I have read that given a closed set $G$ with zero-volume boundary, any continuous function $f$ on $G$ is Riemann integrable.
My question is: why does $G$ have to be closed? What if it's open?
The origin of this question came from a homework problem I have. It asked to prove, given a bounded domain $G$ and a continuous function $f : \bar{G} \to (0, \infty)$, that $1/f$ is integrable on $G$. I was wondering, what if $f$ was on $G$, just a bounded set, and not $\bar{G}$, its closure?
I have tried to think of a counterexample, ie a continuous function $f$ that is integrable on $G$ but not on $\bar{G}$, where $G$ is bounded and not closed.
The first paragraph is not entirely true: $G$ needs to be bounded and closed (or equivalently compact).
$G$ needs to be compact, otherwise $f$ is not necessarily bounded on $G$. For example, $f(x)=\frac{1}{x}$ is continuous on $(0, 1)$ but not Rieman integrable for not being bounded.