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Math412 Notes Select Topics in Mathematical Analysis of Functions of One or Several Variables

Zheng-Chao Han

Section 6.2 Further Riemann Integrability Criteria

Subsection 6.2.1 Riemann Integrability Criterion in terms of the Oscillation of the Integrand

To find a more easily checkable criterion for (6.1.2) , we first make the following definition.
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Definition 6.2.1.

Let \(f\) be defined on \(R\text{.}\) The oscillation of a function \(f\) over the set \(S \subset R\) is defined to be
\begin{equation*} \sup_{S}f - \inf_{S}f = M(f, S)-m(f, S) \end{equation*}
and is denoted as \(\osc(f, S)\text{.}\)
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The oscillation of a function \(f\) at a point \(\bx\) is defined to be
\begin{equation*} \lim_{r\searrow 0} \osc(f, R\cap B(\bx, r)), \end{equation*}
and is denoted as \(\osc(f)(\bx)\text{.}\) Here \(B(\bx, r)\) is the open ball of radius \(r\) centered at \(\bx\text{.}\)
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Note that, in the definition of \(\osc(f)(\bx)\text{,}\) we could replace the open ball \(B(\bx, r)\) by closed ball or rectangles. Furthermore,
\begin{equation*} \lim_{r\searrow 0} \osc(f, R\cap B(\bx, r))=\lim_{s\searrow 0} \osc(f, R\cap R(\bx, s)) =\lim_{s\searrow 0} \osc(f, R\cap \bar{R}(\bx, s))\text{,} \end{equation*}
where \(R(\bx, s)\) denotes the open rectangle centered at \(\bx\) with \(2s\) as its side length. This follows from the relation
\begin{equation*} \osc(f, U) \le \osc(f, V) \text{ whenever } U\subset V, \end{equation*}
and \(B(\bx, r) \subset R(\bx, r) \subset B(\bx, \sqrt{n}r)\text{.}\)
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Note also that \(f\) is continuous at \(\bx\) iff \(\osc(f)(\bx)=0\text{.}\)
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Exercise 6.2.2.

Is it true that for any \(\bx \in U \subset R\) there holds \(\osc(f)(\bx)\le \osc(f, U)\text{?}\)
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Exercise 6.2.3.

Is the following statement true: if \(\bx\) is in a rectangle \(S\text{,}\) and \(\osc(f)(\bx)\ge \epsilon\text{,}\) then \(M_{S}(f)-m_{S}(f)\ge \epsilon\)?
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Proof.

For any real number \(a\text{,}\) if \(\bx_{0}\in \{\bx: \osc(f)(\bx) < a\}\text{,}\) then there exists some \(r>0\) such that \(\osc(f, B(\bx_{0}, r)) < a\text{.}\) For any \(\bx \in B(\bx_{0}, r)\text{,}\) we observe that \(\osc(f)(\bx)\le \osc(f, B(\bx_{0}, r)) < a\text{.}\) Thus \(B(\bx_{0}, r) \subset \{\bx: \osc(f)(\bx) < a\}\text{,}\) proving that the latter is open.
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Note that
\begin{equation} \{\bx: f \text{ discontinuous at } \bx\}= \cup_{k\in \bbN} \{\by: \osc(f)(\by)\ge \frac 1k\},\tag{6.2.1} \end{equation}
namely, \(\{\bx: f \text{ discontinuous at } \bx\}\) is a countable union of of the closed sets \(\{\by: \osc(f)(\by)\ge \frac 1k\}\text{.}\)
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Proof.

