Let $D_{1},D_{2}$ be a bounded subset of $\mathbb{R}^{n}$ and $ \color{red}{ \partial D_{1},\partial D_{2}}$ $\color{red}{\text{are both of }\textit{Lebesgue measure zero}}$ (that is to say:$D_{1},D_{2}$ are $\color{blue}{\textit{Jordan measurable}}$). Also, let $f:D_{1}\cup D_{2}=D\rightarrow \mathbb{R}$ be a bounded function. then $f$ is $\textbf{Riemann integrable}$ over $D_{1}$,over $D_{2}.\Leftrightarrow $ $f$ is $\textbf{Riemann integrable}$ over $D=D_{1}\cup D_{2}.$

The proof of above result is not difficult.The following is my question:

$\textbf{1.}$

If we remove the condition:$ \color{red}{ \partial D_{1},\partial D_{2}}$ $\color{red}{\text{are both of }\textit{Lebesgue measure zero}}$ from the above statement ,

$f$ is $\textbf{Riemann integrable}$ over $D=D_{1}\cup D_{2}.\Rightarrow $ $f$ is $\textbf{Riemann integrable}$ over $D_{1}$,over $D_{2}. $ will be not correct ,there is a counterexample to illustrate:

Let $D=[0,1]^{2},D_{1}=\mathbb{Q}^{2}\cap [0,1]^{2},D_{2}=[0,1]^{2}\backslash (\mathbb{Q}^{2}\cap [0,1]^{2})$. $f\equiv1:D\rightarrow \mathbb{R}$.

Obviously, $f$ is $\textbf{Riemann integrable}$ over $D=D_{1}\cup D_{2}.$ But $$f\cdot \chi _{\small{D_{1}}}(x,y)=\begin{cases} 1 ,& \text{ as }\quad (x,y)\in D_{1} ,\\ 0,& \text{ as }\quad (x,y)\in D_{2}. \end{cases}$$ it is not $\textbf{Riemann integrable}$ over $[0,1]^{2}$,so $f$ is not $\textbf{Riemann integrable}$ over $D_{1}$.

$\textbf{2.}$

If we remove the condition:$ \color{red}{ \partial D_{1},\partial D_{2}}$ $\color{red}{\text{are both of }\textit{Lebesgue measure zero}}$ from the above statement ,

By my intuition, $f$ is $\textbf{Riemann integrable}$ over $D_{1}$,over $D_{2}.\Rightarrow $ $f$ is $\textbf{Riemann integrable}$ 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 !