How to prove if Riemann integrable then J-integrable

[ianchenmu]

Homework Statement

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.

Homework Equations

n/a

The Attempt at a Solution

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 $\jmath$.

 

[lippyka]

I think that $J$-integrable implies Riemann integrable (by taking $\jmath$ to be a grid on $E$) but I am not sure how to prove the other way.