Suppose that \(f\) is Riemann integrable on \(R\) and that \(\epsilon>0\) is given. Our proof of Proposition 6.1.8 gives us a partition \(\cP:=\{R_{\alpha}\}\) such that
\begin{equation*} \int_{R}\, f -\frac{\epsilon}{2} < L(f, \cP)\le \int_{R}\, f \le U (f, \cP) < \int_{R}\, f + \frac{\epsilon}{2}, \end{equation*}
which implies that
\begin{equation*} \sum_{\alpha} \osc(f,R_{\alpha})\vert R_{\alpha} \vert =U (f, \cP)-L (f, \cP) < \epsilon. \end{equation*}
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Suppose that (6.2.2) holds. Then \(U (f, \cP)-L (f, \cP) < \epsilon,\) and
\begin{equation*} 0\le \upint_{R}\, f -\lowint_{\;R} \, f \le U (f, \cP)-L (f, \cP) < \epsilon. \end{equation*}
Since \(\epsilon >0\) is arbitrary, it follows that \(\upint_{R}\, f -\lowint_{\;R} \, f=0\text{,}\) and \(f\) is Riemann integrable on \(R\text{.}\)
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Finally, suppose that \(f\) is continuous on the closed rectangle \(R\text{,}\) then it is uniformly continuous on \(R\text{.}\) For any given \(\epsilon >0\text{,}\) there exists \(\delta >0\) such that for any partition \(\cP:=\{R_{\alpha}\}\) of \(R\) with \(\lambda (P) < \delta\text{,}\) we have \(\osc(f, R_{\alpha}) < \epsilon/\vert R\vert\) for all \(R_{\alpha} \in \cP\text{.}\) It then follows that
\begin{equation*} \sum_{\alpha} \osc(f,R_{\alpha})\vert R_{\alpha} \vert \le \epsilon, \end{equation*}
proving the Riemann integrability of \(f\) on \(R\text{.}\)
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Note that when \(f\) is continuous on the closed rectangle \(R\text{,}\) we achieve (6.2.2) by constructing a partition \(\cP:=\{R_{\alpha}\}\) of \(R\) such that \(\osc(f, R_{\alpha}) < \epsilon/\vert R\vert\) for all \(R_{\alpha} \in \cP\text{.}\) We can also achieve (6.2.2) if we can construct a partition \(\cP:=\{R_{\alpha}\}\) of \(R\) such that the sum of volume of those rectangles \(R_{\alpha}\text{,}\) for which \(\osc(f, R_{\alpha})\ge \epsilon/(2\vert R\vert)\text{,}\) can be made smaller than \(\epsilon/(2 \osc(f, R))\) , as the factors \((f, R_{\alpha})\) have a uniform upper bound \(\osc(f, R)\) for a given bounded function \(f\) on \(R\text{.}\) We will exploit some properties of \(\osc(f)(\bx)\) for this.
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Exercises Exercises

1. Integrability of a real valued function in terms of the integrability of its positive and negative parts.
Let \(f^{+}(x)=\max (f(x), 0), f^{-}(x)=\max(-f(x), 0)\) denote the positive and negative parts of \(f(x)\) respectively. If \(f\) is Riemann integrable on \(C\text{,}\) is it true that \(f^{+}\) and \(f^{-}\) are also Riemann integrable on \(C\text{?}\) If both \(f^{+}\) and \(f^{-}\) are Riemann integrable on \(C\text{,}\) is it true that \(f\) is Riemann integrable on \(C\text{?}\)
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2.
Suppose that \(f\) is Riemann integrable on \(R\text{.}\) Prove that \(|f|\) is Riemann integrable on \(R\text{.}\) Is the converse true?
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3.
Define \(f(x, y)\) on \([0, 1]\times [0, 1]\) by
\begin{equation*} f(x, y)=\begin{cases} \frac 1q \amp x \text{ rational and $y=\frac pq$ with $p, q$ co-prime;} \\ 0 \amp \text{otherwise} \end{cases}\text{.} \end{equation*}
Show that \(f(x, y)\) is Riemann integrable on \([0, 1]\times [0, 1]\) and \(\int_{[0, 1]\times [0, 1]} \, f(x, y)\, =0\text{.}\)
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Subsection 6.2.2 Sets of Content \(0\) and of Measure \(0\)

Definition 6.2.6.

A set \(S\subset \bbR^{n}\) is said to have content \(0\text{,}\) if for any \(\epsilon >0\text{,}\) there exists a finite cover \(\{R_{i}\} \) of \(S\) by rectangles such that
\begin{equation*} \sum_{i}\vert R_{i}\vert < \epsilon. \end{equation*}
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A set \(S\subset \bbR^{n}\) is said to have measure \(0\text{,}\) if for any \(\epsilon >0\text{,}\) there exists an at most countable cover \(\{R_{i}\} \) of \(S\) by rectangles such that
\begin{equation*} \sum_{i}\vert R_{i}\vert < \epsilon. \end{equation*}
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Exercise 6.2.7. Set of measure \(0\) using covers of open rectangles.

Verify that in the definition of a set of measure \(0\text{,}\) open rectangles can be used instead of closed rectangles. Watch out for places later on where this modification is needed.
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Proof.

