MHB Riemann integrable then J-integrable

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In the discussion, it is established that if a bounded function f is Riemann integrable over a closed Jordan domain E in R^n, then it is also J-integrable on E. The relationship between Riemann and J-integrability is highlighted through the definitions of upper and lower R-sums, W(f; j) and w(f; j), respectively. By constructing a sequence of partitions corresponding to the Riemann integrability, it is shown that the limits of these sums converge to the same integral value. Consequently, the infimum of the upper sums equals the supremum of the lower sums, confirming J-integrability. This proof effectively demonstrates the connection between the two types of integrability.
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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 $\jmath$.
 
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Since $f$ is Riemann-integrable, then by definition, there exists a sequence of partitions $\left \{ P_k \right \}_{k\in\mathbb{N}}$ such that $\lim_{k\rightarrow\infty}R(f;P_k)=\int_E f(x)dx$ and $\lim_{k\rightarrow\infty}L(f;P_k)=\int_E f(x)dx$. Let $\jmath_k$ be the set of $J$-domains that corresponds to the partition $P_k$, so that for each element $x\in P_k$ there is a unique $J\in\jmath_k$ such that $x\in J$. Then we have $$\lim_{k\rightarrow\infty}W(f; \jmath_k)=\lim_{k\rightarrow\infty}R(f;P_k)=\int_E f(x)dx$$and $$\lim_{k\rightarrow\infty}w(f; \jmath_k)=\lim_{k\rightarrow\infty}L(f;P_k)=\int_E f(x)dx$$Hence, we have $$\overline{vol}(f;E)=\lim_{k\rightarrow\infty}W(f; \jmath_k)=\lim_{k\rightarrow\infty}w(f; \jmath_k)=\underline{vol}(f;E)$$Therefore, $f$ is $J$-integrable on $E$.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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