Let $D_{1},D_{2}$$D_1,D_2$ be a bounded subset of $\mathbb{R}^{n}$$\mathbb{R}^n$ and $ \color{red}{ \partial D_{1},\partial D_{2}}$ $\color{red}{\text{are both of }\textit{Lebesgue measure zero}}$$\partial D_1,\partial D_2$ are both of Lebesgue measure zero (that is to say:$D_{1},D_{2}$ $D_1,D_2$ are $\color{blue}{\textit{Jordan measurable}}$Jordan measurable). Also, let $f:D_{1}\cup D_{2}=D\rightarrow \mathbb{R}$$f:D_1\cup D_2=D\rightarrow \mathbb{R}$ be a bounded function. then Then ($f$ is $\textbf{Riemann integrable}$Riemann integrable over $D_{1}$,over and over $D_{2}.\Leftrightarrow $$D_{2}$) $\Leftrightarrow$ ($f$ is $\textbf{Riemann integrable}$Riemann integrable over $D=D_{1}\cup D_{2}.$$D=D_1\cup D_2$).
The proof of above result is not difficult.The 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 ,
- 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:
$f$ is $\textbf{Riemann integrable}$ over $D=D_{1}\cup D_{2}.\Rightarrow $ $f$ is Let $\textbf{Riemann integrable}$ over$D=[0,1]^2$ and $D_{1}$$D_1=\mathbb{Q}^2\cap [0,1]^2$,over $D_{2}. $ will be not correct ,there is a counterexample to illustrate:
Let $D=[0,1]^{2},D_{1}=\mathbb{Q}^{2}\cap [0,1]^{2},D_{2}=[0,1]^{2}\backslash (\mathbb{Q}^{2}\cap [0,1]^{2})$$D_2=[0,1]^2\setminus D_1$. $f\equiv1:D\rightarrow \mathbb{R}$.
Obviously, $f$ is $\textbf{Riemann integrable}$ overRiemann integrable over $D=D_{1}\cup D_{2}.$$D$. But $$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}$$ it is$$ 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} $$ is not $\textbf{Riemann integrable}$ overRiemann integrable over $[0,1]^{2}$$[0,1]^2$,so $f$ is not $\textbf{Riemann integrable}$ overRiemann integrable over $D_{1}$$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 !
- 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 !
