3.1: Definition of the Integral

The integral that you studied in calculus is the {}, named after the German mathematician , who provided a rigorous formulation of the integral to

replace the intuitive notion due to and . Since Riemann’s time, other kinds of integrals have been defined and studied; however, they are all generalizations of the Riemann integral, and it is hardly possible to understand them or appreciate the reasons for developing them without a thorough understanding of the Riemann integral. In this section we deal with functions defined on a finite interval \([a,b]\). A {} is a set of subintervals \[\begin{equation} \label{eq:3.1.1} [x_0,x_1],\ [x_1,x_2], \dots, [x_{n-1},x_n], \end{equation} \nonumber \] where \[\begin{equation} \label{eq:3.1.2} a=x_0<x_1\cdots<x_n=b. \end{equation} \nonumber \] Thus, any set of \(n+1\) points satisfying defines a partition \(P\) of \([a,b]\), which we denote by \[ P=\{x_0,x_1, \dots,x_n\}. \nonumber \] The points \(x_0\), \(x_1\), , \(x_n\) are the {} of \(P\). The largest of the lengths of the subintervals is the {} of \(P\), written as \(\|P\|\); thus, \[ \|P\|=\max_{1\le i\le n}(x_i-x_{i-1}). \nonumber \] If \(P\) and \(P'\) are partitions of \([a,b]\), then \(P'\) is a {} if every partition point of \(P\) is also a partition point of \(P'\); that is, if \(P'\) is obtained by inserting additional points between those of \(P\). If \(f\) is defined on \([a,b]\), then a sum \[ \sigma=\sum_{j=1}^n f(c_j)(x_j-x_{j-1}), \nonumber \] where \[ x_{j-1}\le c_j\le x_j,\quad 1\le j\le n, \nonumber \] is a . (Occasionally we will say more simply that \(\sigma\) is a Riemann sum of \(f\) over \([a,b]\).) Since \(c_j\) can be chosen arbitrarily in \([x_j,x_{j-1}]\), there are infinitely many Riemann sums for a given function \(f\) over a given partition \(P\).

We leave it to you (Exercise~) to show that \(\int_a^b f(x)\,dx\) is unique, if it exists; that is, there cannot be more than one number \(L\) that satisfies Definition~.

For brevity we will say integrable'' andintegral’‘when we mean Riemann integrable'' andRiemann integral.’’ Saying that \(\int_a^b f(x)\,dx\) exists is equivalent to saying that \(f\) is integrable on \([a,b]\).

-.3em An important application of the integral, indeed, the one invariably used to motivate its definition, is the computation of the area bounded by a curve \(y=f(x)\), the \(x\)-axis, and the lines \(x=a\) and \(x=b\) (``the area under the curve’’), as in Figure~.

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For simplicity, suppose that \(f(x)>0\). Then \(f(c_j)(x_j-x_{j-1})\) is the area of a rectangle with base \(x_j-x_{j-1}\) and height \(f(c_j)\), so the Riemann sum \[ \sum_{j=1}^n f(c_j)(x_j-x_{j-1}) \nonumber \] can be interpreted as the sum of the areas of rectangles related to the curve \(y=f(x)\), as shown in Figure~.

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An apparently plausible argument, that the Riemann sums approximate the area under the curve more and more closely as the number of rectangles increases and the largest of their widths is made smaller, seems to support the assertion that \(\int_a^b f(x)\,dx\) equals the area under the curve. This argument is useful as a motivation for Definition~, which without it would seem mysterious. Nevertheless, the logic is incorrect, since it is based on the assumption that the area under the curve has been previously defined in some other way. Although this is true for certain curves such as, for example, those consisting of line segments or circular arcs, it is not true in general. In fact, the area under a more complicated curve is {} to be equal to the integral, if the integral exists. That this new definition is consistent with the old one, where the latter applies, is evidence that the integral provides a useful generalization of the definition of area.

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We will show that if \(f\) is unbounded on \([a,b]\), \(P\) is any partition of \([a,b]\), and \(M>0\), then there are Riemann sums \(\sigma\) and \(\sigma'\) of \(f\) over \(P\) such that \[\begin{equation} \label{eq:3.1.7} |\sigma-\sigma'|\ge M. \end{equation} \nonumber \] We leave it to you (Exercise~) to complete the proof by showing from this that \(f\) cannot satisfy Definition~.

Let \[ \sigma=\sum_{j=1}^nf(c_j)(x_j-x_{j-1}) \nonumber \] be a Riemann sum of \(f\) over a partition \(P\) of \([a,b]\). There must be an integer \(i\) in \(\{1,2, \dots,n\}\) such that \[\begin{equation} \label{eq:3.1.8} |f(c)-f(c_i)|\ge \frac{M }{ x_i-x_{i-1}} \end{equation} \nonumber \] for some \(c\) in \([x_{i-1}x_i]\), because if there were not so, we would have \[ |f(x)-f(c_j)|<\frac{M}{ x_j-x_{j-1}},\quad x_{j-1}\le x\le x_j,\quad 1\le j\le n. \nonumber \] Then \[\begin{eqnarray*} |f(x)|\ar=|f(c_j)+f(x)-f(c_j)|\le|f(c_j)|+|f(x)-f(c_j)|\\ \ar\le |f(c_j)|+\frac{M}{ x_j-x_{j-1}},\quad x_{j-1}\le x\le x_j,\quad 1\le j\le n. \end{eqnarray*} \nonumber \] which implies that \[ |f(x)|\le\max_{1\le j\le n}|f(c_j)|+\frac{M}{ x_j-x_{j-1}}, \quad a\le x \le b, \nonumber \] contradicting the assumption that \(f\) is unbounded on \([a,b]\).

Now suppose that \(c\) satisfies , and consider the Riemann sum \[ \sigma'=\sum_{j=1}^nf(c'_j)(x_j-x_{j-1}) \nonumber \] over the same partition \(P\), where \[ c'_j=\left\{\casespace\begin{array}{ll} c_j,&j \ne i,\\ c,&j=i.\end{array}\right. \nonumber \]

Since \[ |\sigma-\sigma'|=|f(c)-f(c_i)|(x_i-x_{i-1}), \nonumber \] implies .

Because of Theorem~, we consider only bounded functions throughout the rest of this section.

To prove directly from Definition~ that \(\int_a^b f(x)\,dx\) exists, it is necessary to discover its value \(L\) in one way or another and to show that \(L\) has the properties required by the definition. For a specific function it may happen that this can be done by straightforward calculation, as in Examples~ and . However, this is not so if the objective is to find general conditions which imply that \(\int_a^b f(x)\,dx\) exists. The following approach avoids the difficulty of having to discover \(L\) in advance, without knowing whether it exists in the first place, and requires only that we compare two numbers that must exist if \(f\) is bounded on \([a,b]\). We will see that \(\int_a^b f(x)\,dx\) exists if and only if these two numbers are equal.

