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.
Theorem 5
Let $S$ be bounded set in $\Bbb R^n$; let $f,g:S\rightarrow\Bbb R$ be bounded functions. Suppose $f$ and $g$ are integrable over $S$. If $f(x)\le g(x)$ for $x\in S$, then $$ \int_S f\le\int_S g $$
So using these results I want prove the following result.
Theorem
If $f$ and $g$ are real function that agree in $S$ then if $f$ is integrable over $S$ even $g$ is there integrable and the two integrals are equal.
Proof. By the definition 3 it is sufficent to prove the statement for the case where $S$ is a rectangle. So if $P$ is a partition of $Q$ then $$ f(x)=g(x) $$ for any $x\in R$ where $R$ is a subrectangles of $P$ so that $m_R(f)=m_R(g)$ and $M_R(f)=M_R(g)$ thus $L(f,P)=L(g,P)$ and $U(f,P)=U(g,P)$. Now if $f$ is integrable over $Q$ then by theorem 2 for any $\epsilon$ there exist a partition $P$ such that $$ U(g,P)-L(g,P)=U(f,P)-L(f,P)<\epsilon $$ so that by the same theorem we conclude that $g$ is integrable in $Q$. Finally inverting the role of $f$ and $g$ in the theorem 5 we conclude that $$ \int_S f=\int_S g $$ and so the statement 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?