Show that the graph of a function is measurable. Let $f:\mathbb R\to\mathbb R$ a measurable functionI have to show that $\Gamma=\{(x,f(x))\mid x\in\mathbb R\}$ is measurable and that $m(\Gamma)=0$. (I work with Lebesgue measure).
My attempt
If $f=1_F$ where $f$ is an interval, we have that $\Gamma= I^c\times \{0\}\cup I\times \{1\}$ and thus it's measurable since it's union and product of measurable set. And we get $$m(\Gamma)=0\cdot m(I)+0\cdot m(I^c)=0.$$
If $f$ is a step function, i.e. $f=\sum_{i=1}^n a_i 1_{I_i}$ where $I_i$ are interval, we have that $$\Gamma=\left(\bigcap_{i=1}^n I_i^c\times \{0\}\right)\cup\left(\bigcup_{i=1}^n I_i\times \{a_i\}\right)$$ and thus $\Gamma$ is measurable since it's finite union, product and intersection of measurable set. And we get $$m(\Gamma)\leq 0\cdot m\left(\bigcap_{i=1}^n I_i\right)+0\sum_{i=1}^n m(I_i)=0$$
and thus $m(\Gamma)=0$.
Now if $f\geq 0$ is measurable, there is a sequence of step function $\varphi_k\nearrow f$. I would like to write $$\Gamma_n=\{(x,\varphi_n(x))\mid x\in\mathbb R\}$$
and say that $\Gamma_n\nearrow \Gamma$, but since $\Gamma_n\cap\Gamma_m$ for all $n\neq m$, I don't really see how to interpret this. Any idea how to write it rigorously ? Then we would have that $m(\Gamma_n)\to m(\Gamma)$, and since $m(\Gamma_n)=0$ for all $n$, we would have the result. 
 A: You've made a good start, and I can suggest a way to deal with your convergence issue. But first let me clear up a misapprehension.  It is not the case that a non-negative measurable function is always the increasing limit of a sequence of step functions -- think of $f=1_C$ where $C\subset [0,1]$ is the usual Cantor set. The only non-negative step function $g$ with $g\le 1_C$ is identically $0$ on $[0,1]$. What is true is a non-negative measurable function is the pointwise limit of an increasing sequence of non-negative simple functions, i.e.,  functions of the form $\sum_{k=1}^n a_k1_{E_k}$, with $a_k\ge 0$ and $E_k$ measurable.
Following the suggestion of @DanielFischer, let's assume that $0\le f\le M$. We'll get at the graph of $f$ indirectly, by first looking at the epigraph
$\hat\Gamma(f)=\{(x,y): y\ge f(x)\}$. Your argument shows that $\hat\Gamma(f)$ is measurable if $f$ is (i) the indicator of a measurable set, (ii) a non-negative simple function, and finally (iii) if $0\le f_n\uparrow f$ (pointwise) then clearly $\hat\Gamma(f)=\cap_n\hat\Gamma(f_n)$, so by simple function approximation the epigraph of any non-negative measurable $f$ is measurable.
By the same token, because $M-f$ is a non-negative measurable function, its epigraph $\hat\Gamma(M-f)$ is measurable. But the latter set is a continuous transformation (vertical reflection and vertical translation)  of the hypograph $\bar\Gamma(f)=\{(x,y): y\le f(x)\}$, which is therefore also measurable. The intersection $\hat\Gamma(f)\cap \bar\Gamma(f)$ is the graph of $f$.  
A little work remains, to remove the boundedness hypotheses, but that should be easy.
