Let $f:\mathbb R\longrightarrow \mathbb R$ a funcion. I try to prove that $$\text{Graph}(f)=\bigcap_{n\in\mathbb N}\bigcup_{i\in\mathbb Z}f^{-1}\left(\left[\frac{i}{2^n},\frac{i+1}{2^n}\right)\right)\times \left[\frac{i}{2^n},\frac{i+1}{2^n}\right),$$
but I have difficulty to prove the $\supset$ inclusion. So let $$(x,y)\in \bigcap_{n\in\mathbb N}\bigcup_{i\in\mathbb Z}f^{-1}\left(\left[\frac{i}{2^n},\frac{i+1}{2^n}\right)\right)\times \left[\frac{i}{2^n},\frac{i+1}{2^n}\right),$$ then, for all $n$, there is $i_n$ s.t. $(x,y)\in f^{-1}\left(\left[\frac{i_n}{2^n},\frac{i_n+1}{2^n}\right)\right)\times \left[\frac{i_n}{2^n},\frac{i_n+1}{2^n}\right)$. In particular, $f(x)\in \left[\frac{i_n}{2^n},\frac{i_n+1}{2^n}\right)$ for all $n$, i.e. $f(x)=\bigcap_{n\in\mathbb N}\left[\frac{i_n}{2^n},\frac{i_n+1}{2^n}\right)$. Now, I would like to prove that $$\bigcap_{n\in\mathbb N}\left[\frac{i_n}{2^n},\frac{i_n+1}{2^n}\right)=\{*\},$$ is a singleton, but I can't.
In one hand, $f(x)\in \bigcap_{n\in\mathbb N}I_n$. In the other hand, if $y\in \bigcap_{n\in\mathbb N}I_n$, then $|y-f(x)|\leq \frac{1}{2^n}\to 0$ and thus $f(x)=y$. The claim follow.