Let $S$ be a bounded set in $\mathbb{R}^n$; let $f:S\rightarrow\mathbb{R}$ be a bounded continuous function; let $A$ be the interior of $S$. Give an example where $\int_Af$ exists and $\int_Sf$ does not.
There's a theorem that says if $\int_Sf$ does not exist, then for a set $E$ of measure nonzero of points $x_0$ on the boundary of $S$, the condition $\lim_{x\rightarrow x_0}f(x)=0$ fails to hold.
So we must select $S$ at least to meet that condition. In particular, the boundary must have measure nonzero. But whenever we pick a connected region in the plane, for example, the boundary has measure zero. So I'm not sure which set $S$ to pick here.
Best Answer
That is not true, only you can't visualise sets where it isn't the case well, these beasts are very complicated.
Let $C \subset [0,1]$ a fat Cantor set, and $U = [0,1]\setminus C$. Then $A = U\times[0,1] \cup [0,1]\times U$ is connected and its boundary $C\times C$ has positive measure. Take any non-measurable subset $N$ of $C\times C$, and let $S = A \cup N$. Let $f \equiv 1$.
then $\int_A f$ exists ($A$ is open, and hence measurable, it is contained in $[0,1]^2$ so has finite measure), but $\int_S f$ does not exist because $S$ is not measurable.