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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!

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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 $x$.

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  • $\begingroup$ "passing any given x". I'm sorry, I didn't get this... $\endgroup$ – Nenne Jan 18 '17 at 15:15
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    $\begingroup$ I'm thinking of animating these intervals by letting $n\to\infty$ and watching the intervals step further and further to the right. The intervals are $[1,2],[2,3],[3,4],\ldots$ and as you move out into the tail of this list of intervals the location they cover moves further and further to the right (because $1<2<3<\cdots$). You should try to visualize what is happening; geometric reasoning is usually the best way to start such a proof, and you can "firm it up" with more rigor once you have visualized what is happening. A picture isn't a proof, but it is a starting point for a proof. $\endgroup$ – MPW Jan 18 '17 at 15:23
  • $\begingroup$ Your reply is clear, but I can't imagine why x is not included in these sets while moving out to infinity... Shouldn't x be infinity itself? $\endgroup$ – Nenne Jan 18 '17 at 15:29
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    $\begingroup$ You're taking a limit pointwise. So when you write $f=\lim_{n\to\infty}f_n$, $f$ is the function that assigns to a SPECIFIC FIXED $x$ the value $\lim_{n\to\infty}f_n(x)$; it's the limit of the sequence $f_1(x),f_2(x),f_3(x),\ldots$. Note that $x$ isn't changing, you fix $x$ and let $n$ run to infinity. Here's a specific example: $f(7.1) = \lim_{n\to\infty}f_n(7.1)$. In your case, your "$f_n$"s are the functions $1_{[n,n+1]}$. You haven't given the limiting function a name. But if you call it $f$, then $f(7.1)=\lim_{n\to\infty}1_{[n,n+1]}(7.1)=0$. $\endgroup$ – MPW Jan 18 '17 at 16:57
  • $\begingroup$ Also, in speaking of these things, you should be very careful to distinguish between "$f$" and "$f(x)$". People often talk about "the function $f(x)$". This is sloppy and imprecise. The name of the function is not "$f(x)$", it is "$f$". The symbol "$f(x)$" is the value of the function $f$ at the point $x$; it is a point in the range of $f$. The function isn't a value, it is the ensemble of the domain, the range, and the rule of assignment. You wouldn't speak of "the function $f(7.1)$", so don't speak of "the function $f(x)$" -- it's just as wrong. This may be confusing you. $\endgroup$ – MPW Jan 18 '17 at 17:02
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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 theorem that goes:

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

Thus

$$\to \lim_n 1_{[n,n+1]} = 1_{\lim [n,n+1] = \emptyset} = 0$$

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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.

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