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:

1. 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 $D_1$.

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 !