Let $f_{n}(x)=\chi_{[n,\infty)}(x) = \begin{cases} 1 & \text{if}\,x\in[n,\infty)\\ 0 & \text{if}\, x \notin [n,\infty) \end{cases}$.
According to my textbook, as $n \to \infty$, $f_{n}(x) \to 0$ pointwise.
Could someone please explain to me (and if you could provide me with a proof, that would be even better) why this is?
Probabilistically, I can kind of see why: as $n$ gets larger, the intervals $[n,\infty)$ "shrink" (if you can really say such a thing when one of the endpoints is infinite), so there is less likelihood that a given point $x$ will fall into that interval, thus forcing the value of $\chi_{[n,\infty)}(x)$ to be $0$ the further out in the sequence we go.
However, this is a straight-up analysis course I'm taking, not a probability course, and I would like to understand this better in a more "analysisy" way.
Pointwise convergence $f_n \to 0$ is per definitionem the following: $f_n(x) \to 0$ for all $x \in \mathbf R$.
So, let $x \in \mathbf R$. Choose $N \in \mathbf N$ with $N > x$ (such $N$ exist), then for all $n \ge N$ we have that $x \not\in [n,\infty)$, hence $f_n(x) = 0$. So $f_n(x) = 0$ for all $n \ge N$. This implies $f_n(x) \to 0$.