Let $\{A_n\}_{n=1}^{\infty}$ be a sequence of sets, and define the indicator function $\mathbf{1}_{A_n}(x)$ to equal $1$ if $x \in A_n$ and $0$ otherwise. According to set-theoretic limits we have that:
$$
\liminf_{n\to\infty} A_n = \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k,
$$
or that $x \in \liminf_n A_n$ if and only if there is some $n$ such that $x$ is in all $A_k$ for $k\ge n$. Now the equivalent definition using indicator functions is:
$$
\liminf_{n \to \infty}A_n = \{x \mid \liminf_{n\to\infty}\mathbf{1}_{A_n}(x) = 1 \}. \qquad(\star)
$$
My question is: why is the condition on the right hand side of $(\star)$ written as $\liminf_{n\to\infty}\mathbf{1}_{A_n}(x)=1$, as opposed to having $\lim_{n\to\infty}\mathbf{1}_{A_n}(x)=1$? It seems to me that if $\liminf_{n \to\infty}\mathbf{1}_{A_n}(x) =1$, it must be the case that $\lim_{n\to\infty}\mathbf{1}_{A_n}(x)=1$ since the sequence in question is comprised of zeros and ones.
If not, can someone demonstrate a binary sequence such that $\liminf_{n\to\infty}b_n = 1$ but $\lim_{n\to\infty}b_n \ne 1$?
Best Answer
You're right: it could be written as a limit, for the reasons you stated.
However, using the limit-inferior has the slight benefit that the limit-inferior exists by virtue of the sequence being bounded (which is always true). If you use limits instead, then you may find $x$ such that $\mathbf{1}_{A_n}(x)$ has no limit, and some may argue that even writing $\lim_{n\to \infty} \mathbf{1}_{A_n} (x)$ is less than correct.
There's also potentially a pedagogical benefit of connecting the limit-inferior of sets to the limit-inferior of functions.