Let $D_1,D_2$ be a bounded subset of $\mathbb{R}^n$ and $\partial D_1,\partial D_2$ are both of Lebesgue measure zero (that is to say: $D_1,D_2$ are Jordan measurable). Also, let $f:D_1\cup D_2=D\rightarrow \mathbb{R}$ be a bounded function. Then ($f$ is Riemann integrable over $D_{1}$ and over $D_{2}$) $\Leftrightarrow$ ($f$ is Riemann integrable over $D=D_1\cup D_2$).
The proof of above result is not difficult. The following is my question:
If we remove the condition "$\partial D_1,\partial D_2$ are both of Lebesgue measure zero" from the above statement, then the result ($f$ is Riemann integrable over $D=D_1\cup D_2$) $\Rightarrow$ ($f$ is Riemann integrable over $D_{1}$ and over $D_{2}$) will be not correct. There is a counterexample to illustrate:
Let $D=[0,1]^2$ and $D_1=\mathbb{Q}^2\cap [0,1]^2$, $D_2=[0,1]^2\setminus D_1$. $f\equiv1:D\rightarrow \mathbb{R}$.
Obviously, $f$ is Riemann integrable over $D$. But $$ f\cdot \chi _{\small{D_{1}}}(x,y)=\begin{cases} 1 ,& \text{ as }\quad (x,y)\in D_{1} ,\\ 0,& \text{ as }\quad (x,y)\in D_{2}. \end{cases} $$ is not Riemann integrable over $[0,1]^2$,so $f$ is not Riemann integrable over $D_1$.
If we remove the condition: "$\partial D_1,\partial D_2$ are both of Lebesgue measure zero" from the above statement, by my intuition, ($f$ is Riemann integrable over $D_1$ and over $D_2$) $\Rightarrow$ ($f$ is Riemann integrable over $D=D_1\cup D_2$) is also not correct ! But until now I have as yet neither found a counterexample to illustrate my intuition nor given a proof to support it correct !