If \(m\le f(x)\le M\) for all \(x\) in \([a,b]\), then \[ m(b-a)\le s(P)\le S(P)\le M(b-a) \nonumber \] for every partition \(P\); thus, the set of upper sums of \(f\) over all partitions \(P\) of \([a,b]\) is bounded, as is the set of lower sums. Therefore, Theorems~ and imply that \(\overline{\int_a^b}f(x)\,dx\) and \(\underline{\int_a^b}f(x)\,dx\) exist, are unique, and satisfy the inequalities \[ m(b-a)\le\overline{\int_a^b} f(x)\,dx\le M(b-a) \nonumber \] and \[ m(b-a)\le \underline{\int_a^b}f(x)\,dx\le M(b-a). \nonumber \]

If \(P=\{x_0,x_1, \dots,x_n\}\), then \[ S(P)=\sum_{j=1}^n M_j(x_j-x_{j-1}), \nonumber \] where \[ M_j=\sup_{x_{j-1}\le x\le x_j}f(x). \nonumber \] An arbitrary Riemann sum of \(f\) over \(P\) is of the form \[ \sigma=\sum_{j=1}^n f(c_j)(x_j-x_{j-1}), \nonumber \] where \(x_{j-1}\le c_j\le x_j\). Since \(f(c_j)\le M_j\), it follows that \(\sigma\le S(P)\).

Now let \(\epsilon>0\) and choose \(\overline c_j\) in \([x_{j-1},x_j]\) so that \[ f(\overline c_j) > M_j -\frac{\epsilon}{ n(x_j-x_{j-1})},\quad 1\le j\le n. \nonumber \] The Riemann sum produced in this way is \[ \overline \sigma=\sum_{j=1}^n f(\overline c_j)(x_j-x_{j-1})>\sum_{j=1}^n\left[M_j-\frac{\epsilon}{ n(x_j-x_{j-1})})\right](x_j-x_{j-1})=S(P)-\epsilon. \nonumber \] Now Theorem~ implies that \(S(P)\) is the supremum of the set of Riemann sums of \(f\) over \(P\).

Exercise~.

\begin{example} \rm Let \[ f(x)=\left\{\casespace\begin{array}{ll} 0&\mbox{ if $x$ is irrational}, \\ 1&\mbox{ if $x$ is rational},\end{array}\right. \nonumber \] and \(P=\{x_0,x_1, \dots,x_n\}\) be a partition of \([a,b]\). Since every interval contains both rational and irrational numbers (Theorems~ and ), \[ m_j=0 \mbox{\quad and \quad} M_j=1,\quad 1\le j\le n. \nonumber \] Hence, \[\begin{eqnarray*} S(P)\ar=\sum_{j=1}^n1\cdot(x_j-x_{j-1})=b-a\\ \arraytext{and}\\ s(P)\ar=\sum_{j=1}^n0\cdot(x_j-x_{j-1})=0. \end{eqnarray*} \nonumber \] Since all upper sums equal \(b-a\) and all lower sums equal \(0\), Definition~ implies that \[ \overline{\int_a^b}f(x)\,dx=b-a \mbox{\quad and \quad} \underline{\int_a^b}f(x)\,dx=0. \nonumber \] \end{example}

The motivation for Definition~ can be seen by again considering the idea of area under a curve. Figure~ shows the graph of a positive function \(y=f(x)\), \(a\le x\le b\), with \([a,b]\) partitioned into four subintervals.

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The upper and lower sums of \(f\) over this partition can be interpreted as the sums of the areas of the rectangles surmounted by the solid and dashed lines, respectively. This indicates that a sensible definition of area \(A\) under the curve must admit the inequalities \[ s(P)\le A\le S(P) \nonumber \] for every partition \(P\) of \([a,b]\). Thus, \(A\) must be an upper bound for all lower sums and a lower bound for all upper sums of \(f\) over partitions of \([a,b]\). If \[\begin{equation} \label{eq:3.1.11} \overline{\int_a^b}f(x)\,dx=\underline{\int_a^b}f(x)\,dx, \end{equation} \nonumber \]

there is only one number, the common value of the upper and lower integrals, with this property, and we define \(A\) to be that number; if does not hold, then \(A\) is not defined. We will see below that this definition of area is consistent with the definition stated earlier in terms of Riemann sums.

The {} is an important generalization of the Riemann integral. We define it here, but confine our study of it to the exercises in this and other sections of this chapter.

\begin{exerciselist}

Show that there cannot be more than one number \(L\) that satisfies Definition~.

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Suppose that \(\int_a^bf(x)\,dx\) exists and there is a number \(A\) such that, for every \(\epsilon>0\) and \(\delta>0\), there is a partition \(P\) of \([a,b]\) with \(\|P\|<\delta\) and a Riemann sum \(\sigma\) of \(f\) over \(P\) that satisfies the inequality \(|\sigma-A|<\epsilon\). Show that \(\int_a^b f(x)\,dx=A\).

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Prove directly from Definition~ that \[ \int_a^b x^2\,dx=\frac{b^3-a^3}{3}. \nonumber \] Do not assume in advance that the integral exists. The proof of this is part of the problem.

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Generalize the proof of Exercise~ to show directly from Definition~ that \[ \int_a^b x^m\,dx=\frac{b^{m+1}-a^{m+1}}{ m+1} \nonumber \] if \(m\) is an integer \(\ge0\).

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Prove directly from Definition~ that \(f(x)\) is integrable on \([a,b]\) if and only if \(f(-x)\) is integrable on \([-b,-a]\), and, in this case, \[ \int_a^b f(x)\,dx=\int_{-b}^{-a}f(-x)\,dx. \nonumber \]

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Let \(f\) be bounded on \([a,b]\) and let \(P\) be a partition of \([a,b]\). Prove: The lower sum \(s(P)\) of \(f\) over \(P\) is the infimum of the set of all Riemann sums of \(f\) over \(P\).

2pt Let \(f\) be defined on \([a,b]\) and let \(P=\{x_0,x_1, \dots,x_n\}\) be a partition of \([a,b]\).

Find \(\underline{\int_0^1}f(x)\,dx\) and \(\overline{\int_0^1}f(x)\,dx\) if

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Given that \(\int_a^be^x\,dx\) exists, evaluate it by using the formula \[ 1+r+r^2+\cdots+r^n=\frac{1-r^{n+1}}{1-r} \quad (r \ne1) \nonumber \] to calculate certain Riemann sums.

