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If $\frac{1}{n}\sum_{i=1}^{n}X_i \to \mu$ almost surely, for $E(X_i) = \mu$, does $\frac{1}{n}\sum_{i=1}^{n}X_{i+n} \to \mu$ almost surely as well?

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    If we write $\bar{X}_n = \frac{1}{n}(X_1 + \cdots + X_n)$, can you check that $\frac{1}{n}(X_{n+1} + \cdots + X_{2n}) = 2\bar{X}_{2n} - \bar{X}_n$? What can you conclude from this? – 2017-01-09
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    Got it thanks!!, so basically the result asymptotically is $2\mu - \mu = \mu$ and so since almost sure convergence implies probability, Slutsky yields the answer? – 2017-01-09
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    No need for Slutsky, obviously. – 2017-01-09

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This seems to just be a true analysis fact. Let $S_n = x_1+\dots+x_n$ and $S_n' = x_{n+1}+\dots+x_{2n}$. Then, for large $n$, $\frac{S_{2n}}{2n}$ is close to $\mu$. But $\frac{S_{2n}}{2n} = \frac{S_n}{2n}+\frac{S_n'}{2n}$. But since $\frac{S_n}{2n}$ is close to $\frac{\mu}{2}$, we see $\frac{S_n'}{2n}$ is close to $\frac{\mu}{2}$, which is what you want. I'll let you fill in the $\epsilon$'s.

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another way would be to see that S_n/n converges to u and S_2n/n converges to 2u almost surely so (S_2n-S_n)/n converges to u almost surely