Cezaro mean convergence implies regular convergence outside of a rare set I have a bounded sequence of non-negative real numbers ($0 \leq a_i < C$), which Cezaro mean converges to $0$:
$$
  \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n - 1} a_i = 0
$$
How do I proof that $\exists J \subset \mathbb{N}$ such that
$$
  \lim_{i \to \infty, \ i \not\in J} a_i = 0
$$
and $J$ is rare, namely
$$
  \lim_{n \to \infty} \frac{|\{ k \ | \ k \in J, 0 \leq k < n \}|}{n} = 0
$$
My attempts to prove by contradiction turned out to be non rigorous. I suspect something from measure theory should be involved, but have no solid ideas.
 A: Similar question was answered there.
It's possible to construct $J$, which has required properties from the original sequence $a_i$. Firstly:
$$
  I^k = \{i \ \big| \ i \in \mathbb{N}, \ a_i \geq \frac{1}{k} \}, \quad I_n^k  = I^k \ \cap \ [1;n) = \{i \ \big| \ i \in \mathbb{N}, \ i < n, \ a_i \geq \frac{1}{k} \}
$$
For fixed $k$:
$$
    \frac{|I^k_n|}{n} = \frac{k}{n} \cdot
    \sum_{i=1}^{n - 1} \left\{\begin{aligned}
        0, \quad \text{if} \; a_i < \frac{1}{k} \\
        \frac{1}{k}, \quad \text{if} \; a_i \geq \frac{1}{k}
    \end{aligned}\right\}
\leq k \cdot \frac{1}{n} \sum_{i=1}^{n - 1} a_i \underset{n \to \infty}{ \longrightarrow} 0
$$
It implies that $\frac{|I^k_n|}{n} \to 0$, ie for any $\varepsilon > 0$ exist $n_\varepsilon$, such that $\forall n \geq n_\varepsilon$ satisfies $\frac{|I^k_n|}{n}<\varepsilon$. For $k = 1, 2, ...$ choose $\varepsilon = \frac{1}{k}$ and denote $n_{\varepsilon}$ as $n_k$, $n_0 = 1$. $n \geq n_k \ \Rightarrow \ \frac{ | I^{k}_n  | }{n} < \frac{1}{k}$. $n_k$ may be choosen as nondecreasing. It is not bounded (or $a_i$ already has ordinary convergence). Then $J = \bigcup_{k = 0}^\infty \Big( I^{k+1} \ \cap \ [n_k; n_{k+1}) \Big)$ has the required properties. $a_i$ approaches zero outside of the set, because for any $\varepsilon > 0$ exists $m = n_{ \lceil1/\varepsilon\rceil}$ and all of the sequence members with indicies $i \geq m$ which are greater than $\varepsilon$ already lie in $J$.
$$
    J_n = J \cap [1; n) = \bigcup_{i = 0}^\infty \Big( I^{i} \ \cap \ [1;n) \ \cap \ [n_i; n_{i+1} \Big) =
    \bigcup_{i = 0}^{k(n)} \Big( I^{i}_n \cap [n_i; n_{i+1}) \Big) \subset I^{k(n)}_n
$$
$$
    \frac{|J_n|}{n} \leq \frac{|I_n^{k(n)}|}{n} \leq \frac{1}{k(n)}
$$
$k(n)$ increases to infinity with $n \to \infty$, then $\frac{|J_n|}{n} \to 0$.
