Why does the following sequence of functions converge pointwise? Why does
$\lim_{n\to ∞} \frac{1}{\sqrt x} \mathbb{1}_{[2^{-n-1},2^{-n}]}$
converge pointwise to $0$? I thought that the function isn't defined for $0$ when ${n\to ∞}$, because the set of the indicator function is equal to $0$ then. I would greatly appreciate it, if someone could explain this to me. Thank you very much!
 A: There are two issues here : how the functions are defined and whether they converge pointwise.
Definition of the function
As @TonyK points out, the expression given for $f_n(x) = \frac{1}{x} 1_{[2^{-n-1},2^{-n}]}(x)$ is ill-defined if $x\leq 0$. However, because there is the indicator function whose support is included in $(0,+\infty)$, it is understood implicitly that $f_n(x) =0$ when $x\leq 0$. It is a convention.
Is it rigorous ? Strictly speaking, no, but this kind of implicit convention are everywhere in math. IMHO, what matters is that the reader with the appropriate math background is able to make it fully rigorous himself.
Pointwise convergence
If $x$ is real, then $f_n(x)$ is non zero iff $2^{-n-1} \leq x \leq 2^{-n}$ ie iff $-\log_2(x)-1 \leq n \leq -\log_2(x) $. Therefore, for $x$ fixed and $n$ large enough, we have $f_n(x)= 0$. In particular:
$$\forall x\in \mathbb R, \lim_{n\to +\infty} f_n(x) = 0$$
This is what is meant by "$(f_n)$ converges pointwise to the zero function".
A: You are absolutely right $-$ $1/\sqrt x$ is undefined for $x\le 0$, so $\frac{1}{\sqrt x} \mathbb{1}_{[2^{-n-1},2^{-n}]}$ is undefined also.
It is true that the meaning is clear, but that doesn't make it correct. However, you have to go to some lengths to make it correct. For instance:

Define $f_n(x)=0$ if $x\le 0$, and $f_n(x)=\frac{1}{\sqrt x}
> \mathbb{1}_{[2^{-n-1},2^{-n}]}$ otherwise. Then $f_n(x)$ converges
pointwise to $0$ as $n\to\infty$.

So it is easy to see why the author didn't bother. But it is still incorrect!
A: Here are two answers to two different questions since I started responding before the question was edited
Let $f(x)=\frac{1}{\sqrt{x}}$. For each $n \in \mathbb{N}$, we define $f_n:=f \chi_{[2^{n-1}, 2^n]}$.
Let $\varepsilon>0$ be given. Take $\gamma=\min \{\varepsilon, 1 \} $.
By the Archimedean Principle, there is $N \in \mathbb{N}$ so that $0<-2\log\left( \gamma\right)+\log(2)<N\log(2)$. This means we have
$$0<\frac{1}{\gamma^2}<2^{N-1} \iff \frac{1}{\sqrt{2^{N-1}}}<\gamma.$$
So if $n \geq N$, then we have $f_n(x)=0$ on $\left(0,2^{N-1}\right)$ and
$$0 \leq f_n(x)\leq f(x)<\gamma \leq \varepsilon \;\text{ on }\;  \left[2^{N-1}, \infty\right).$$
This shows $f_n \to 0$ uniformly on $(0, \infty)$.

Let $g(x)=\frac{1}{\sqrt{x}}$. For each $n \in \mathbb{N}$ we define $g_n:=g \chi_{[2^{-n-1}, 2^{-n}]}$. Let $x \in (0,\infty)$.
Case 1:  Suppose $x > 2^{-1}$. Then $g_n(x)=0$ for all $n \in \mathbb{N}$. And so $\lim_{n \to \infty} g_n(x)=0$.
Case 2: Suppose $0<x\leq 2^{-1}$. By the Archimedean Principle, there is $N \in \mathbb{N}$ so that $0<-\log(x)<N\log(2)$. This means we have
$$2^{-n}<x \text{ whenever } n \geq N.$$
In other words, $x \notin [2^{-n-1}, 2^{-n}]$ whenever $n \geq N$. So $\lim_{n \to \infty} g_n(x)=0$ as $g_n(x)=0$ for all $n \geq N$.
This shows $g_n \to 0$ pointwise on $(0, \infty)$.
