If I have correctly understood and been able to generalise the proof that I found in Kolmogorov-Fomin's Элементы теории функций и функционального анализа for the case of $n=1$, I know that, if $R=\prod_{i=1}^n[a_i,b_i]\subset\mathbb{R}^n$ and $f:R\to\mathbb{R}$ is Riemann integrable on the bounded set $R$, then it also is Lebesgue integrable and the two integrals are the same:$$(R)\int_R f(x_1,\ldots,x_n)dx_1\ldots dx_n=\int_R fd\mu$$where $\mu$ is the usual Lebesgue measure defined on measurable subsets of $\mathbb{R}^n$.
$f:D\to\mathbb{R}$ is defined as Riemann integrable on a generical domain $D\subset\mathbb{R}^n$ when the function $\bar{f}:R\to\mathbb{R}$ (with $R$ as above and such that $D\subset R$) defined by $$\bar{f}(\boldsymbol{x}) = \begin{cases} f(\boldsymbol{x}), & \boldsymbol{x}\in D \\ 0, &\boldsymbol{x}\in R\setminus D \end{cases}$$is Riemann integrable on $R$. My question is: if $f$ is Riemann integrable on a generical domain $D$, can we say that it also is Lebesgue integrable on $D$? I think the only obstacle to overcome is veryfying whether $D$ is $\mu$-measurable, because Lebesgue integrals are only defined on measurable subsets, as far as I know, but I cannot overcome that. A search in this site gives many related results, but I have only found cases where $D=[a,b]\in\mathbb{R}$. I thank any answerer very much!
It is a matter of definition.
Let's decide, as for Riemann, that $f$ is called $\,$ L-integrable over $D\,$ if $\,\bar f \,$ is $\,$L-integrable over $R$ (according to the usual definition) and that $$\int_D f=\int_{R} \bar f$$ Then the statement (R-integrability $\Rightarrow$ L-integrability) is true also for arbitrary (bounded) sets.
The set $D$ doesn't need to be measurable !
All this because of the validity of the following statement:
Let $E$ and $E'$ be measurable sets with $E \subset E'$. Let $f$ be a function on $E$ and let $g$ be the function defined on $E'$ by $$g(\mathbf x) = \begin{cases} f(\mathbf x)\quad\text {if} \quad \mathbf x\in E \\ 0 \;\,\qquad\text {if}\quad\mathbf x\in E'\setminus E \end{cases}$$ then $f$ is L-integrable on $E\,$ iff $\,g$ is L-integrable on $E'$ and, if so, $$\int_E f=\int_{E'} g$$ So the above definition is independent of $R$ and reduces to the usual one if $D$ is measurable.