The first and second properties are obvious. For the third one, suppose \(S=\cup_{i}S_{i}\text{,}\) where each \(S_{i}\) is a set of measure \(0\text{.}\) For any \(\epsilon>0\text{,}\) there exists an at most countable cover \(\{R_{i\,j}\}_{j=1}^{N_{i}}\) (\(N_{i}\) could be \(\infty\)) of \(S_{i}\) such that \(\sum_{j=1}^{N_{i}}\vert R_{i\,j}\vert < \frac{\epsilon}{2^{i}}\text{.}\) Then \(\{R_{i\,j}: i\in \bbN, 1\le j \le N_{i}\}\) is an at most countable cover of \(S\text{,}\) and
\begin{equation*} \sum_{i}\sum_{j=1}^{N_{i}}\vert R_{i\,j}\vert < \sum_{i} \frac{\epsilon}{2^{i}}=\epsilon, \end{equation*}
proving that \(S\) is a set of measure \(0\text{.}\)
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The fourth property is a direct consequence of the third property. For the last property, suppose \(S\) is a compact set of measure \(0\text{.}\) For \(\epsilon>0\) we can choose a cover of \(S\) by open rectangles \(\{R_{i}\} \) such that \(\sum_{i}\vert R_{i}\vert < \epsilon\text{.}\) Using the compactness of \(S\text{,}\) we can then select a finite sub cover which satisfies the desired property.
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Remark 6.2.9.

The rectangles in the finite cover in the definition of a set of content \(0\) are allowed to have non-empty overlaps and are not necessarily the cells of a partition, but we have the following
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Observation: If \(S\) has content \(0\text{,}\) then for any \(\epsilon >0\text{,}\) there exists a partition \(\cP\) of a rectangle containing \(S\) such that
\begin{equation} \sum_{R_{\alpha}\in \cP, R\cap S \ne \emptyset} |R_{\alpha}| \lt \epsilon.\tag{6.2.3} \end{equation}
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This can be seen by first finding an open cover \(\{ U_{\beta}\}\) of the closure \(\bar S\) of \(S\) such that \(\sum_{\beta} |U_{\beta}| \lt \epsilon\text{.}\) Then for any \(\bx \in \bar S\text{,}\) there exists some \(r_{\bx}>0\) such that the open hypercube \(Q(\bx, 4 r_{\bx})\) with side lengths \(4 r_{\bx}\) and centered at \(\bx\) is contained in \(\cup_{\beta} U_{\beta}\text{.}\) Considering the open cover \(\{ Q(\bx, r_{\bx}): \bx \in \bar S\}\) of \(\bar S\) and using the compactness of \(\bar S\text{,}\) we find a finite cover \(\{ Q(\bx_{i}, r_{\bx_{i}}): 1\le i \le N\}\text{.}\) Let \(\delta =\min\{r_{\bx_{i}}: 1\le i \le N\}\text{.}\) Then \(\delta >0\) and for any partition \(\cP\) of a rectangle \(R\) containing \(\bar S\) such that \(\lambda (\cP) \lt \delta\text{,}\) if any rectangle \(R_{\alpha}\) of \(\cP\) satisfies \(R_{\alpha}\cap \bar S \ne \emptyset\text{,}\) then taking any \(\bx \in R_{\alpha}\cap \bar S\text{,}\) there exists some \(\bx_{i}\) such that \(\bx \in Q(\bx_{i}, r_{\bx_{i}})\text{.}\) This then implies that any point \(\by \in R_{\alpha}\) also lies in \(Q(\bx_{i}, 4r_{\bx_{i}})\subset \cup_{\beta} U_{\beta}\text{.}\) Thus \(\{R_{\alpha}: R_{\alpha}\cap \bar S \ne \emptyset\}\) is a finite number of hypercubes in the partition \(\cP\) that covers \(\bar S\text{,}\) and \(\cup_{R_{\alpha}\cap \bar S \ne \emptyset} R_{\alpha} \subset \cup_{\beta} U_{\beta}\text{,}\) therefore
\begin{equation*} \sum_{R_{\alpha}\in \cP, R\cap \bar S \ne \emptyset} |R_{\alpha}| \lt \epsilon. \end{equation*}
(6.2.3) then follows from this.
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For a set \(S\) of content \(0\text{,}\) as a consequence of (6.2.3), we have
\begin{equation} U(\chi_{S},\cP)= \sum_{R_{\alpha}\in \cP, R_{\alpha}\cap S \ne \emptyset} |R_{\alpha}| \lt \epsilon.\tag{6.2.4} \end{equation}
This then implies that \(\int_{R} \chi_{S} =0\text{.}\)
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Remark 6.2.10.

