Given a function $f:D\subset\mathbb{R}^n\to\mathbb{R}$ let $\bar{f}:R\to\mathbb{R}$, where $R$ is a hyperrectangle such that $D\subset R$, be defined by $$\bar{f}(\boldsymbol{x}) = \begin{cases} f(\boldsymbol{x}), & \boldsymbol{x}\in D \\ 0, & \boldsymbol{x}\in R\setminus D \end{cases}$$
If there is a number $I\in\mathbb{R}$ such that, for any partition $P$ (whose mesh is $\delta_P$) of $R$ into hyperrectangles $R_i$ of volume $\Delta V_i$, $$\forall\varepsilon>0\quad\exists\delta>0\,:\,\delta_P<\delta\Rightarrow\left|\sum_i \bar{f}(\boldsymbol{x}_i^\ast)\Delta V_i-I\right|<\varepsilon$$ no matter how we chose $\boldsymbol{x}_i^\ast\in R_i$, then we define $$\int_D f(\boldsymbol{x})\,d^nx:=I.$$
This is the most common definition of Riemann integral that I have found. I have always been convinced that $f$ cannot be unbounded if it is integrable according to this definition because, if it were unbounded, $\bar{f}(\boldsymbol{x}_i^\ast)$ could be chosen with an arbitrarily big absolute value, making $I$ impossible to exist, at least finite.
Am I right in believeing that, according to the above explained definition of Riemann integral, if $f$ is integrable, it must be bounded?
What causes me some doubts is that I read this definition of Riemann integral saying that the integrand function $f$ may be unbounded on a set of null Jordan measure, although I suspect that the two definitions are not equivalent... Is that so? I heartily thank any answerer.
You are running into a technicality here. Consider first the one-dimensional case.
Boundedness is a necessary condition for $f:[a,b] \to \mathbb{R}$ to be Riemann integrable. A necessary and sufficient condition is that the set of discontinuity points has measure zero.
If $f$ is unbounded on $[a,b]$, you can show that there exists some $\epsilon_0$ such that for any $I \in \mathbb{R}$ and any partition $P$ there is a refinement $P'$ and a Riemann sum $S(P',f)$ such that
$$|S(P',f) - I| \geqslant \epsilon_0.$$
Consider
$$f_1(x) = \begin{cases}x &\mbox{if } x \in [0,1] \setminus \{\frac1{2}\} \\ 10 &\mbox{if } x = \frac1{2}\end{cases} \\ f_2(x) = \begin{cases}x &\mbox{if } x \in [0,1] \setminus \{\frac1{2}\} \\ +\infty &\mbox{if } x = \frac1{2}\end{cases}\\ f_3(x) = \begin{cases}x^{-1} &\mbox{if } x \in (0,1] \\0 &\mbox{if } x = 0\end{cases} \\ $$
Now $f_1$ is Riemann integrable because the discontinuity is in a set of measure zero.
The function $f_3$ is also almost everywhere continuous, but it is not Riemann integrable because it is unbounded -- regardless of how we assign a value at $x = 0$.
Is the function $f_2$ Riemann integrable? Technically it is unbounded if we work in the extended real numbers. Clearly no upper Riemann sum is finite. In this sense, it is not technically Riemann integrable. However the discontinuity is removable, so a simple modification restores inegrability. This could never happen for $f_3$, which is unbounded on a set of positive measure.
In the multidimensional case, it is also the case that Riemann integrability requires boundedness. But again you can make some arbitrary assignments (including $\infty$) on any subset of measure (or content) $0$ and still argue that the function is integrable.