The zero function is integrable in $\pmb{ANY}$ set and its integral is zero. 
Definition 1
Let $Q$ a rectangle; let $f:Q\rightarrow\Bbb R$ be a bounded function. As $P$ ranges over all partitions of $Q$, define
$$
\underline{\int_Q}f:=\underset{P}\sup\{L(f,P)\}\,\,\,\text{and}\,\,\,\overline{\int_Q}f:=\underset{P}\inf\{U(f,P)\}.
$$
These numbers are called the lower integral and upper integral, respectively, of $f$ over $Q$.
Theorem 2
Let $Q$ be a rectangle; let $f:Q\rightarrow\Bbb R$ a bounded function. Then
$$
\underline{\int_Q}f\le\overline{\int_Q}f;
$$
equality holds if and only if given $\epsilon>0$, there exist a corresponding partition $P$ of $Q$ for which
$$
U(f,P)-L(f,P)<\epsilon
$$
Definition 3
Let $S$ be a bounded set in $\Bbb R^n$; let $f:S\rightarrow\Bbb R$ be a bounded function. Define $f_S:\Bbb R^n\rightarrow\Bbb R$ by the equation
$$
f_S(x):=\begin{cases}f(x),\,\,\,\text{for}\,\,\,x\in S\\0,\,\,\,\text{otherwise}\end{cases}.
$$
Choose a rectangle $Q$ containing $S$. We define the integral of $f$ over $S$ by the equation
$$
\int_S f:=\int_Q f_S
$$
provided the latter integral exists.
Lemma 4
Let $Q$ and $Q'$ be two rectangles in $\Bbb R^n$. If $f:\Bbb R^n\rightarrow R$ is a bounded function that vanishes outside $Q\cap Q'$, then
$$
\int_Q f=\int_{Q'} f;
$$
one integral exists if and only if the other does.

So using the preceding results I want prove the following two things.
Lemma
The zero function $\pmb{0}$ is integrable in any rectangle $Q$ and is integral is zero.
Proof. So if $P$ is a partition of $Q$ then clearly
$$
m_R(\pmb{0})\le 0\le M_R(\pmb{0})
$$
for any subrectangle $R$. So if
$$
m_R(\pmb{0})<0<M_R(\pmb{0})
$$
for some rectangle $R$ then by properties of the infimum and supremum for any $\epsilon>0$ there exist $x,y\in Q$ such that
$$
0=\pmb{0}(x)<m_R(\pmb{0})+\epsilon\,\,\,\text{and}\,\,\, 0=\pmb{0}(y)>M_R(\pmb{0})-\epsilon
$$
and clearly this is impossible. So we conlcude that
$$
L(f,P)=0=U(f,P)
$$
for any partition $P$ of $Q$ so that the positive and negative number are respectively an upper bound of $\{L(f,P)\}$ and a lower bound of $\{U(f,P)\}$ so that there exist the lower integral and the upper integral of $\pmb 0$ function. So for any $\epsilon>0$ then
$$
U(f,P)-L(f,P)=0<\epsilon
$$
for any partition $P$ so that by the  theorem 2 we conclude that $\pmb{0}$ is integrable over $Q$.  Now if $\int_Q\pmb{0}\neq 0$ then $\underset{P}\sup\{L(f,P)\}>0$ so that by the property of supremum for any $\epsilon\in\big(0,\underset{P}\sup\{L(f,P)\}\big)$ there exist a partition $P$ such that
$$
0<\underset{P}\sup\{L(f,P)\}-\epsilon<L(f,P)=0
$$
and this is clearly impossible. So the corollary holds.
Theorem
If $S$ is a bounded set in $\Bbb R^n$ then the zero function $\pmb 0$ is there integrable and its integral is zero.
Proof. So if $Q$ is a rectangle containing $S$ then the function $0_S$ fulfills the hypothesis of the preceding lemma so that with the same argument (formally it would be necessary to repeat it!) it is possible to prove that $0_S$ is integrable over $Q$ and its integral is zero thus the theorem holds.
So I ask if the statement of the question is true and in particular if the proof I gave is correct: I realise that this could be a trivial result but unfortunately I see that never text prove it although it is used in many proofs. So could someone help me, please?
 A: Because the result is intuitively obvious it is easy to give a proof with true statements that are not fully justified.  If your objective is to be really precise, then I would make the following improvements.
(1) Given $m_R(\pmb{0})\le 0\le M_R(\pmb{0})$ and assuming that $m_R(\pmb{0})< 0\ < M_R(\pmb{0})$ you assert that for any $\epsilon > 0$ there exist $x,y \in Q$ such that
$$0=\pmb{0}(x)<m_R(\pmb{0})+\epsilon\,\,\,\text{and}\,\,\, 0=\pmb{0}(y)>M_R(\pmb{0})-\epsilon$$
and this is clearly is impossible.  As it may hold for some $\epsilon $, produce a specific example where it fails to hold.  For example, with $\epsilon = -m_R(\mathbf{0})/2 > 0$ we get the contradiction $0=\pmb{0}(x)<m_R(\pmb{0})/2 <0$.
(2) The proof that $\int_Q \mathbf{0} = 0$ is indirect and a little cumbersome.  Why not simply say that for all $P$ we have
$$0 = L(f,P) \leqslant \int_Q \mathbf{0} \leqslant U(f,P) = 0$$
(3) A minor detail is in proving that $\int_S \mathbb{0} =0 $ for any bounded set $S$, you are starting with the definition
$$\int_S \mathbb{0} := \int_Q\mathbb{0}_S,$$
where $Q$ can be any rectangle containing $S$.  I would add that $\mathbf{0}_S$ is everywhere continuous and, therefore,  integrable on $Q$ regardless of the content of boundary $\partial S$.
