Prove that the graph of a measurable function is measurable.

Hint:

1. Show that the mapping $$(x,y) \mapsto T(x,y) := f(x)-y$$ is measurable.
2. Conclude that $\Gamma(f) = T^{-1}(\{0\})$ is measurable.

Hint

One can remark that $$\Gamma(f)=\bigcap_{n\in\mathbb N}\bigcup_{m\in\mathbb Z} f^{-1}\left(\left[\frac{m}{2^n},\frac{m+1}{2^{n}}\right]\right)\times \left[\frac{m}{2^n},\frac{m+1}{2^{n}}\right].$$