Q: Convergence of an event in Probability to zero Let $A_1,...A_N$ be sets with $A_i \subset \mathbb{R}$ for all $ 1 \leq i \leq N$, $ A = \bigcup_{i=1}^NA_i$ and let $ \mathbb{P}$ be a probability measure on an appropriate probability space. The author now wants to prove the following lemma.
For every $ \epsilon >0 $ there exists a positive constant $ C_{\epsilon} < \infty$ such that
$$ \mathbb{P}(A) \leq C_{\epsilon} \cdot N^{- \epsilon} $$
Now he argues as follows: 
$ \mathbb{P}(A) = \mathbb{P}({\bigcup_{i=1}^N}A_i) \leq \sum_{i=1}^N \mathbb{P}(A_i) $ holds due to the sigma subadditivity of $\mathbb{P}$. Further as it is assumed that the $A_i´s$ are indistinguishable we have $ \sum_{i=1}^N \mathbb{P}(A_i) = N \cdot \mathbb{P}(A_1).$ So all in all:
$$ \mathbb{P}(A)\leq N \cdot \mathbb{P}(A_1) $$
Until here everything is fine but he now claims that it is sufficient to prove that for every $\epsilon >0$ there exists a $C_{\epsilon}$ such that $ \mathbb{P}(A_1)\leq C_{\epsilon} N^{-\epsilon} $
But in my opinion this would yield: 
$$ P(A) \leq N \mathbb{P}(A_1) \leq C_{\epsilon}\cdot N^{-\epsilon +1} $$
which could tend to infinity if $ \epsilon < 1$. But it is the goal of this lemma to prove convergence to zero. My first thought was that he could just define a new epsilon $\tilde{\epsilon}=-\epsilon +1$ and the lemma would be fulfilled. But this doesnt feel true. Any suggestions where my error in reasoning is?
