$\liminf_\limits{n\to\infty}1_{A_n}(x)=1$ $\implies$ $\lim_\limits{n\to\infty}1_{A_n}(x)=1$?

Question

Source:

Partial proof from textbook:

I've omitted the case where $x\in A^c$ as it's not relevant. I've also highlighted the part I'm having trouble with in blue.

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Here is my attempt at explaining the highlighted in blue section:

Since $1_{A_n}(x)=1$ for all but finitely many $n\in\mathbb{N}$, that is also the same as saying that there is a $N\in\mathbb{N}$ such that for all $N'\geq N, 1_{A_{N'}}=1$. This also means that for all $N'\geq N$, I can trivially find any $\epsilon>0$ such that $|1_{A_{N'}}(x)-1|<\epsilon$ which is the definition of $\lim_\limits{n\to\infty}1_{A_n}(x)$.

What I'm having trouble with is that so far in the text, I have been told that to show a limit exists, we have to show that the $\limsup$ and $\liminf$ are equal in value. It doesn't seem like we have done it here...

Answer 1

First, to show a limit of a sequence $(a_n)$ exists, multiple methods are possible. One, is to show that $\limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n$. Another is to show that $(a_n)$ converges to some limit $L$ in the "for every $\epsilon>0$, there exists an $N$..." sense. This is the approach being used in the textbook. Since $x \in A = \liminf_{n\to\infty} A_n = \limsup_{n\to\infty} A_n$, $x$ is in $A_n$ for all but finitely many $n$, so by the definition of the limit, $\lim_{n\to\infty} 1_{A_n}(x) = 1$. There are many more ways of showing a sequence of real numbers converges (showing it is a Cauchy sequence, showing it has finitely many upcrossing for every pair of rational numbers,...).

To answer the question in your title more directly, a classical way to show $\limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n$--and thus the existence of a limit--is to use the fact that we always know that $ \liminf_{n\to\infty} a_n \le \limsup_{n\to\infty} a_n$. Thus, if we can show that for some real number $L$ that $L\le \liminf_{n\to\infty} a_n \le \limsup_{n\to\infty} a_n\le L$, then it must follow that $\liminf_{n\to\infty} a_n = \limsup_{n\to\infty} a_n= L$, and the sequence converges with limit $L$. Since the indicator function is at most $1$, this exact argument works in your case with $L = 1$.