# Extracting a subsequence Cesàro converging to the limsup of the Cesàro sums

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 $$N \to \infty$$?

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 $$$$X_i = \chi_{(0,1/2)},\quad m_k \leq i < m_{k+1}$$$$ if $$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*}$$
• Hm why is $Y = 1$ on $(\frac{1}{2}, 1)$? It seems I get $Y = 1$ on $(0, \frac{1}{2})$ and $0$ otherwise. Commented May 4 at 12:42
• 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. Commented May 4 at 12:49
• If you need more details I can write down all the $\epsilon, \delta$ 's Commented May 4 at 12:50