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?
Best Answer
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 $\partial S$.