[Math] Prove that the graph of a measurable function is measurable.

Question

How can I prove that the graph of a measurable function is measurable. I recall that the graph of $f:\mathbb R\longrightarrow \mathbb R$ $$\Gamma(f)=\{(x,f(x))\mid x\in \mathbb R\}.$$

Attempt

1) if $f=1_{[a,b]}$ then $\Gamma(f)=[a,b]\times \{1\}\cup[a,b]^c\times \{0\}$, and thus $\Gamma(f)$ is measurable.

2) If $f$ is a step function, same thing.

3) If $f\geq 0$, then there are step function $f_n$ s.t. $f_n\nearrow f$. Now, I would like to have $\Gamma(f)=\bigcup_{n\in\mathbb N}\Gamma(f_n)$, but unfortunately it doesn't look to be the case. Any idea ?

4) Same for $f$ measurable. I have that $f=f^+-f^-$, but I don't think that $\Gamma(f)=\Gamma(f^+)\cup \Gamma(f^-)$. Any idea ?

Answer 1

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.