Limit of an indicator function? I'm having some problem in the computation of limits of indicator functions. 
Why the following limit takes values $0$?
$$\lim_{n\to\infty}1_{[n,n+1]}$$
Which are the steps I should follow to compute every kind of limit of such functions? I started thinking about applying the definition of limit as the x-values go to infinity, but at some point I get really confused because of the set of the indicator function... Does it work like the domain for other functions?
How does this work for limsup and liminf of an indicator function?
Thank you! 
 A: Answering your first question: For a given $x$, there are at most two values of $n$ for which $1_{[n,n+1]}$ is nonzero. As $n\to\infty$, the interval which supports $1_{[n,n+1]}$ will "fly off" to the right, passing any given $x$. So in the tail of the sequence, $1_{[n,n+1]}(x)$ is eventually zero for any give $x$.
A: Our conjecture is that it approaches zero. Why? The thing is $1_{[n,n+1]} = 0$ for all but only a few $n$'s. Consider the following:
$$1_{[3,3+1]}(5) = 0$$
$$1_{[4,4+1]}(5) = 1$$
$$1_{[5,5+1]}(5) = 1$$
$$1_{[6,6+1]}(5) = 0$$
$$1_{[7,7+1]}(5) = 0$$
$$\vdots$$
$$1_{[n,n+1]}(5) = 0 \ \text{unless} \ n=4,5$$
Also,
$$1_{[3,3+1]}(5.5) = 0$$
$$1_{[4,4+1]}(5.5) = 0$$
$$1_{[5,5+1]}(5.5) = 1$$
$$1_{[6,6+1]}(5.5) = 0$$
$$1_{[7,7+1]}(5.5) = 0$$
$$\vdots$$
$$1_{[5,5+1]}(5.5) = 0 \ \text{unless} \ n=5$$
Thus, we must (try to) show that for any $x$,
$$\forall \varepsilon > 0, \exists N \in \mathbb N : |1_{[n,n+1]}-0| < \varepsilon \leftarrow n \ge N$$
For $x<1$, $1_{[n,n+1]} = 0$, so we can pick any $N$.
For $x\ge1$, $1_{[n,n+1]} = 0$ for $n \ge N=\text{ceiling}(x)$.
Actually the following would be true as well:
$$\forall \varepsilon > 0, |1_{[n,n+1]}-0| < \varepsilon \ \forall \ \text{but finitely many} \ n \ge 1$$

This may be advanced but since you know liminf and limsup of numbers, I guess you could understand liminf and limsup of sets.
Observe that
$$\limsup_n [n,n+1] := \bigcap_{m=1}^{\infty} \bigcup_{n=m}^{\infty} [n,n+1] = \bigcap_{m=1}^{\infty} [m,m+1] \cup [m+1,m+2] \cup \cdots$$
$$ = \bigcap_{m=1}^{\infty} [m, \infty)$$
$$ = [1,\infty) \cap [2,\infty) \cap [3,\infty) \cap \cdots = [2,\infty) \cap [3,\infty) \cap \cdots = \emptyset$$
Now $\liminf_n [n,n+1] \subseteq \limsup_n [n,n+1]$ where
$$\liminf_n [n,n+1] := \bigcup_{m=1}^{\infty} \bigcap_{n=m}^{\infty} [n,n+1]$$
We could conclude $\liminf_n [n,n+1] = \emptyset$ or simply compute explicitly:
$$\liminf_n [n,n+1] = \bigcup_{m=1}^{\infty} [m,m+1] \cap [m+1, m+2] \cap [m+2, m+3] \cap \cdots$$
$$ = \bigcup_{m=1}^{\infty} \{m+1\} \cap [m+2, m+3] \cap \cdots$$
$$ = \bigcup_{m=1}^{\infty} \emptyset \cap \cdots$$
$$ = \bigcup_{m=1}^{\infty} \emptyset = \emptyset$$
Now if $$\liminf = \limsup$$, which is the case, define $$\lim := \liminf := \limsup$$
Now there's a folklore that goes:

$$\to \lim_n 1_{A_n} = 1_{\lim A_n}$$

Thus
$$\to \lim_n 1_{[n,n+1]} = 1_{\lim [n,n+1] = \emptyset} = 0$$
A: Take $x\in\mathbb{R}$ arbitrarily and choose $x<N\in\mathbb{N}$ such that $x\notin [N,N+1]$ then
$\chi_{[n,n+1]}(x)=0$ for all $n\geq N$
and thus $\chi_{[n,n+1]}(x)\rightarrow 0$ for $n\rightarrow \infty$ pointwise.
But $\sup_{x\in\mathbb{R}}|\chi_n(x)-0|=1$ for all $n\in\mathbb{N}$. So there is no convergence in the $\sup$-norm.