Given that \(\int_0^b\sin x \,dx\) exists, evaluate it by using the identity \[ \cos(j-1)\theta-\cos(j+1)\theta=2\sin\theta\sin j\theta \nonumber \] to calculate certain Riemann sums.

Given that \(\int_0^b\cos x\,dx\) exists, evaluate it by using the identity \[ \sin(j+1)\theta-\sin(j-1)\theta=2\sin\theta\cos j \theta \nonumber \] to calculate certain Riemann sums.

Show that if \(g(x)=x+c\) (\(c\)=constant), then \(\int_a^b f(x)\,dg(x)\) exists if and only if \(\int_a^bf(x)\,dx\) exists, in which case \[ \int_a^b f(x)\,dg(x)=\int_a^bf(x)\,dx. \nonumber \]

Suppose that \(-\infty<a<d<c<\infty\) and \[ g(x)=\left\{\casespace\begin{array}{ll} g_1,&a<x<d,\\ g_2,&d<x<b,\end{array}\right. \mbox{($g_1,g_2=$ constants), \quad} \nonumber \] and let \(g(a)\), \(g(b)\), and \(g(d)\) be arbitrary. Suppose that \(f\) is defined on \([a,b]\), continuous from the right at \(a\) and from the left at \(b\), and continuous at \(d\). Show that \(\int_a^b f(x)\,dg(x)\) exists, and find its value.

Suppose that \(-\infty<a=a_0<a_1<\cdots<a_p=b<\infty\), let \(g(x)=g_m\) (constant) on \((a_{m-1},a_m)\), \(1\le m\le p\), and let \(g(a_0)\), \(g(a_1)\), , \(g(a_p)\) be arbitrary. Suppose that \(f\) is defined on \([a,b]\), continuous from the right at \(a\) and from the left at \(b\), and continuous at \(a_1\), \(a_2\), , \(a_{p-1}\). Evaluate \(\int_a^bf(x)\,dg(x)\).

For the case where \(g\) is nondecreasing and \(f\) is bounded on \([a,b]\), define upper and lower Riemann–Stieltjes integrals in a way analogous to Definition~.

\end{exerciselist}

The following lemma is the starting point for our study of the integrability of a bounded function \(f\) on a closed interval \([a,b]\).

We will prove and leave the proof of to you (Exercise~). First suppose that \(r=1\), so \(P'\) is obtained by adding one point \(c\) to the partition \(P=\{x_0,x_1, \dots,x_n\}\); then \(x_{i-1}<c<x_i\) for some \(i\) in \(\{1,2, \dots,n\}\). If \(j \ne i\), the product \(M_j(x_j-x_{j-1})\) appears in both \(S(P)\) and \(S(P')\) and cancels out of the difference \(S(P)-S(P')\). Therefore, if \[ M_{i1}=\sup_{x_{i-1}\le x\le c}f(x) \mbox{\quad and \quad} M_{i2}= \sup_{c\le x\le x_i}f(x), \nonumber \] then \[\begin{equation} \label{eq:3.2.4} \begin{array}{rcl} S(P)-S(P')\ar=M_i(x_i-x_{i-1})-M_{i1}(c-x_{i-1})-M_{i2}(x_i-c) \\[2\jot] \ar=(M_i-M_{i1})(c-x_{i-1})+(M_i-M_{i2})(x_i-c). \end{array} \end{equation} \nonumber \] Since implies that \[ 0\le M_i-M_{ir}\le2M,\quad r=1,2, \nonumber \] implies that \[ 0\le S(P)-S(P')\le2M(x_i-x_{i-1})\le2M\|P\|. \nonumber \] This proves for \(r=1\).

Now suppose that \(r>1\) and \(P'\) is obtained by adding points \(c_1\), \(c_2\), , \(c_r\) to \(P\). Let \(P^{(0)}=P\) and, for \(j\ge1\), let \(P^{(j)}\) be the partition of \([a,b]\) obtained by adding \(c_j\) to \(P^{(j-1)}\). Then the result just proved implies that \[ 0\le S(P^{(j-1)})-S(P^{(j)})\le2M\|P^{(j-1)}\|,\quad 1\le j\le r. \nonumber \]

Adding these inequalities and taking account of cancellations yields \[\begin{equation} \label{eq:3.2.5} 0\le S(P^{(0)})-S(P^{(r)})\le2M(\|P^{(0)}\|+\|P^{(1)}\|+\cdots+\|P^{(r-1)}\|). \end{equation} \nonumber \] Since \(P^{(0)}=P\), \(P^{(r)}=P'\), and \(\|P^{(k)}\|\le\|P^{(k-1)}\|\) for \(1\le k\le r-1\), implies that \[ 0\le S(P)-S(P')\le 2Mr\|P\|, \nonumber \] which is equivalent to .

Suppose that \(P_1\) and \(P_2\) are partitions of \([a,b]\) and \(P'\) is a refinement of both. Letting \(P=P_1\) in and \(P=P_2\) in shows that \[ s(P_1)\le s(P') \mbox{\quad and \quad} S(P')\le S(P_2). \nonumber \] Since \(s(P')\le S(P')\), this implies that \(s(P_1)\le S(P_2)\). Thus, every lower sum is a lower bound for the set of all upper sums. Since \(\overline{\int_a^b}f(x)\,dx\) is the infimum of this set, it follows that \[ s(P_1)\le\overline{\int_a^b}f(x)\,dx \nonumber \] for every partition \(P_1\) of \([a,b]\). This means that \(\overline{\int_a^b} f(x)\,dx\) is an upper bound for the set of all lower sums. Since \(\underline{\int_a^b} f(x)\,dx\) is the supremum of this set, this implies .

We prove that \(\overline{\int_a^b}f(x)\,dx=\int_a^bf(x)\,dx\) and leave it to you to show that \(\underline{\int_a^b}f(x)\,dx=\int_a^bf(x)\,dx\) (Exercise~).

