Section 7.2 Properties of Riemann Integrals

The Darboux definition gives a practical criterion for integrability: a bounded function is integrable exactly when some partition makes the upper and lower sums arbitrarily close. We now use this criterion to prove the basic algebraic and order properties of the integral, and then to identify some important classes of integrable functions.

Subsection 7.2.1 Basic Properties

Theorem 7.2.1. Basic Properties of the Integral.

Let \(f,g \in R[a,b]\text{,}\) and let \(c \in \R\text{.}\)

The functions \(f+g\) and \(cf\) belong to \(R[a,b]\text{,}\) and

\begin{equation*} \int_a^b (f+g)=\int_a^b f+\int_a^b g, \qquad \int_a^b (cf)=c\int_a^b f. \end{equation*}
If \(f(x) \le g(x)\) for every \(x \in [a,b]\text{,}\) then

\begin{equation*} \int_a^b f \le \int_a^b g. \end{equation*}
If \(a \lt c \lt b\text{,}\) then the restrictions of \(f\) to \([a,c]\) and \([c,b]\) are integrable, and

\begin{equation*} \int_a^b f=\int_a^c f+\int_c^b f. \end{equation*}
If \(|f(x)| \le M\) for every \(x \in [a,b]\text{,}\) then

\begin{equation*} \left|\int_a^b f\right| \le M|b-a|. \end{equation*}

Proof.

We begin with part (1). Let \(I_f=\int_a^b f\) and \(I_g=\int_a^b g\text{,}\) and fix \(\varepsilon \gt 0\text{.}\) By the Integrability Criterion, there are partitions \(P_f\) and \(P_g\) of \([a,b]\) such that

\begin{equation*} U(P_f,f)-L(P_f,f) \lt \frac{\varepsilon}{2}, \qquad U(P_g,g)-L(P_g,g) \lt \frac{\varepsilon}{2}. \end{equation*}

Let \(P=P_f \cup P_g\text{.}\) Since refinement decreases upper sums and increases lower sums,

\begin{equation*} U(P,f)-L(P,f) \lt \frac{\varepsilon}{2}, \qquad U(P,g)-L(P,g) \lt \frac{\varepsilon}{2}. \end{equation*}

For each subinterval \(I_i\) of \(P\text{,}\) let \(m_i^f,M_i^f\) and \(m_i^g,M_i^g\) be the infima and suprema of \(f\) and \(g\) on \(I_i\text{.}\) Then

\begin{equation*} m_i^f+m_i^g \le \inf_{x \in I_i}(f(x)+g(x)), \qquad \sup_{x \in I_i}(f(x)+g(x)) \le M_i^f+M_i^g. \end{equation*}

Summing over the partition gives

\begin{equation*} L(P,f)+L(P,g) \le L(P,f+g) \le U(P,f+g) \le U(P,f)+U(P,g). \end{equation*}

Hence

\begin{equation*} U(P,f+g)-L(P,f+g) \le \bigl(U(P,f)-L(P,f)\bigr)+\bigl(U(P,g)-L(P,g)\bigr) \lt \varepsilon. \end{equation*}

So \(f+g\) is integrable.

Also, because \(I_f\) lies between \(L(P,f)\) and \(U(P,f)\text{,}\) and similarly for \(I_g\text{,}\) we have

\begin{equation*} I_f+I_g-\varepsilon \lt L(P,f)+L(P,g) \le \int_a^b (f+g) \le U(P,f)+U(P,g) \lt I_f+I_g+\varepsilon. \end{equation*}

Since \(\varepsilon\) is arbitrary, \(\int_a^b (f+g)=I_f+I_g\text{.}\)

For scalar multiples, let \(P=\{x_0,\dots,x_n\}\) be any partition, and let \(m_i,M_i\) be the infima and suprema of \(f\) on \([x_{i-1},x_i]\text{.}\) If \(c \ge 0\text{,}\) then the infimum and supremum of \(cf\) on the same interval are \(cm_i\) and \(cM_i\text{.}\) If \(c \lt 0\text{,}\) they are \(cM_i\) and \(cm_i\text{.}\) In either case,

\begin{equation*} U(P,cf)-L(P,cf)=|c|\bigl(U(P,f)-L(P,f)\bigr). \end{equation*}

Hence \(cf\) is integrable. When \(c \ge 0\text{,}\)

\begin{equation*} L(P,cf)=cL(P,f), \qquad U(P,cf)=cU(P,f), \end{equation*}

and the same squeeze argument as above gives \(\int_a^b (cf)=c\int_a^b f\text{.}\) When \(c \lt 0\text{,}\) the same conclusion follows from the identities

\begin{equation*} L(P,cf)=cU(P,f), \qquad U(P,cf)=cL(P,f). \end{equation*}

For part (2), if \(f(x) \le g(x)\) on \([a,b]\text{,}\) then on every subinterval of every partition \(P\) the infimum and supremum of \(f\) are at most the corresponding infimum and supremum of \(g\text{.}\) Thus

\begin{equation*} L(P,f) \le L(P,g) \qquad\text{and}\qquad U(P,f) \le U(P,g). \end{equation*}

