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Let $X_n$ be a sequence of uniformly bounded random variables — that is, there exists some $K > 0$ such that $|X_n| \leq K$ almost surely for all $n \in \mathbb N$.

Write $\bar X_N := \frac{1}{N} \sum_{n = 1}^N X_n$, and $Y := \limsup_{N \to \infty} \bar X_N$.

Question: Does there exist some subsequence $X_{n_k}$ such that $\frac{1}{N} \sum_{k=1}^{N} X_{n_k}$ converges to $Y$ almost surely as $N \to \infty$?

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No I do not think this is always possible. Take for example a subsequence of $\mathbb{N}$, call it $\{ m_k \}$ and consider the random variables \begin{equation} X_i = \chi_{(0,1/2)},\quad m_k \leq i < m_{k+1} \end{equation} if $k$ is even and let $X_i = \chi_{(1/2,1)}$ otherwise. Then if $m_k$ is growing sufficiently fast we have $Y \equiv 1$ (with your notation). So suppose that for some subsequence $n_k$ you have that $\frac{1}{N}\sum_{n=1}^{N} X_{n_k} =:Z_k $ converges almost everywhere to $1$, by Fatou's lemma you would have \begin{equation*} 1 = \int_0^1 \liminf Z_k \leq \liminf_k \int_0^1 Z_k = \frac{1}{2}. \end{equation*}

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  • $\begingroup$ Hm why is $Y = 1$ on $(\frac{1}{2}, 1)$? It seems I get $Y = 1$ on $(0, \frac{1}{2})$ and $0$ otherwise. $\endgroup$
    – Nate River
    Commented May 4 at 12:42
  • $\begingroup$ Its a bit messy to write down, but the idea is that if the interval $ [m_k, m_{k+1}) $ is much longer than the interval $[1,m_k)$, when you average the only part of the sequence that counts is in the r.v. having index in the longer interval. $\endgroup$ Commented May 4 at 12:49
  • $\begingroup$ If you need more details I can write down all the $\epsilon, \delta$ 's $\endgroup$ Commented May 4 at 12:50
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    $\begingroup$ Oh never mind. I misread the construction. What you suggested works. Nice construction! $\endgroup$
    – Nate River
    Commented May 4 at 12:55

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