Suppose that \(P\) is a partition of \([a,b]\) and \(\sigma\) is a Riemann sum of \(f\) over \(P\). Since \[\begin{eqnarray*} \overline{\int_a^b}f(x)\,dx-\int_a^b f(x)\,dx\ar= \left(\overline{\int_a^b}f(x)\,dx-S(P)\right)+(S(P)-\sigma) \\[2\jot] &&+\left(\sigma-\int_a^b f(x)\ dx\right), \end{eqnarray*} \nonumber \]

the triangle inequality implies that \[\begin{equation} \label{eq:3.2.7} \begin{array}{rcl} \dst{\left|\overline{\int_a^b}f(x)\,dx-\int_a^b f(x)\,dx \right|}\ar\le \dst{\left|\overline{\int_a^b}f(x)\,dx-S(P)\right|+|S(P)-\sigma|} \\[2\jot] &&+\dst{\left|\sigma-\int_a^b f(x)\ dx\right|}. \end{array} \end{equation} \nonumber \] Now suppose that \(\epsilon>0\). From Definition~, there is a partition \(P_0\) of \([a,b]\) such that \[\begin{equation} \label{eq:3.2.8} \overline{\int_a^b} f(x)\,dx\le S(P_0)< \overline{\int_a^b}f(x)\,dx+\frac{\epsilon}{3}. \end{equation} \nonumber \] From Definition~, there is a \(\delta>0\) such that \[\begin{equation} \label{eq:3.2.9} \left|\sigma-\int_a^bf(x)\,dx\right|<\frac{\epsilon}{3} \end{equation} \nonumber \] if \(\|P\|<\delta\). Now suppose that \(\|P\|<\delta\) and \(P\) is a refinement of \(P_0\). Since \(S(P)\le S(P_0)\) by Lemma~, implies that \[ \overline{\int_a^b} f(x)\,dx\le S(P)< \overline{\int_a^b}f(x)\,dx+\frac{\epsilon}{3}, \nonumber \] so \[\begin{equation} \label{eq:3.2.10} \left|S(P)-\overline{\int_a^b}f(x)\,dx\right|<\frac{\epsilon}{3} \end{equation} \nonumber \] in addition to . Now , , and imply that \[\begin{equation} \label{eq:3.2.11} \left|\overline{\int_a^b} f(x)\,dx-\int_a^b f(x)\,dx\right|< \frac{2\epsilon}{3}+|S(P)-\sigma| \end{equation} \nonumber \] for every Riemann sum \(\sigma\) of \(f\) over \(P\). Since \(S(P)\) is the supremum of these Riemann sums (Theorem~), we may choose \(\sigma\) so that \[ |S(P)-\sigma|<\frac{\epsilon}{3}. \nonumber \] Now implies that \[ \left|\overline{\int_a^b} f(x)\,dx-\int_a^b f(x)\,dx \right|< \epsilon. \nonumber \] Since \(\epsilon\) is an arbitrary positive number, it follows that \[ \overline{\int_a^b}f(x)\,dx=\int_a^b f(x)\,dx. \nonumber \] -6.5ex6.5ex

We show that holds if \(\|P\|\) is sufficiently small, and leave the rest of the proof to you (Exercise~).

The first inequality in follows immediately from Definition~. To establish the second inequality, suppose that \(|f(x)|\le K\) if \(a\le x\le b\). From Definition~, there is a partition \(P_0= \{x_0,x_1, \dots,x_{r+1}\}\) of \([a,b]\) such that \[\begin{equation} \label{eq:3.2.13} S(P_0)<\overline{\int_a^b}f(x)\,dx+\frac{\epsilon}{2}. \end{equation} \nonumber \] If \(P\) is any partition of \([a,b]\), let \(P'\) be constructed from the partition points of \(P_0\) and \(P\). Then \[\begin{equation} \label{eq:3.2.14} S(P')\le S(P_0), \end{equation} \nonumber \] by Lemma~. Since \(P'\) is obtained by adding at most \(r\) points to \(P\), Lemma~ implies that \[\begin{equation} \label{eq:3.2.15} S(P')\ge S(P)-2Kr\|P\|. \end{equation} \nonumber \] Now , , and imply that \[\begin{eqnarray*} S(P)\ar\le S(P')+2Kr\|P\|\\ \ar\le S(P_0)+2Kr\|P\|\\ &<&\overline{\int_a^b} f(x)\,dx+\frac{\epsilon}{2}+2Kr\|P\|. \end{eqnarray*} \nonumber \] Therefore, holds if \[ \|P\|<\delta=\frac{\epsilon}{4Kr}. \nonumber \] -4.5ex4.5ex

If \(\epsilon>0\), there is a \(\delta>0\) such that \[\begin{equation} \label{eq:3.2.18} \underline{\int_a^b}f(x)\,dx-\epsilon<s(P)\le S(P)< \overline{\int_a^b}f(x)\,dx+\epsilon \end{equation} \nonumber \] if \(\|P\|<\delta\) (Lemma~). If \(\sigma\) is a Riemann sum of \(f\) over \(P\), then \[ s(P)\le \sigma\le S(P), \nonumber \] so and imply that \[ L-\epsilon<\sigma<L+\epsilon \nonumber \] if \(\|P\|<\delta\). Now Definition~ implies .

Theorems~ and imply the following theorem.

The next theorem translates this into a test that can be conveniently applied.

We leave it to you (Exercise~) to show that if \(\int_a^b f(x)\,dx\) exists, then holds for \(\|P\|\) sufficiently small. This implies that the stated condition is necessary for integrability. To show that it is sufficient, we observe that since \[ s(P)\le \underline{\int_a^b}f(x)\,dx\le\overline{\int_a^b}f(x)\,dx\le S(P) \nonumber \] for all \(P\), implies that \[ 0\le\overline{\int_a^b} f(x)\,dx-\underline{\int_a^b}f(x)\,dx< \epsilon. \nonumber \] Since \(\epsilon\) can be any positive number, this implies that \[ \overline{\int_a^b} f(x)\,dx=\underline{\int_a^b} f(x)\,dx. \nonumber \] Therefore, \(\int_a^b f(x)\,dx\) exists, by Theorem~.

The next two theorems are important applications of Theorem~.

Let \(P=\{x_0,x_1, \dots,x_n\}\) be a partition of \([a,b]\). Since \(f\) is continuous on \([a,b]\), there are points \(c_j\) and \(c'_j\) in \([x_{j-1},x_j]\) such that \[ f(c_j)=M_j=\sup_{x_{j-1}\le x\le x_j}f(x) \nonumber \] and \[ f(c'_j)=m_j=\inf_{x_{j-1}\le x\le x_j}f(x) \nonumber \] (Theorem~). Therefore, \[\begin{equation} \label{eq:3.2.20} S(P)-s(P)=\sum_{j=1}^n\left[f(c_j)-f(c'_j)\right](x_j-x_{j-1}). \end{equation} \nonumber \] Since \(f\) is uniformly continuous on \([a,b]\) (Theorem~), there is for each \(\epsilon>0\) a \(\delta>0\) such that \[ |f(x')-f(x)|<\frac{\epsilon}{ b-a} \nonumber \] if \(x\) and \(x'\) are in \([a,b]\) and \(|x-x'|<\delta\). If \(\|P\|<\delta\), then \(|c_j-c'_j|<\delta\) and, from , \[ S(P)-s(P)<\frac{\epsilon}{ b-a} \sum_{j=1}^n(x_j-x_{j-1})=\epsilon. \nonumber \] Hence, \(f\) is integrable on \([a,b]\), by Theorem~.

