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In these lecture notes, on page 33, it says:

$$\exists N \in \mathbb{N}: \frac{S_N}{N} > \epsilon \iff \bar{S}(x) > \epsilon.$$ At the bottom of page 32, we get the following definition: $$\bar{S}(x):= \limsup_{n->\infty} \frac{S_N}{N}.$$

I can't see why that $\iff$ holds. To work out the limit superior, we disregard the first $n$ terms, for $n\rightarrow \infty$. In particular, it means disregarding the first $N$ terms. So how do we know that, among the remaining terms, the there will still be an $N$ such that $\frac{S_N}{N}>\epsilon$?

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    Look at the middle and rightmost terms in the equation right after (4.5) - it says $\frac{S_N}{N}$ – 2017-01-06
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    Ok, saw it. But it states $S^\epsilon/N=0$ if $\bar{S}<\epsilon$. Hence we have $\Rightarrow$ in the equation there (yeah, the arrow is at the wrong spot, but that shouldn't matter). So I can clearly see the $\iff$ if you just consider $S^\epsilon...$ and if that holds, also this single existence holds. The only arrow which seems to be off now is the one you stated. Maybe we know $S_N/N$ is strictly increasing or we don't need this implication and should ignore it as a typo. – 2017-01-06

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