Since \(\int_a^b f \le U(P,f)\) and \(L(P,g) \le \int_a^b g\) for every partition \(P\text{,}\) it follows that \(\int_a^b f \le \int_a^b g\text{.}\)

For part (3), let \(I=\int_a^b f\) and fix \(\varepsilon \gt 0\text{.}\) By the Integrability Criterion, there is a partition \(P\) of \([a,b]\) such that \(U(P,f)-L(P,f) \lt \varepsilon/2\text{.}\) If \(c \notin P\text{,}\) adjoin \(c\) to obtain a refinement; the gap does not increase. Write \(P_1\) and \(P_2\) for the induced partitions of \([a,c]\) and \([c,b]\text{.}\) Then

\begin{equation*} L(P,f)=L(P_1,f)+L(P_2,f), \qquad U(P,f)=U(P_1,f)+U(P_2,f), \end{equation*}

\begin{equation*} \bigl(U(P_1,f)-L(P_1,f)\bigr)+\bigl(U(P_2,f)-L(P_2,f)\bigr) = U(P,f)-L(P,f) \lt \frac{\varepsilon}{2}. \end{equation*}

Since each summand is nonnegative, both are less than \(\varepsilon/2\text{.}\) Therefore the restrictions of \(f\) to \([a,c]\) and \([c,b]\) are integrable.

Let \(I_1=\int_a^c f\) and \(I_2=\int_c^b f\text{.}\) Because \(I_1\) and \(I_2\) lie between the corresponding lower and upper sums, we obtain

\begin{gather*} I_1+I_2-\varepsilon \lt L(P_1,f)+L(P_2,f) = L(P,f) \le I,\\ I \le U(P,f) = U(P_1,f)+U(P_2,f) \lt I_1+I_2+\varepsilon. \end{gather*}

Hence \(I=I_1+I_2\text{.}\)

Finally, for part (4), the hypothesis gives \(-M \le f(x) \le M\) on \([a,b]\text{.}\) By part (2),

\begin{equation*} \int_a^b (-M) \le \int_a^b f \le \int_a^b M. \end{equation*}

Since constant functions are integrable and

\begin{equation*} \int_a^b (-M)=-M(b-a), \qquad \int_a^b M=M(b-a), \end{equation*}

we conclude that

\begin{equation*} -M(b-a) \le \int_a^b f \le M(b-a). \end{equation*}

Therefore \(\left|\int_a^b f\right| \le M(b-a)=M|b-a|\text{.}\)

Subsection 7.2.2 Step Functions

A function \(s \colon [a,b] \to \R\) is a step function if there is a partition \(P=\{x_0,\dots,x_n\}\) of \([a,b]\) and real numbers \(c_1,\dots,c_n\) such that

\begin{equation*} s=\sum_{i=1}^n c_i 1_{I_i}, \end{equation*}

where

\begin{equation*} I_i=[x_{i-1},x_i) \qquad (1 \le i \lt n), \qquad I_n=[x_{n-1},x_n]. \end{equation*}

If \(J\) is any subinterval of \([a,b]\) with endpoints \(c \le d\text{,}\) then the indicator function \(1_J\) is Riemann integrable and

\begin{equation*} \int_a^b 1_J=d-c. \end{equation*}

Indeed, let \(\varepsilon \gt 0\text{.}\) One may choose a partition that isolates the endpoints of \(J\) inside two subintervals of lengths less than \(\varepsilon/2\) each. Outside those short endpoint intervals the function is constant, so the upper and lower sums differ by at most \(\varepsilon\text{.}\) Thus \(1_J\) is integrable, and its integral is the length of \(J\text{.}\)

Proposition 7.2.2.

Every step function on \([a,b]\) is Riemann integrable. More precisely, if

\begin{equation*} s=\sum_{i=1}^n c_i 1_{I_i} \end{equation*}

for a partition \(P=\{x_0,\dots,x_n\}\) and the associated subintervals \(I_i\text{,}\) then

\begin{equation*} \int_a^b s=\sum_{i=1}^n c_i(x_i-x_{i-1}). \end{equation*}

Proof.