Let \(P=\{x_0,x_1, \dots,x_n\}\) be a partition of \([a,b]\). Since \(f\) is nondecreasing, \[\begin{eqnarray*} f(x_j)\ar=M_j=\sup_{x_{j-1}\le x\le x_j}f(x)\\ \arraytext{and}\\ f(x_{j-1})\ar=m_j=\inf_{x_{j-1}\le x\le x_j}f(x). \end{eqnarray*} \nonumber \] Hence, \[ S(P)-s(P)=\sum_{j=1}^n(f(x_j)-f(x_{j-1})) (x_j-x_{j-1}). \nonumber \] Since \(0<x_j-x_{j-1}\le \|P\|\) and \(f(x_j)-f(x_{j-1})\ge0\), \[\begin{eqnarray*} S(P)-s(P)\ar\le \|P\| \sum_{j=1}^n(f(x_j)-f(x_{j-1})) \\ \ar=\|P\|(f(b)-f(a)). \end{eqnarray*} \nonumber \]

Therefore, \[ S(P)-s(P)<\epsilon\mbox{\quad if \quad} \|P\|(f(b)-f(a))<\epsilon, \nonumber \] so \(f\) is integrable on \([a,b]\), by Theorem~.

The proof for nonincreasing \(f\) is similar.

We will also use Theorem~ in the next section to establish properties of the integral. In Section~3.5 we will study more general conditions for integrability.

We now use the results of Sections~3.1 and 3.2 to establish the properties of the integral. You are probably familiar with most of these properties, but not with their proofs.

4pt

Any Riemann sum of \(f+g\) over a partition \(P=\{x_0,x_1, \dots,x_n\}\) of \([a,b]\) can be written as \[ \begin{array}{rcl} \sigma_{f+g}\ar=\dst{\sum_{j=1}^n}\,[f(c_j)+g(c_j)](x_j-x_{j-1})\\[2\jot] \ar=\dst{\sum_{j=1}^n}\,f(c_j)(x_j-x_{j-1})+ \dst{\sum_{j=1}^n}\,g(c_j)(x_j-x_{j-1})\\[2\jot] \ar=\sigma_f+\sigma_g, \end{array} \nonumber \] where \(\sigma_f\) and \(\sigma_g\) are Riemann sums for \(f\) and \(g\). Definition~ implies that if \(\epsilon>0\) there are positive numbers \(\delta_1\) and \(\delta_2\) such that \[\begin{eqnarray*} \left|\sigma_f-\int_a^b f(x)\,dx\right|\ar<\frac{\epsilon}{2} \mbox{\quad if\quad}\|P\|<\delta_1\\ \arraytext{and}\\ \left|\sigma_g-\int_a^b g(x)\,dx\right|\ar<\frac{\epsilon}{2} \mbox{\quad if\quad}\|P\|<\delta_2. \end{eqnarray*} \nonumber \] If \(\|P\|<\delta=\min(\delta_1,\delta_2)\), then \[\begin{eqnarray*} \left|\sigma_{f+g}-\int_a^b f(x)\,dx-\int_a^b g(x)\,dx\right| \ar=\left|\left(\sigma_f-\int_a^b f(x)\,dx\right)+ \left(\sigma_g-\int_a^b g(x)\,dx\right)\right|\\ \ar\le \left|\sigma_f-\int_a^b f(x)\,dx\right|+ \left|\sigma_g-\int_a^b g(x)\,dx\right|\\ &<&\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon, \end{eqnarray*} \nonumber \] so the conclusion follows from Definition~.

The next theorem also follows from Definition~ (Exercise~).

Theorems~ and and induction yield the following result (Exercise~).

Since \(g(x)-f(x)\ge0\), every lower sum of \(g-f\) over any partition of \([a,b]\) is nonnegative. Therefore, \[ \underline{\int_a^b}(g(x)-f(x))\,dx\ge0. \nonumber \] Hence, \[\begin{equation}\label{eq:3.3.2} \begin{array}{rcl} \dst\int_a^b g(x)\,dx-\int_a^b f(x)\,dx\ar=\dst\int_a^b (g(x)-f(x))\,dx\\[2\jot] \ar=\dst\underline{\int_a^b}(g(x)-f(x))\,dx\ge0, \end{array} \end{equation} \nonumber \] which yields . (The first equality in follows from Theorems~ and ; the second, from Theorem~.)

Let \(P\) be a partition of \([a,b]\) and define \[\begin{eqnarray*} M_j\ar=\sup\set{f(x)}{x_{j-1}\le x\le x_j},\\ m_j\ar= \inf\set{f(x)}{x_{j-1}\le x\le x_j},\\ \overline{M}_j\ar=\sup\set{|f(x)|}{x_{j-1}\le x\le x_j},\\ \overline{m}_j\ar=\inf\set{|f(x)|}{x_{j-1}\le x\le x_j}. \end{eqnarray*} \nonumber \] Then \[\begin{equation} \label{eq:3.3.4} \begin{array}{rcl} \overline{M}_j-\overline{m}_j\ar= \dst\sup\set{|f(x)|-|f(x')|}{x_{j-1}\le x,x'\le x_j}\\ \ar\le \dst\sup\set{|f(x)-f(x')|}{x_{j-1}\le x,x'\le x_j}\\ \ar=M_j-m_j. \end{array} \end{equation} \nonumber \] Therefore, \[ \overline{S}(P)-\overline{s}(P)\le S(P)-s(P), \nonumber \] where the upper and lower sums on the left are associated with \(|f|\) and those on the right are associated with \(f\). Now suppose that \(\epsilon>0\). Since \(f\) is integrable on \([a,b]\), Theorem~ implies that there is a partition \(P\) of \([a,b]\) such that \(S(P)-s(P)<\epsilon\). This inequality and imply that \(\overline S(P)-\overline s(P)<\epsilon\). Therefore, \(|f|\) is integrable on \([a,b]\), again by Theorem~.

Since \[ f(x)\le|f(x)|\mbox{\quad and \quad}-f(x)\le|f(x)|,\quad a\le x\le b, \nonumber \]

Theorems~ and imply that \[ \int_a^b f(x)\,dx\le\int_a^b|f(x)|\,dx\mbox{\quad and } -\int_a^b f(x)\,dx\le\int_a^b|f(x)|\,dx, \nonumber \] which implies .

