Prove that, the graph of a measurable function is measurable and has Lebesgue measure zero. 
Prove that, the graph of a measurable function is measurable and has Lebesgue measure zero.

I saw some proofs in the internet, if the function is continuous. What is the relation between a continuous function and a measurable function, must they be equal $\mu-a.e.$, or is this approach useless. 
Must it have countable discontinuities ? then we could show it like in the continuous case or can you give any hints ?
Thanks in advance
 A: Assume that $f:\mathbb{R}^d \to \mathbb{R}$ is measurable. We prove that $\left \{ (x,f(x)):x\in \mathbb{R}^d \right \}$ is a Lebesgue null set.
It is enough to prove the claim when $f:B\to \mathbb{R}$ is measurable and $B\subset \mathbb{R}^d$ is a box (a product of intervals). Then, since $\mathbb{R}^{d}$ is a countable union of boxes, the claim will follow.
Let $G:=\left \{ (x,f(x)):x\in B \right \}$, and for $n\in \mathbb{Z}$, let $F_n:= G\bigcap (B\times[n,n+1])$. It is enough to prove that $F_n$ is null for every $n\in \mathbb{Z}$, so without loss of generality, we show it for $F:=F_0$.
Fix some $k\in \mathbb{N}$, and let $I_j:=[\frac{j}{k},\frac{j+1}{k}], j=0,\ldots,k-1$. Then,
$$m(G\cap (B \times I_j))\le m(f^{-1}(I_j)\times I_j)\le \frac{1}{k}m(f^{-1}(I_j)) $$
$$\implies  m(F)=m\left(\bigcup_j G\cap (B \times I_j) \right)=\sum_j m(G\cap (B \times I_j))\le\frac{1}{k} \sum_j m(f^{-1}(I_j)) $$
$$=\frac{1}{k}m\left(\bigcup_j f^{-1}(I_j)\right)=\frac{1}{k}m(f^{-1}([0,1]))\le\frac{1}{k}m(B). $$
Since $k$ was arbitrary we are done.     
A: Hint: since you know it for continuous functions, you may find Lusin's theorem useful.
A: Let $F=\{\,(x,f(x))\mid x\in \mathbb R\,\}$. First show that $F\cap([a,b)\times[c,d))$ has measure zero for all $a<b, c<d$.
Indeed for $n\in \mathbb N$, $1\le i\le n$, let $$F_i=F\cap\left([a,b)\times[c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\right).$$
Then $$\begin{align}\mu(F_i)&\le \mu\bigl([c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\bigr)\cdot\mu\bigl(f^{-1}([c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\bigr)\\
&=\frac{\mu\bigl([c,d)\bigr)}{n}\cdot\mu\bigl(f^{-1}([c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\bigr)\end{align}$$ 
and hence $$\begin{align}\mu\bigl(F\cap([a,b)\times[c,d))\bigr)&=\sum_{i=1}^n\mu(F_i)\\&\le \frac{\mu\bigl([c,d)\bigr)}n\sum_{i=1}^n\mu\left(f^{-1}([c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\right)\\
&=\frac{\mu\bigl([c,d)\bigr)}n\mu\bigl((f^{-1}([c,d))\bigr)\\
&\le \frac1n{\mu\bigl([c,d)\bigr)}\mu\bigl([a,b)\bigr).\end{align}$$
As $n$ is arbitrary and the two intervals have finite measure, we conclude
$$\mu\bigl(F\cap([a,b)\times[c,d))\bigr)=0.$$
We can cover $\mathbb R\times \mathbb R$ with countabley many such rectangles, hence also $\mu(F)=0$.