Each indicator function \(1_{I_i}\) is integrable, with \(\int_a^b 1_{I_i}=x_i-x_{i-1}\text{.}\) Therefore part (1) of Basic Properties of the Integral shows that any linear combination \(\sum_{i=1}^n c_i1_{I_i}\) is integrable and that

\begin{equation*} \int_a^b s = \sum_{i=1}^n c_i \int_a^b 1_{I_i} = \sum_{i=1}^n c_i(x_i-x_{i-1}). \end{equation*}

Subsection 7.2.3 Continuous Functions of Integrable Functions

Theorem 7.2.3. Continuous Compositions Preserve Integrability.

Suppose \(f \in R[a,b]\text{,}\) let \(m,M \in \R\) satisfy \(f([a,b]) \subseteq [m,M]\text{,}\) and let \(\phi \colon [m,M] \to \R\) be continuous. Then \(\phi \circ f \in R[a,b]\text{.}\)

A proof of this can be found in [2]. The same result holds for the more general Riemann-Stieltjes integral.

The order of composition matters. There exist a Riemann integrable function \(f\) and a continuous function \(\phi\) such that \(f \circ \phi\) is not Riemann integrable; see [10].

Proposition 7.2.4.

If \(f,g \in R[a,b]\text{,}\) then \(fg \in R[a,b]\text{.}\)

Proof.

By the Basic Properties of the Integral, the functions \(f+g\text{,}\) \(f\text{,}\) and \(g\) are integrable. The function \(t \mapsto t^2\) is continuous on every closed bounded interval, so Continuous Compositions Preserve Integrability shows that \((f+g)^2\text{,}\) \(f^2\text{,}\) and \(g^2\) are integrable. Since

\begin{equation*} fg=\frac12\bigl((f+g)^2-f^2-g^2\bigr), \end{equation*}

another application of Basic Properties of the Integral gives \(fg \in R[a,b]\text{.}\)

Proposition 7.2.5.

If \(f \in R[a,b]\text{,}\) then \(|f| \in R[a,b]\text{.}\) Moreover,

\begin{equation*} \left|\int_a^b f\right| \le \int_a^b |f|. \end{equation*}

Proof.

The function \(t \mapsto |t|\) is continuous, so Continuous Compositions Preserve Integrability shows that \(|f|\) is integrable.

Since \(-|f(x)| \le f(x) \le |f(x)|\) on \([a,b]\text{,}\) part (2) of Basic Properties of the Integral gives

\begin{equation*} \int_a^b (-|f|) \le \int_a^b f \le \int_a^b |f|. \end{equation*}

Part (1) of the same theorem implies \(\int_a^b (-|f|)=-\int_a^b |f|\text{.}\) Therefore

\begin{equation*} -\int_a^b |f| \le \int_a^b f \le \int_a^b |f|, \end{equation*}

and hence \(\left|\int_a^b f\right| \le \int_a^b |f|\text{.}\)

Subsection 7.2.4 Large Classes of Integrable Functions

For \(n \ge 1\text{,}\) the uniform \(n\)-partition of \([a,b]\) is the partition

\begin{equation*} P_n([a,b])=\left\{a=x_0 \lt x_1 \lt \cdots \lt x_n=b\right\}, \end{equation*}

where

\begin{equation*} \Delta x_i = x_i-x_{i-1}=\frac{b-a}{n} \qquad (1 \le i \le n). \end{equation*}

When the interval is clear from the context, we simply write \(P_n\text{.}\)

Theorem 7.2.6. Continuous Functions Are Integrable.

Every continuous function on a closed bounded interval is Riemann integrable.

Proof.

Let \(f \colon [a,b] \to \R\) be continuous, and let \(\varepsilon \gt 0\text{.}\) By Theorem 4.5.3, \(f\) is uniformly continuous on \([a,b]\text{.}\) Therefore there exists \(\delta \gt 0\) such that

\begin{equation*} |f(x)-f(x')| \lt \frac{\varepsilon}{b-a} \end{equation*}

whenever \(x,x' \in [a,b]\) and \(|x-x'| \lt \delta\text{.}\)

Choose \(n\) so large that \((b-a)/n \lt \delta\text{,}\) and let \(P_n=\{x_0,\dots,x_n\}\) be the uniform \(n\)-partition of \([a,b]\text{.}\) For each subinterval \(I_i=[x_{i-1},x_i]\text{,}\) the restriction of \(f\) to \(I_i\) is continuous, so the Extreme Value Theorem gives points \(x_i^*,x_i^{**} \in I_i\) such that

\begin{equation*} M_i=f(x_i^*), \qquad m_i=f(x_i^{**}), \end{equation*}

where \(M_i\) and \(m_i\) are the supremum and infimum of \(f\) on \(I_i\text{.}\)

Since \(|x_i^*-x_i^{**}| \le (b-a)/n \lt \delta\text{,}\) we have

\begin{equation*} M_i-m_i = |f(x_i^*)-f(x_i^{**})| \lt \frac{\varepsilon}{b-a}. \end{equation*}

Therefore

\begin{equation*} U(P_n,f)-L(P_n,f) = \sum_{i=1}^n (M_i-m_i)\Delta x_i \lt \frac{\varepsilon}{b-a}\sum_{i=1}^n \Delta x_i = \varepsilon. \end{equation*}

By the Integrability Criterion, \(f\) is integrable.

