limit of the indicator function Show that $\lim_{n \to \infty} \mathbb{1}_{\mathbb{R}^d \setminus B(0,n)} = 0$.
I don’t know exactly how to explain this mathematically. I would say that if n gets bigger than the ball $B(0,n)$ gets bigger and covers $\mathbb{R}^d$, thus near $\infty$ it equals $\mathbb{1}_\emptyset$ which is $0$. Is there a more mathematical way to express this?
 A: I think the comments above solves your question. However, I can write a proof for you if you are unfamiliar with calculus.
Proof.
For $\forall x \in \mathbb{R}^d$, if $n>\|x\|_2$, then
$$1_{\mathbb{R}^d\setminus  B(0,n)}(x)=0.$$
Hence,
$$\lim_{n\rightarrow\infty} 1_{\mathbb{R}^d\setminus  B(0,n)}(x)=0$$
A: The result is true and not true, depending on the definition of $\lim$. For functions, there are many different kinds of limits.
$\newcommand{\one}{\unicode{x1d7d9}}$

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*$\one_{B(0,n)^c}$ converges pointwise  to $0$: unpacking the defintion of $\one$,
$$ \one_{B(0,n)^c}(x) = \begin{cases} 1 & |x|>n\\ 0 &|x|<n\end{cases}$$
So for a fixed $x_0$, for all $n$ larger than $|x_0|$, we're in the bottom case, and therefore $\one_{B(0,n)^c}(x_0) = 0$. This shows the stronger statement that for each $x$, $\one_{B(0,n)^c}(x)$ is eventually equal to zero as $n\to\infty$.


*$\one_{B(0,n)^c}$ does not converge uniformly to $0$ (or to anything else). This is because it is not Cauchy in the space of bounded functions with the supremum norm $\|f\|_{\infty}:=\sup_{x\in\mathbb R^d} |f(x)|$: for $n<m$ there are always points $x$ such that $\one_{B(0,n)^c}(x) - \one_{B(0,m)^c} = 1$ (take any point such that $|x| \in (n,m)$.) However, a variant of the first argument shows that $\one_{B(0,n)^c}$ converges "locally uniformly" to $0$; considered as a function on any bounded $K\subset \mathbb R^d$, $\one_{B(0,n)^c}$ is eventually the zero function.
A: Choose any $\epsilon>0$, take any $x \in \mathbb{R}^d$ and let $m$ be the smallest positive integer such that $m>\|x\|$. Then for all $n>m$, $|1_{\mathbb{R}^d\setminus B(0,n)}(x)-0|=0<\epsilon$. Thus the pointwise limit of the said function is $0$ (notice that $m$ is dependent on $x$).
$\sup_{x \in \mathbb{R}^d}|1_{\mathbb{R}^d\setminus B(0,n)}(x)-0|=1 \nrightarrow 0$ as $n \rightarrow \infty$. So the convergence isn't uniform. [Here we use:  Let  $f_n,f:E \rightarrow \mathbb{R}$, $M_n=\sup_{x \in E}|f_n(x)-f(x)|$. Then $f_n \rightarrow f$ uniformly iff $M_n \rightarrow 0$ as $n \rightarrow \infty$].
Now if $\{1_{\mathbb{R}^d\setminus B(0,n)}\}_n$  uniformly converges to some other function then $\{1_{\mathbb{R}^d\setminus B(0,n)}(x)\}_n$ would be a Cauchy sequence for any $x \in \mathbb{R^d}$. Choose $\epsilon=0.5~~~\exists~N \in \mathbb{N}$ such that $m,n>N$, $x \in \mathbb{R}^d$ implies $|1_{\mathbb{R}^d\setminus B(0,n)}(x)-1_{\mathbb{R}^d\setminus B(0,m)}(x)|<0.5$. Let  $p,q>N$ and take an $x_0 \in \mathbb{R}^d$ such that $p<\|x_0\|<q$, then $|1_{\mathbb{R}^d\setminus B(0,p)}(x_0)-1_{\mathbb{R}^d\setminus B(0,q)}(x_0)|=1$, a contradiction. Thus $\{1_{\mathbb{R}^d\setminus B(0,n)}\}_n$ isn't uniformly convergent to any point.
