# p-adic analogue of the Strong Law of Large Numbers

Is there a $p$-adic analogue of the Strong Law of Large Numbers? In particular, suppose that $f_i: \mathbb{Z}_p \longrightarrow \mathbb{Q}_p$ for $i = 1,2,\ldots$ is an sequence of random variables that is i.i.d. with respect to the $\mathbb{Q}_p$-valued normalized Haar measure $\mu$ on $\mathbb{Z}_p$. Must one have $$\int_{\mathbb{Z}_p} f_1 d \mu = \lim_{n \rightarrow \infty} \frac{f_1(x) + f_2(x) + \cdots + f_n(x)}{n}$$ for almost all $x \in \mathbb{Z}_p$? If not, is there a partial result in this direction?

• Jesse, be more precise about what you mean by a $p$-adic random variable and integration wrt the $p$-adic-valued Haar "measure." Such integration is not well-defined in general, as a limit of Riemann sums can depend on the choice of sample points. See p. 39 of Koblitz's $p$-adics book (2nd ed.) for a problem with the simple example $f(x) = x$. If you want to define this integral using sample points $0,1,\ldots,p^n-1$ at "level $p^n$" then the integral exists for a restricted class of continuous functions. See Robert's $p$-adics book for "Volkenborn integral." – KConrad Apr 23 '15 at 9:17
• I'm using Chapter II Sections 3 and 5 of Koblitz as a reference. – Jesse Elliott Apr 23 '15 at 9:26
• For i.i.d. of course you have to assume the integrals exist. Is continuity necessary? I don't know even if it's sufficient. – Jesse Elliott Apr 23 '15 at 9:36
• The $p$-adic Haar measure does not work well for the integration developed in his book, since it is not a bounded distribution. I've never heard of a robust $p$-adic version of the LLN. – KConrad Apr 23 '15 at 9:46
• Ok. The functions $f_i$ I was trying to apply it to are at least locally constant, so the question makes sense in that context at least. – Jesse Elliott Apr 23 '15 at 18:08

I think this can't work. If you look at the simplest possible case, where the $f_i$ take values 0 and 1 with probabilities $1-1/p$ and $1/p$, say, then you are effectively looking at the average of the digit sums of $x$.
To make sense (since your $f_i(x)$ generally take values in $\mathbb Q_p$), you should look for a limit in $\mathbb Q_p$. But now, the numerator is an integer close to $n/p$. By standard random walk techniques, you know that the ratio will be exactly $1/p$ infinitely often (with $p$-valuation $-1$).
On the other hand, for $n$ not a multiple of $p$, the $p$-valuation of the ratio will be (at least) $0$. Hence there is no pointwise convergence.