2
$\begingroup$

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 $D_1$.

  2. 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 !

$\endgroup$
6
  • $\begingroup$ Please don't use italic text inside colours inside maths environments for emphasis. Your message is not clearer for using many colours and fonts. $\endgroup$
    – David Roberts
    Jun 5, 2015 at 2:43
  • $\begingroup$ @David Roberts: Thanks for your mention. $\endgroup$
    – Elliot
    Jun 5, 2015 at 2:45
  • $\begingroup$ (2) is true. Use that a function is R integrable precisely if it is bounded and continuous a.e. $\endgroup$ Jun 6, 2015 at 1:07
  • $\begingroup$ @Christian Remling:My friend,your comments may not make sure (2) is true. $int(D_{1})\cup int(D_{2})\subseteq int(D_{1}\cup D_{2})$ .Maybe some interior points of $D_{k}(k=1,2)$ at which $f$ is contionuous will become the discontionuos points of $f $ on $D_{1}\cup D_{2}$. $\endgroup$
    – Elliot
    Jun 6, 2015 at 3:11
  • $\begingroup$ @Q.L.Kwai: Express $\chi_{D_1\cup D_2}f$ in terms of $\chi_{D_j}f$. $\endgroup$ Jun 6, 2015 at 6:14

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.