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I am trying to find: $$\lim_{n\to\infty}\inf \ f_n$$

where $f_n = \mathbb{1}_{[n,n+1]}$ is the indicator function that takes value $1$ in the set $[n,n+1]$ and $0$ elsewhere. It seems intuitively obvious to me that the liminf should be zero. However, when I picture the graph of $f_n$, it seems that for $n \geq 1$, it is just a horizontal line at $1$ that goes off to infinity. Can anyone tell me what I am missing here? Thanks!

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  • $\begingroup$ The graph of $f_n$ does not go off to infinity. It hovers above an interval of lenght $1$, and then it is zero everywhere else. $\endgroup$ – RustyStatistician Nov 25 '15 at 6:37
  • $\begingroup$ Please use \liminf_{n\to\infty} or, even better, \liminf\limits_{n\to\infty}, but not \lim_{n\to\infty}\inf. $\endgroup$ – Did Nov 25 '15 at 8:58
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Hint: For any fixed $x$, there exists $N \in \mathbb{N}$ such that $N > x$ so that $I_{[n, n + 1]}(x) = 0$ for all $n > N$. So in fact the statement can be strengthened as $$\lim_{n \to \infty} f_n(x) = 0,$$ which of course implies that $\liminf_{n \to \infty} f_n(x) = 0$.


A more general way to see it. Let $A_n$ a sequence of subsets of $X$. Define $$\liminf_{n\to\infty} A_n:=\bigcup_{n\in\mathbb N}\bigcap_{m\ge n} A_m\\\limsup_{n\to\infty} A_n:=\bigcap_{n\in\mathbb N}\bigcup_{m\le n} A_m$$

It holds the following caracterization:

$$\liminf_{n\to\infty}A_n=\{x\in X\,:\,x\in A_n\text{ for all }n\text{ sufficiently large}\}\\ \limsup_{n\to\infty}A_n=\{x\in X\,:\,x\in A_n\text{ for infinitely many values of }n\}$$

And, not so surprisingly,

$$1_{\liminf_{n\to\infty} A_n}=\liminf_{n\to\infty}\ 1_{A_n}\\ 1_{\limsup_{n\to\infty} A_n}=\limsup_{n\to\infty}\ 1_{A_n}$$

This should help you solve it, and maybe provide you with some insight on the problem.

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