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Riemann integrable then J-integrable

ianchenmu

Member
Feb 3, 2013
74
Let $E\subset\mathbb{R}^n$ be a closed Jordan domain and $f:E\rightarrow\mathbb{R}$ a bounded function. We adopt the convention that $f$ is extended to $\mathbb{R}^n\setminus E$ by $0$.
Let $\jmath$ be a finite set of Jordan domains in $\mathbb{R}^n$ that cover $E$.


Define $M_J=sup\left \{ f(x)\;|\;x\in J \right \}$, $m_J=inf\left \{ f(x)\;|\;x\in J \right \}$




$W(f;\jmath )=\sum_{J\in\jmath ,J\cap E\neq \varnothing }M_JVol(J)\;\;\;\;\;\;\;\;\;\;$(upper R-sum)
$w(f;\jmath )=\sum_{J\in\jmath ,J\cap E\neq \varnothing }m_JVol(J)\;\;\;\;\;\;\;\;\;\;$(lower R-sum)


Define $\overline{vol}(f;E)=inf\left \{ W(f;\jmath ) \right \}\;$, $\;\underline{vol}(f;E)=sup\left \{ w(f;\jmath ) \right \}$.


Say that $f$ is $J$-integrable on $E$ if $\overline{vol}(f;E)=\underline{vol}(f;E)$.


**Prove** that if $f$ is Riemann integrable on $E$ then it is $J$-integrable.




How to relate this? The definition of Riemann integrable has only a difference that $\jmath$ is an n-dimensional rectangle and $J$ is a grid on $\jmath$.