Theorem 7.2.7. Monotone Functions Are Integrable.

Every monotone function on a closed bounded interval is Riemann integrable.

Proof.

First suppose that \(f\) is increasing on \([a,b]\text{.}\) Let \(P_n=\{x_0,\dots,x_n\}\) be the uniform \(n\)-partition of \([a,b]\text{.}\) Then on each subinterval \([x_{i-1},x_i]\text{,}\)

\begin{equation*} M_i=f(x_i), \qquad m_i=f(x_{i-1}). \end{equation*}

Hence

\begin{equation*} U(P_n,f)-L(P_n,f) = \sum_{i=1}^n (f(x_i)-f(x_{i-1}))\frac{b-a}{n} = \frac{b-a}{n}\bigl(f(b)-f(a)\bigr). \end{equation*}

The right-hand side tends to \(0\) as \(n \to \infty\text{,}\) so the Integrability Criterion shows that \(f\) is integrable.

If \(f\) is decreasing, then \(-f\) is increasing, so the argument above shows that \(-f\) is integrable. By part (1) of Basic Properties of the Integral, \(f=-(-f)\) is integrable as well.

Example 7.2.8.

The Thomae function\(h \colon [0,1] \to \R\) is defined by

\begin{equation*} h(x)= \begin{cases} \dfrac{1}{n} & \text{if } x=\dfrac{m}{n} \text{ in lowest terms},\\ 0 & \text{if } x \in [0,1]\setminus \Q. \end{cases} \end{equation*}

This function is integrable on \([0,1]\text{,}\) with \(\int_0^1 h=0\text{,}\) but it is neither continuous nor monotone on any nondegenerate subinterval of \([0,1]\text{.}\)

A plot of Thomae’s function on [0,1], with taller points at rationals having smaller denominators. — Figure 7.2.9. A plot of the Thomae function on \([0,1]\text{,}\) showing sample values at rational points with denominator at most \(10\text{.}\)

By Proposition 1.2.7 and the argument from Example 7.1.6, every nondegenerate subinterval of \([0,1]\) contains both rational and irrational numbers. Thus, if \(\alpha \lt q \lt \beta\) with \(\alpha,\beta\) irrational and \(q\) rational, then

\begin{equation*} h(\alpha)=0, \qquad h(q) \gt 0, \qquad h(\beta)=0, \end{equation*}

so \(h\) cannot be increasing or decreasing on that interval. Also, if \(q \in [0,1] \cap \Q\text{,}\) choose irrational numbers \(\alpha_n \to q\text{.}\) Then \(h(\alpha_n)=0 \to 0\text{,}\) while \(h(q) \gt 0\text{,}\) so \(h\) is not continuous at \(q\text{.}\) Since rationals are dense, \(h\) is not continuous on any nondegenerate subinterval.

Every nondegenerate subinterval of \([0,1]\) contains irrational numbers, so the infimum of \(h\) on each such interval is \(0\text{.}\) Hence every lower sum of \(h\) is \(0\text{.}\)

To control the upper sums, fix \(N \in \N\) and let

\begin{equation*} H_N=\{x \in [0,1] : h(x) \ge 1/N\}. \end{equation*}

If \(x \in H_N\text{,}\) then \(x=m/n\) in lowest terms with \(1 \le n \le N\text{.}\) Therefore \(H_N\) is finite. Choose pairwise disjoint closed intervals \(J_x \subseteq [0,1]\text{,}\) one for each \(x \in H_N\text{,}\) such that \(x \in J_x\) and the sum of their lengths is less than \(1/N\text{.}\) Let \(P\) be a partition containing all endpoints of these intervals.

Any subinterval of \(P\) that is disjoint from \(\bigcup_{x \in H_N} J_x\) contains no point of \(H_N\text{,}\) so the supremum of \(h\) there is less than \(1/N\text{.}\) On the other subintervals, the supremum is at most \(1\text{.}\) It follows that

\begin{equation*} U(P,h) \le \sum_{x \in H_N} |J_x| + \frac{1}{N} \left(1-\sum_{x \in H_N} |J_x|\right) \lt \frac{1}{N}+\frac{1}{N} = \frac{2}{N}. \end{equation*}

Since \(N\) is arbitrary, the Integrability Criterion shows that \(h\) is integrable. Because every lower sum is \(0\text{,}\) we have \(\int_0^1 h=0\text{.}\)

Prev Top Next