In the book of Mathematical Analysis II by Zorich, at page 123, it is asked that
Give necessary and sufficient conditions on a function $f : E → \mathbb{R}$ defined on a bounded (but not necessarily Jordan-measurable) set $E$ under which the Riemann integral of $f$ over $E$ exists.
My question is that, if $E$ is open, than would $f$ be discontinuous on $\partial E$ ? I mean $f$ is not even defined on $\partial E$, so I assume that it wouldn't, but then I recall the definition of Riemann integral over any subset of $\mathbb{R}^n$, that is
$$\int_T f *dx := \int_{T \subset I}f *\gamma_T (x)* dx, \quad where \quad \gamma_T (x) = \begin{cases} 1 & x \in T\\ 0 & otherwise \end{cases}\quad and \quad I \text{ is any interval containing $E$}$$ and Zorich points out $\int_{T \subset I} \gamma_T (x)* dx$ is defined only for sets $T$ whose boundary is measure zero since $\gamma_T(x)$ has discontinuities on $\partial T$.
Therefore, I'm sort of confused because $\gamma_T$ is defined on the boundary of $E$, but $f$ is not.
I mean even if we (somehow) say that $f$ is discontinues on $\partial E$, then $\partial E$ has to be Jordan measure zero and $int (E)$ has to be Lebesgue measure zero, right ?
Edit: For additional questions, please see the comments on the given answer.
A function $f$ is Riemann integrable on a bounded rectangle $Q$ if and only if it is bounded and the set of discontinuity points is of measure $0$.
On a bounded set $E$ the integral exists when $f \chi_E$ is integrable over any rectangle $Q$ with $E \subset Q$ and is given by $\int_E f = \int_Q f \chi_E$.
Necessary and sufficient conditions are $f$ is almost everwhere continuous in $E$ and either $\partial E$ is of measure $0$ or $f(x) \to 0$ as $x \to p$ for all points $p \in \partial E$ except perhaps in a subset of $\partial E$ of measure $0$. This second condition takes care of the case where $m(\partial E) \neq 0$.