Definition 6.4.1.
A bounded subset \(G\) of \(\bbR^{n}\) is called Jordan measurable if its boundary \(\partial G\) has content \(0\text{.}\)
Section 6.4 Integration on a Jordan Measurable Set
We extend the Riemann integration from a rectangular region to a more general region called Jordan measurable.
Remark 6.4.2.
Note that the characteristic function \(\chi_{G}(\bx)\) of \(G\) is discontinuous precisely on \(\partial G\text{,}\) so if \(\partial G\) has content \(0\text{,}\) then \(\chi_{G}(\bx)\) is Riemann integrable. Conversely, if \(G\) is bounded and \(\chi_{G}(\bx)\) is Riemann integrable, then \(\partial G\) has measure \(0\text{.}\) But \(\partial G\) is now a bounded closed set, so it has content \(0\text{,}\) making \(G\) Jordan measurable.
But this definition makes certain open sets not Jordan measurable, while it seems natural to define the length of any open set of \(\bbR\) to be \(\sum_{k}|I_{k}|\text{,}\) where \(\cup_{k}I_{k}\) is the canonical decomposition of an open set \(\mathcal O\) in \(\mathbb R\) as the non-overlapping union of open intervals.
Definition 6.4.3.
Suppose that \(G\) of \(\bbR^{n}\) is a Jordan measurable set and \(f(\bx)\) is a bounded function on \(G\text{.}\) We say that \(f\) is integrable on \(G\) if \(\chi_{G}(\bx)f(\bx)\) is a Riemann integrable function (on a rectangular box containing \(G\)). In such a case, we call \(\int \chi_{G}(\bx)f(\bx)\) the integral of \(f\) on \(G\) and denote it as \(\int_{G}f\text{.}\)
Remark 6.4.4.
The above definition is legitimate due to the following property: if \(S_{1}, S_{2}\) are two rectangles containing \(G\) and \(\chi_{G}(\bx)f(\bx)\) is Riemann integrable in \(S_{1}\text{,}\) then it is Riemann integrable in \(S_{2}\) and \(\int_{S_{1}} \chi_{G}(\bx)f(\bx)=\int_{S_{2}} \chi_{G}(\bx)f(\bx)\text{.}\)
Proposition 6.4.5. Basic Properties of Jordan Measurable Sets and Riemann Integrable Functions.
Suppose that \(A\) and \(B\) are Jordan measurable sets in \(\bbR^{n}\text{.}\) Then
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\(A\cup B\text{,}\) \(A\cap B\text{,}\) and \(A\setminus B\) are Jordan-measurable.
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If \(A\) is Jordan-measurable and has measure \(0\text{,}\) then any bounded function \(f\) on \(A\) is Riemann integrable on \(A\) and \(\int_{A} f =0\text{.}\)
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If \(f, g\) are Riemann integrable on \(A\text{,}\) then for any constants \(a, b\text{,}\) \(a f+b g\) is Riemann integrable on \(A\) and\begin{equation*} \int_{A}\left(a f+b g \right)=a \int_{A} f + b \int_{A} g. \end{equation*}
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If \(A, B\) are Jordan-measurable and \(A\cap B\) has measure \(0\text{,}\) then\begin{equation*} \int_{A\cup B}f= \int_{A}f +\int_{B}f\text{.} \end{equation*}
Proof.
(a) follows by using \(\partial (A\cup B)\subset \partial A \cup \partial B\text{,}\) \(\partial (A\cap B)\subset \partial A \cap \partial B\text{,}\) and a similar one for \(\partial (A\setminus B)\text{.}\)
When \(A\) is Jordan-measurable and has measure \(0\text{,}\) its interior must be empty so \(A\subset \partial A\text{.}\) Since \(\partial A\) has content \(0\text{,}\) it follows that \(\bar A=\partial A\) has content \(0\text{.}\) Then, as in proving (6.2.4) and (6.2.4), for any \(\epsilon >0\text{,}\) \(\bar A\) can be covered by a finite number of open rectangles \(\{ S_{i} \}\) such that \(\sum_{i}|S_{i}| \lt \epsilon\text{,}\) so there exists some \(\delta >0\) such that if any partition \(\cP\) satisfies \(\lambda (\cP) \lt \delta\text{,}\) then any rectangle \(R_{\alpha}\) of \(\cP\) satisfying \(R_{\alpha} \cap \bar A \ne \emptyset\) must satisfy \(R_{\alpha} \subset \cup_{i} S_{i}\text{.}\) Therefore
\begin{equation*}
\sum_{R_{\alpha} \cap \bar A \ne \emptyset} |R_{\alpha} | \le \sum_{i}|S_{i}| \lt \epsilon\text{.}
\end{equation*}
It further follows that
\begin{equation*}
- C \epsilon \le
- C \sum_{R_{\alpha} \cap \bar A \ne \emptyset} |R_{\alpha} |\le
L(\chi_{A}f, \cP)\le U(\chi_{A}f, \cP) \le C \sum_{R_{\alpha} \cap \bar A \ne \emptyset} |R_{\alpha} |
\lt C \epsilon\text{,}
\end{equation*}
where \(C>0\) is such that \(| f(\bx) |\le C\) for all \(\bx \in A\text{.}\) Since \(\epsilon >0\) in these inequalities, they show that \(\chi_{A}f\) is Riemann integrable with \(\int_{A}f=\int \chi_{A}f =0\text{.}\)
Remark 6.4.6.
The most common domains of integration are built on the following kinds: \(A\) is a Jordan-measurable set in \(\mathbb R^{n-1}\text{,}\) and \(f(\bx)\le g(\bx)\) for all \(\bx\in A\) are two continuous functions on \(A\text{,}\) then the graph region \(G_{A; f, g} :=\left\{ (\bx, y): \bx\in A, f(\bx)\le y \le g(\bx)\right\}\) is Jordan-measurable in \(\mathbb R^{n}\text{.}\)
If \(h(\bx, y)\) is a continuous function defined on \(G_{A; f, g}\text{,}\) then Fubini’s Theorem applied to \(h(\bx, y)\chi_{G_{A; f, g}}(\bx, y)\) would take the form of
\begin{equation*}
\int_{G_{A; f, g}} h(\bx, y) = \int_{A} \left(\int_{f(\bx)}^{g(\bx)} h(\bx, y)\, dy \right) \, d\bx\text{.}
\end{equation*}
