If a function $f:A\to\mathbb{R}$, bounded, is integrable then its 'plot' has volume zero. What about the reciprocal? 
If a function $f:A\to\mathbb{R}$, bounded in the rectangle $A\subset
 \mathbb{R}^m$, is integrable then its 'plot' (its graph) has volume
  zero. What about the reciprocal?

I'm trying first to prove and then think about the recriprocal. Why does the graph of an integrable function have volume zero? This question has some hints, but I was unable to finish. 
We have to prove that the set $X = \{(x,f(x)); x\in A, f(x)\in\mathbb{R}\}$ has volume $0$, that is, the integral of its characteristic function is $0$. What's the characteristic function of this set? Let's call $\chi_X(x)$. We know that $\chi_X(x) = 1, $ if $x\in X$, and $\chi_X = 0,$ if $x\notin X$.
I think the solution is to think about the volume of $A$ and the volume of $im(f)$. Then we're gonna have $vol \ X = vol\ A \cdot vol \ im(f)$. I'm trying to follow the hint:

In general you need to approximate f by finitely many pieces such that
  f is constant on all the pieces (or the piece is small itself).

But how should I break these pieces of $f$? It seems counter intuitive for me, because I can always see a $m$-dimensional block as a $m-1$ dimensional block times a $1$ dimensional line. I could break this line in many constant pieces and so prove that the volume of the entire block is $0$, which is an absurd. 
For the reciprocal, I have to prove that if the volume is $0$, then the plot is not integrable... The condition for not being integrable is that for every $\epsilon>0$ (bla bla bla), $\sum_{B\in P} w_B\cdot vol B <\epsilon$. Can you see a connection?
 A: If I understand correctly, your question concerns this statement:

Suppose $A\subset\mathbb R^n$ is a rectangle and $f\colon A\to\mathbb R$ is bounded. If the graph $G_f\subset A\times\mathbb R$ has zero measure, then $f$ is integrable.

This requires some interpretation.
First, what does zero measure mean?
I will interpret that the meaning is that the Lebesgue outer measure of $G_f$ is zero.
The set $G_f$ might not be Borel or even measurable by just assuming $f$ to be bounded, so I recommend being careful with the kind of "volume zero" you mean.
All bounded sets have a well defined Lebesgue outer measure, so it's a safe choice.
Second, what does "integrable" mean?
Riemann integrable or Lebesgue integrable?
The answer to both versions turns out to be the same, but again precision is recommended.
The statement above is false.
In the Riemann integral case, you can simply take $f$ to be the characteristic function of $\mathbb Q^n$ or any other countable dense set.
This $f$ is bounded and the graph is contained in $A\times\{0,1\}$ so it has zero outer measure in $\mathbb R^{n+1}$.
In the Lebesgue case you just need to pick a more complicated set for the characteristic function.
Take any non-measurable set (which exist if you assume the axiom of choice, for example).
Then the characteristic function is not Lebesgue integrable, but the graph is again contained in the set $A\times\{0,1\}$ of zero outer measure.