We consider the case where \(f\) and \(g\) are nonnegative, and leave the rest of the proof to you (Exercise~). The subscripts \(f\), \(g\), and \(fg\) in the following argument identify the functions with which the various quantities are associated. We assume that neither \(f\) nor \(g\) is identically zero on \([a,b]\), since the conclusion is obvious if one of them is.

If \(P=\{x_0,x_1, \dots,x_n\}\) is a partition of \([a,b]\), then \[\begin{equation}\label{eq:3.3.5} S_{fg}(P)-s_{fg}(p)=\sum_{j=1}^n (M_{fg,j}-m_{fg, j})(x_j-x_{j-1}). \end{equation} \nonumber \] Since \(f\) and \(g\) are nonnegative, \(M_{fg,j}\le M_{f,j}M_{g,j}\) and \(m_{fg,j}\ge m_{f,j}m_{g,j}\). Hence, \[\begin{eqnarray*} M_{fg,j}-m_{fg,j}\ar\le M_{f,j}M_{g,j}-m_{f, j}m_{g,j}\\[2\jot] \ar=(M_{f,j}-m_{f,j})M_{g,j}+m_{f,j}(M_{g,j}- m_{g,j})\\[2\jot] \ar\le M_g(M_{f,j}-m_{f,j})+M_f(M_{g,j}-m_{g,j}), \end{eqnarray*} \nonumber \] where \(M_f\) and \(M_g\) are upper bounds for \(f\) and \(g\) on \([a,b]\). From and the last inequality, \[\begin{equation} \label{eq:3.3.6} S_{fg}(P)-s_{fg}(P)\le M_g[S_f(P)-s_f(P)]+M_f[S_g(P)-s_g(P)]. \end{equation} \nonumber \] Now suppose that \(\epsilon>0\). Theorem~ implies that there are partitions \(P_1\) and \(P_2\) of \([a,b]\) such that \[\begin{equation} \label{eq:3.3.7} S_f(P_1)-s_f(P_1)<\frac{\epsilon}{2M_g}\mbox{\quad and\quad} S_g(P_2)-s_g(P_2)<\frac{\epsilon}{2M_f}. \end{equation} \nonumber \] If \(P\) is a refinement of both \(P_1\) and \(P_2\), then and Lemma~ imply that \[ S_f(P)-s_f(P)<\frac{\epsilon}{2M_g}\mbox{\quad and\quad} S_g(P)-s_g(P)<\frac{\epsilon}{2M_f}. \nonumber \] This and yield \[ S_{fg}(P)-s_{fg}(P)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. \nonumber \] Therefore, \(fg\) is integrable on \([a,b]\), by Theorem~.

3pt From Theorem~, \(u\) is integrable on \([a,b]\). Therefore, Theorem~ implies that the integral on the left exists. If \(m=\min\set{u(x)}{a\le x\le b}\) and \(M=\max\set{u(x)}{a\le x\le b}\) (recall Theorem~), then \[ m\le u(x)\le M \nonumber \] and, since \(v(x)\ge0\), \[ mv(x)\le u(x) v(x)\le Mv(x). \nonumber \] Therefore, Theorems~ and imply that 2pt \[\begin{equation} \label{eq:3.3.9} m\int_a^b v(x)\,dx\le\int_a^b u(x)v(x)\,dx\le M\int_a^b v(x)\,dx. \end{equation} \nonumber \] 2pt This implies that holds for any \(c\) in \([a,b]\) if \(\int_a^b v(x)\,dx=0\). If \(\int_a^b v(x)\,dx\ne0\), let 1pt \[\begin{equation} \label{eq:3.3.10} \overline{u}=\frac{\dst\int_a^b u(x)v(x)\,dx}{\dst\int_a^bv(x)\,dx} \end{equation} \nonumber \] 1pt Since \(\int_a^b v(x)\,dx>0\) in this case (why?), implies that \(m\le\overline{u}\le M\), and the intermediate value theorem (Theorem~) implies that \(\overline{u}=u(c)\) for some \(c\) in \([a,b]\). This implies .

1pt If \(v(x)\equiv1\), then reduces to \[\overline{u}=\frac{1}{ b-a}\int_a^b u(x)\,dx, \nonumber \] so \(\overline{u}\) is the average of \(u(x)\) over \([a,b]\). More generally, if \(v\) is any nonnegative integrable function such that \(\int_a^b v(x)\,d x\ne0\), then \(\overline{u}\) in is the {}. Theorem~ says that a continuous function assumes any such weighted average at some point in \([a,b]\).

3pt

Suppose that \(\epsilon>0\). From Theorem~, there is a partition \(P=\{x_0,x_1, \dots,x_n\}\) of \([a,b]\) such that \[\begin{equation} \label{eq:3.3.11} S(P)-s(P)=\sum_{j=1}^n(M_j-m_j)(x_j-x_{j-1})<\epsilon. \end{equation} \nonumber \] We may assume that \(a_1\) and \(b_1\) are partition points of \(P\), because if not they can be inserted to obtain a refinement \(P'\) such that \(S(P')-s(P')\le S(P)-s(P)\) (Lemma~). Let \(a_1=x_r\) and \(b_1=x_s\). Since every term in is nonnegative, \[ \sum_{j=r+1}^s (M_j-m_j)(x_j-x_{j-1})<\epsilon. \nonumber \] Thus, \(\overline{P}=\{x_r,x_{r+1}, \dots,x_s\}\) is a partition of \([a_1,b_1]\) over which the upper and lower sums of \(f\) satisfy \[ S(\overline{P})-s(\overline{P})<\epsilon. \nonumber \] Therefore, \(f\) is integrable on \([a_1,b_1]\), by Theorem~.

We leave the proof of the next theorem to you (Exercise~).

So far we have defined \(\int_\alpha^\beta f(x)\,dx\) only for the case where \(\alpha<\beta\). Now we define \[ \int_\beta^\alpha f(x)\,dx=-\int_\alpha^\beta f(x)\,dx \nonumber \] if \(\alpha<\beta\), and \[ \int_\alpha^\alpha f(x)\,dx=0. \nonumber \] With these conventions, holds no matter what the relative order of \(a\), \(b\), and \(c\), provided that \(f\) is integrable on some closed interval containing them (Exercise~).

Theorem~ and these definitions enable us to define a function $ F(x)=_c^x f(t),dt$, where \(c\) is an arbitrary, but fixed, point in \([a,b]\).

