwise it is not very respectful of people who would make the effort of answering your question. Is mu a real valued positive measure? if so, is the sigma algebra that of Borel sets etc. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Apr 3, 2013 at 16:14

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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.

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