QUESTION a) Let $f:A\rightarrow \mathbb{R}$ be a integrable function defined in a block $A\subset\mathbb{R}^m$. If $A_0\subset\mathbb{R}^m$ is a block such that $A\subset A_0$, then the function $\overline{f}:A_0\rightarrow \mathbb{R}$ given by
\begin{equation*} \overline{f}(x) = \begin{cases} f(x) & \text{if } x \in A,\\ 0 & \text{if } x \in A_0\backslash A. \end{cases} \end{equation*} is integrable (in $A_0$) and $$\int_{A_0}\overline{f}(x) dx= \int_{A} f(x) dx$$
QUESTION b) Let $A=\displaystyle\Pi_{i=1}^{n}[a_i, b_i]$ be a block in $\mathbb{R}^m$, such that $A \subset A_0$, then $$\int_{A_0}\chi_{{}_{A}} dx = \Pi_{i=1}^{m} (b_i-a_i)$$
MY ATTEMPTY:
a) Because $A\subset A_0$ let's divide in two cases:
Supposing that $A_0\cap A=A$ by definition of $f$ we have $$\int_{A_0} \overline{f}(x) dx = \int_{A} f(x) dx.$$
Now, in the set $A_0\backslash A$ we have $\overline{f}(x) = f(x) =0$. Considering $A_0\backslash A \subset \displaystyle\bigcup_{i=1}^{\infty} C_i$, where $C_i's$ are open blocks, by the Lindelöf's theorem has a countable subcover, let's define it as $A_0\backslash A\subset \displaystyle\bigcup_{i=1}^{k} A_i =D$, however all countable set has zero measure (because is a set of points which has zero measure). Then we can write $\{x \in A_0\backslash A; f(x)=0\}= D$. Thus $$\int_{A_0}\overline{f}(x) dx = \int_{A} f(x) dx.$$
b) Supposing $A\subset A_0$, let's divide in two cases:
if $\overline{A_0}=\partial A$, so because $\partial A$ is defined as an edge of the block $A=\Pi_{i=1}^{n}[a_i, b_i]\subset \mathbb{R}^m$, then by definition has zero measure. Thus in this case, $$\int_{\overline{A_0}} \chi_{{}_{A}}(x) dx = \int_A \chi_{{}_{A}}(x) dx= \Pi_{i=1}^{m}(b_i-a_i).$$
if $A$ is a proper subset of $A_0$, then we have by definition of characteristic function that $\chi_{{}_{A_{0}\backslash A}}(x) = 0$. Therefore, by the question one, we have $$\int_{A_0} \chi{{}_{A}} dx = \int_{A} \chi{{}_{A}} dx = \Pi_{i=1}^{m}(b_i-a_i).$$
MY DOUBTS: Are my answers right? Would you correct for me or help me to improve it, please?
Your proof is fine, but it might be simpler just to observe that when you take upper and lower sums for the Jordan measure of a set, it's the same as taking upper and lower Darboux sums for a Riemann integral, so your integrals are simply Riemann integrals over cubes. And since it's given that $f$ is integrable, and $\chi_{A}$ certainly is, then $g:=\overline f\cdot \chi_{A}$ is also integrable, and this gives $a).$ For $b).\ $ we go the other way, that is, integrals of the constant function $f=1$ over cubes give Jordan content, so we have $\int_{A_0} \chi{{}_{A}} dx=\int_{A} dx=\operatorname{vol} A=\Pi_{i=1}^{m}(b_i-a_i).$