The effect of requiring a finite cover in defining a set of content \(0\) vs a possibly countably infinite cover in defining a set of measure \(0\) can be seen through the following examples.
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The set \(\bbQ\) of rationals in \(\bbR\) is a set of measure \(0\text{,}\) but not a set of content \(0\text{;}\) its closure, \(\bbR\text{,}\) is not a set of measure \(0\text{.}\) The set \(\bbZ\) of integers is a closed set of measure \(0\text{,}\) but not a set of content \(0\text{.}\)
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Exercises Exercises

1.
Show that if a set has content \(0\text{,}\) then its boundary also has content \(0\text{.}\)
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2.
Give an example of a closed set of measure \(0\) which does not have content \(0\) and an example of a bounded set of measure \(0\) such that its boundary does not have measure \(0\text{.}\)
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3.
Suppose that \(f\) is an increasing function on \(\bbR\text{.}\) Show that the set of points where \(f\) is discontinuous has measure \(0\text{.}\)
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Subsection 6.2.3 Riemann Integrability in terms of the Set of Discontinuity of the Integrand

We are now ready to formulate the following theorem.
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Proof.

We will use (6.2.1) for the only if part.
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Suppose that \(f\) is Riemann integrable on \(R\text{.}\) For each \(k\in \bbN\text{,}\) we will prove that \(D_{k}:=\{\by: \osc(f)(\by)\ge \frac 1k\}\) is a set of content \(0\text{.}\)
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Given any \(\epsilon >0\text{.}\) There exists a partition \(\cP=\{R_{\alpha}\}\) of \(R\) such that
\begin{equation*} \sum_{\alpha} \osc(f, R_{\alpha})\vert R_{\alpha} \vert < \frac{\epsilon}{2k}. \end{equation*}
The rectangles in \(\cP\) are divided into two subgroups: the subgroup \(\cL_{k}\) consisting those \(R_{\alpha}\) such that \(\osc(f, R_{\alpha})\ge \frac{1}{2k}\text{,}\) and the subgroup \(\cS_{k}\) consisting those \(R_{\alpha}\) such that \(\osc(f, R_{\alpha})< \frac{1}{2k}\text{.}\) Then it follows from
\begin{equation*} \frac{1}{2k} \sum_{R_{\alpha}\in \cL_{k} } \vert R_{\alpha} \vert \le \sum_{R_{\alpha}\in \cL_{k} } \osc(f, R_{\alpha})\vert R_{\alpha} \vert < \frac{\epsilon}{2k} \end{equation*}
that
\begin{equation*} \sum_{R_{\alpha}\in \cL_{k} } \vert R_{\alpha} \vert < \epsilon. \end{equation*}
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We now claim that
\begin{equation*} D_{k}\subset \cup_{R_{\alpha}\in \cL_{k} } R_{\alpha}. \end{equation*}
This will show that \(D_{k}\) is a set of content \(0\text{.}\)
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If the claim were not true, there would exist some \(\bx \in D_{k} \setminus \cup_{R_{\alpha}\in \cL_{k} } R_{\alpha}\text{.}\) Thus \(\bx \in \cup_{R_{\alpha}\in \cS_{k} } R_{\alpha}\text{.}\) Since the complement of \(\cup_{R_{\alpha}\in \cL_{k} } R_{\alpha}\) is open, there exists some ball \(B(\bx, r) \subset \cup_{R_{\alpha}\in \cS_{k} } R_{\alpha}\text{.}\) If \(\bx\in \text{ interior}(R_{\alpha})\) for some \(R_{\alpha} \in \cS_{k}\text{,}\) it would force \(\frac 1k \le \osc(f)(\bx)\le \osc(f, R_{\alpha}) < \frac{1}{2k}\text{,}\) which would be a contradiction. So \(\bx\) can only be on the boundary of one or more \(R_{\alpha} \in \cS_{k}\text{.}\) We can choose \(r>0\) small enough such that any \(\by\in B(\bx, r)\) and \(\bx\) will be in one such common rectangle. Therefore, \(\vert f(\by)-f(\bx)\vert < \frac{1}{2k}\text{.}\) This would lead to \(\osc(f)(\bx)\le \osc(f, B(\bx, r)) < \frac 1k\text{,}\) contradicting \(\bx \in D_{k}\text{.}\)
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The proof for the if part will be added.
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Exercises Exercises

1. The product of two Riemann integrable functions is Riemann integrable.
Let \(f, g\) be two Riemann integrable functions on \(R\text{.}\) Prove that their product \(f\cdot g\) is Riemann integrable on \(R\text{.}\)
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2.
Is the characteristic function of the Cantor set in Exercise 6.1.14 Riemann integrable? What about the characteristic function of the complement of this Cantor set? ---Note that this complement is an open set of \(\bbR\text{.}\) Work out the upper and lower integrals of either of these functions.
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