Let $f,g: S \rightarrow \mathbb{R}$ Riemannintegrable functions in a bounded set $S \subset { \mathbb{R} }^{ n }$.
Then $fg: S \rightarrow \mathbb{R}$ it is also integrable in $S$.
In effect: f,g integrables in $S$ implies by definition that
${ f }_{ S }(x)=\begin{cases} f(x)\quad if\quad x\in S \\ 0\quad else \end{cases}$ $\quad$ and $\quad$ ${ g }_{ S }(x)=\begin{cases} g(x)\quad if\quad x\in S \\ 0\quad else \end{cases}$
They are integrable in a rectangle $Q\subset { \mathbb{R} }^{ n }$ such that $S\subset Q$.
By this, the set's ${ D }_{ { f }_{ S } }$ and ${ D }_{ { g }_{ S } }$ of discontinuities of ${ f }_{ S }$ and ${ g }_{ S }$ in $Q$ respectively, have measure zero in ${ \mathbb{R} }^{ n }$.
Now we define ${ f }_{ S }\cdot { g }_{ S }:Q\rightarrow \mathbb{R}$ the product function between ${ f }_{ S }$ and ${ g }_{ S }$.
We have ${ f }_{ S }\cdot { g }_{ S }(x)={ f }_{ S }(x)\cdot { g }_{ S }(x)=\begin{cases} f(x)\cdot g(x)\quad if\quad x\in S\quad \\ 0\quad else \end{cases}:={ (fg) }_{ s }(x)$
and so $dom({ (fg) }_{ s })=dom({ f }_{ S })\cap dom({ g }_{ S })$.
Thus ${ D }_{ { (fg) }_{ s } }={ D }_{ { f }_{ S } }\cap { D }_{ { g }_{ S } }$ where ${ D }_{ { (fg) }_{ s } }$ is the set of discontinuities of ${ (fg) }_{ s }$ in $Q$ and therefore have measure zero in ${ \mathbb{R} }^{ n }$.
Thus ${ (fg) }_{ s }$ is Riemann-integrable in $Q$, and this implies by definition that $fg:S\rightarrow \mathbb{R}$ is Riemann-integrable in $S$ and also: $\int _{ S }^{ }{ fg } =\int _{ Q }^{ }{ { (fg) }_{ s } } $ $\blacksquare $
(My question is whether this proof is successful)