If \(x\) and \(x'\) are in \([a,b]\), then \[ F(x)-F(x')=\int_c^x f(t)\,dt-\int_c^{x'} f(t)\,dt=\int_{x'}^x f(t)\, dt, \nonumber \] by Theorem~ and the conventions just adopted. Since \(|f(t)|\le K\) \((a\le t\le b)\) for some constant \(K\), \[ \left|\int_{x'}^x f(t)\,dt\right|\le K|x-x'|,\quad a\le x,\, x'\le b \nonumber \] (Theorem~), so \[ |F(x)-F(x')|\le K|x-x'|,\quad a\le x,\,x'\le b. \nonumber \] -2em2em

We consider the case where \(a<x_0<b\) and leave the rest to you (Exercise~). Since \[ \frac{1}{ x-x_0}\int_{x_0}^x f(x_0)\,dt=f(x_0), \nonumber \] we can write \[ \frac{F(x)-F(x_0)}{ x-x_0}-f(x_0)=\frac{1}{ x-x_0}\int_{x_0}^x [f(t)-f(x_0)]\,dt. \nonumber \] From this and Theorem~, \[\begin{equation}\label{eq:3.3.13} \left|\frac{F(x)-F(x_0)}{ x-x_0}-f(x_0)\right|\le \frac{1}{ |x-x_0|} \left|\int_{x_0}^x |f(t)-f(x_0)|\,dt\right|. \end{equation} \nonumber \] (Why do we need the absolute value bars outside the integral?) Since \(f\) is continuous at \(x_0\), there is for each \(\epsilon>0\) a \(\delta>0\) such that \[ |f(t)-f(x_0)|<\epsilon\mbox{\quad if\quad} |x-x_0|<\delta \nonumber \] and \(t\) is between \(x\) and \(x_0\). Therefore, from , \[ \left|\frac{F(x)-F(x_0)}{ x-x_0}-f(x_0)\right|<\epsilon \frac{|x-x_0|}{ |x-x_0|}=\epsilon\mbox{\quad if\quad} 0<|x-x_0|<\delta. \nonumber \] Hence, \(F'(x_0)=f(x_0)\).

The next theorem relates integration and differentiation in another way.

If \(P=\{x_0,x_1, \dots,x_n\}\) is a partition of \([a,b]\), then \[\begin{equation}\label{eq:3.3.15} F(b)-F(a)=\sum_{j=1}^n(F(x_j)-F(x_{j-1})). \end{equation} \nonumber \] From Theorem~, there is in each open interval \((x_{j-1},x_j)\) a point \(c_j\) such that \[ F(x_j)-F(x_{j-1})=f(c_j)(x_j-x_{j-1}). \nonumber \]

Hence, can be written as \[ F(b)-F(a)=\sum_{j=1}^nf(c_j)(x_j-x_{j-1})=\sigma, \nonumber \] where \(\sigma\) is a Riemann sum for \(f\) over \(P\). Since \(f\) is integrable on \([a,b]\), there is for each \(\epsilon>0\) a \(\delta>0\) such that \[ \left|\sigma-\int_a^b f(x)\,dx\right|<\epsilon\mbox{\quad if\quad} \|P\|<\delta. \nonumber \] Therefore, \[ \left|F(b)-F(a)-\int_a^b f(x)\,dx\right|<\epsilon \nonumber \] for every \(\epsilon>0\), which implies .

Apply Theorem~ with \(F\) and \(f\) replaced by \(f\) and \(f'\), respectively.

A function \(F\) is an {} of \(f\) on \([a,b]\) if \(F\) is continuous on \([a,b]\) and differentiable on \((a,b)\), with \[ F'(x)=f(x),\quad a<x<b. \nonumber \] If \(F\) is an antiderivative of \(f\) on \([a,b]\), then so is \(F+c\) for any constant \(c\). Conversely, if \(F_1\) and \(F_2\) are antiderivatives of \(f\) on \([a,b]\), then \(F_1-F_2\) is constant on \([a,b]\) (Theorem_{). Theorem} shows that antiderivatives can be used to evaluate integrals.

The function \(F_0(x)=\int_a^x f(t)\,dt\) is continuous on \([a,b]\) by Theorem~, and \(F_0'(x) =f(x)\) on \((a,b)\) by Theorem~. Therefore, \(F_0\) is an antiderivative of \(f\) on \([a,b]\). Now let \(F=F_0+c\) (\(c=\) constant) be an arbitrary antiderivative of \(f\) on \([a,b]\). Then -2pt \[ F(b)-F(a)=\int_a^b f(x)\,dx+c-\int_a^a f(x)\,dx-c=\int_a^b f(x)\,dx. \nonumber \] -2.5em2.5em

When applying this theorem, we will use the familiar notation \[ F(b)-F(a)=F(x)\bigg|^b_a. \nonumber \]

Since \(u\) and \(v\) are continuous on \([a,b]\) (Theorem~), they are integrable on \([a,b]\). Therefore, Theorems~ and imply that the function \[ (uv)'=u'v+uv' \nonumber \] is integrable on \([a,b]\), and Theorem~ implies that \[ \int_a^b[u(x)v'(x)+u'(x)v(x)]\,dx=u(x)v(x)\bigg|^b_a, \nonumber \] which implies .

We will use Theorem~ here and in the next section to obtain other results.

Since \(f\) is differentiable on \([a,b]\), it is continuous on \([a,b]\) (Theorem~). Since \(g\) is continuous on \([a,b]\), so is \(fg\) (Theorem~). Therefore, Theorem~ implies that the integrals in exist. If \[\begin{equation}\label{eq:3.3.18} G(x)=\int_a^x g(t)\,dt, \end{equation} \nonumber \] then \(G'(x)=g(x),\ a<x<b\) (Theorem~). Therefore, Theorem~ with \(u=f\) and \(v=G\) yields \[\begin{equation}\label{eq:3.3.19} \int_a^b f(x)g(x)\,dx=f(x)G(x)\bigg|^b_a-\int_a^b f'(x)G(x)\,dx. \end{equation} \nonumber \] Since \(f'\) is nonnegative and \(G\) is continuous, Theorem~ implies that \[\begin{equation}\label{eq:3.3.20} \int_a^b f'(x)G(x)\,dx=G(c)\int_a^b f'(x)\,dx \end{equation} \nonumber \]

for some \(c\) in \([a,b]\). From Corollary~, \[ \int_a^b f'(x)\,dx=f(b)-f(a). \nonumber \] From this and , can be rewritten as \[ \int_a^b f'(x)G(x)\,dx=(f(b)-f(a))\int_a^c g(x)\,dx. \nonumber \] Substituting this into and noting that \(G(a)=0\) yields \[\begin{eqnarray*} \int_a^b f(x)g(x)\,dx\ar=f(b)\int_a^b g(x)\,dx-(f(b)-f(a)) \int_a^c g(x)\,dx,\\ \ar=f(a)\int_a^c g(x)\,dx+f(b)\left(\int_a^b g(x)\,dx-\int_c^a g(x)\,dx\right)\\ \ar=f(a)\int_a^c g(x)\,dx+f(b)\int_c^b g(x)\,dx. \end{eqnarray*} \nonumber \] -2.5em2.5em

-.4em The following theorem on change of variable is useful for evaluating integrals.

