Find a countable set with the given property. Given any extended-valued $f$ on $(-\infty,+\infty)$, prove that there exists a countable set $D$ with the following property. For each $t\in\mathbb R$, there exist $t_n\in D$, $t_n\to t$ such that $f(t)=\lim_{n\to\infty}{f(t_n)}$. The assertion remains true if "$t_n\to t$" is replaced by "$t_n\downarrow t$" or "$t_n\uparrow t$".
This is an exercise in Kai Lai Chung's A course in Probability Theory, and a hint is given:"Consider the graph $(t,f(t))$ and introduce a metric."
I think it is similar to the fact that $\mathbb R^n$ is separable, but what metric should we define on the graph?
 A: You don’t really need a metric on $G$; you just need to know that if $\Bbb R^*$ is the extended reals with their usual topology, then $\Bbb R\times\Bbb R^*$ is second countable and hence hereditarily separable. If you want to make a metric explicit, observe that $\Bbb R^*$ is homeomorphic to a closed interval; for instance, it’s homeomorphic to $[-1,1]$ via the map
$$h:\Bbb R^*\to[-1,1]:x\mapsto\begin{cases}
-1,&\text{if }x=-\infty\\
\frac2{\pi}\tan^{-1}x,&\text{if }x\in\Bbb R\\
1,&\text{if }x=\infty\;.
\end{cases}$$
Let $G\subseteq\Bbb R\times\Bbb R^*$ be the graph of $f$; i.e., $G=\{\langle t,f(t)\rangle:t\in\Bbb R\}$. $G$ is separable, so there is a countable $E\subseteq G$ such that $\operatorname{cl}E\supseteq G$. Let $D=\{t\in\Bbb R:\langle t,f(t)\rangle\in E\}$.
Let $t\in\Bbb R$ be arbitrary. Then there is a sequence $\langle e_n:n\in\Bbb N\rangle$ in $E$ converging to $\langle t,f(t)\rangle$ in $G$. For $n\in\Bbb N$ let $t_n\in D$ be such that $e_n=\langle t_n,f(t_n)\rangle$; then $f(t)=\lim_nf(t_n)$.