Both integrals in exist: the one on the left by Theorem~, the one on the right by Theorems~ and and the continuity of \(f(\phi(t))\). By Theorem~, the function \[ F(x)=\int_a^x f(y)\,dy \nonumber \] is an antiderivative of \(f\) on \([a,b]\) and, therefore, also on the closed interval with endpoints \(\alpha\) and \(\beta\). Hence, by Theorem~, \[\begin{equation}\label{eq:3.3.22} \int_\alpha^\beta f(x)\,dx=F(\beta)-F(\alpha). \end{equation} \nonumber \] By the chain rule, the function \[ G(t)=F(\phi(t)) \nonumber \]

is an antiderivative of \(f(\phi(t))\phi'(t)\) on \([c,d]\), and Theorem~ implies that \[\begin{eqnarray*} \int_c^d f(\phi(t))\phi'(t)\,dt\ar= G(d)-G(c)=F(\phi(d))-F(\phi(c))\\ \ar=F(\beta)-F(\alpha). \end{eqnarray*} \nonumber \] Comparing this with yields .

These examples illustrate two ways to use Theorem~. In Example~ we evaluated the left side of by transforming it to the right side with a suitable substitution \(x=\phi(t)\), while in Example~ we evaluated the right side of by recognizing that it could be obtained from the left side by a suitable substitution.

The following theorem shows that the rule for change of variable remains valid under weaker assumptions on \(f\) if \(\phi\) is monotonic.

We consider the case where \(f\) is nonnegative and \(\phi\) is nondecreasing, and leave the the rest of the proof to you (Exercises~ and ).

First assume that \(\phi\) is increasing. We show first that \[\begin{equation}\label{eq:3.3.24} \overline{\int^b_a} f(x)\,dx=\overline{\int^d_c} f(\phi(t))\phi'(t)\,dt. \end{equation} \nonumber \] Let \(\overline{P}=\{t_0,t_1, \dots,t_n\}\) be a partition of \([c,d]\) and \(P=\{x_0,x_1, \dots,x_n\}\) with \(x_j=\phi(t_j)\) be the corresponding partition of \([a,b]\). Define \[\begin{eqnarray*} U_j\ar=\sup\set{\phi'(t)}{t_{j-1}\le t\le t_j},\\ u_j\ar=\inf\set{\phi'(t)}{t_{j-1}\le t\le t_j},\\ M_j\ar=\sup\set{f(x)}{x_{j-1}\le x\le x_j},\\ \arraytext{and}\\ \overline{M}_j\ar=\sup\set{f(\phi(t))\phi'(t)}{t_{j-1}\le t\le t_j}. \end{eqnarray*} \nonumber \] Since \(\phi\) is increasing, \(u_j\ge0\). Therefore, \[ 0\le u_j\le\phi'(t)\le U_j,\quad t_{j-1}\le t\le t_j. \nonumber \] Since \(f\) is nonnegative, this implies that \[ 0\le f(\phi(t))u_j\le f(\phi(t))\phi'(t)\le f(\phi(t))U_j,\quad t_{j-1}\le t\le t_j. \nonumber \] Therefore, \[ M_ju_j\le \overline{M}_j\le M_jU_j, \nonumber \]

which implies that \[\begin{equation}\label{eq:3.3.25} \overline{M}_j=M_j\rho_j, \end{equation} \nonumber \] where \[\begin{equation}\label{eq:3.3.26} u_j\le\rho_j\le U_j. \end{equation} \nonumber \]

Now consider the upper sums \[\begin{equation}\label{eq:3.3.27} \overline{S}(\overline{P})=\sum_{j=1}^n\overline{M}_j (t_j-t_{j-1}) \mbox{\quad and\quad} S(P)=\sum_{j=1}^n M_j(x_j-x_{j-1}). \end{equation} \nonumber \] From the mean value theorem, \[\begin{equation}\label{eq:3.3.28} x_j-x_{j-1}=\phi(t_j)-\phi(t_{j-1})=\phi'(\tau_j)(t_j-t_{j-1}), \end{equation} \nonumber \] where \(t_{j-1}<\tau_j<t_j\), so \[\begin{equation}\label{eq:3.3.29} u_j\le\phi'(\tau_j)\le U_j. \end{equation} \nonumber \] From , , and , \[\begin{equation}\label{eq:3.3.30} \overline{S}(\overline{P})-S(P)=\sum_{j=1}^n M_j(\rho_j-\phi'(\tau_j))(t_j-t_{j-1}). \end{equation} \nonumber \]

Now suppose that \(|f(x)|\le M\), \(a\le x\le b\). Then , , and imply that \[ \left|\overline{S}(\overline{P})-S(P)\right|\le M\sum_{j=1}^n (U_j-u_j)(t_j-t_{j-1}). \nonumber \] The sum on the right is the difference between the upper and lower sums of \(\phi'\) over \(\overline{P}\). Since \(\phi'\) is integrable on \([c,d]\), this can be made as small as we please by choosing \(\|\overline{P}\|\) sufficiently small (Exercise~).

From , \(\|P\|\le K\|\overline{P}\|\) if \(|\phi'(t)|\le K\), \(c\le t\le d\). Hence, Lemma~ implies that \[\begin{equation} \label{eq:3.3.31} \left| S(P)-\overline{\int_a^b} f(x)\,dx\right|<\frac{\epsilon}{3} \mbox{\quad and\quad}\left|\overline{S}(\overline{P}) -\overline{\int_c^d} f(\phi(t))\phi'(t)\,dt\right|<\frac{\epsilon}{3} \end{equation} \nonumber \] if \(\|\overline{P}\|\) is sufficiently small. Now \[\begin{eqnarray*} \left|\overline{\int_a^b} f(x)\,dx-\overline{\int_c^d} f\left(\phi (t)\right)\phi'(t)\,dt\right|\ar\le \left|\overline{\int_a^b} f(x)\, dx-S(P)\right| +|S(P)-\overline{S}(\overline{P})|\\&&+\left| \overline{S}(\overline{P})-\overline{\int_c^d} f(\phi(t)) \phi'(t)\,dt\right|. \end{eqnarray*} \nonumber \] Choosing \(\overline{P}\) so that \(|S(P)-\overline{S}(\overline{P}|< \epsilon/3\) in addition to yields \[ \left|\int_a^b f(x)\,dx-\int_c^d f(\phi(t))\phi'(t)\,dt \right|<\epsilon. \nonumber \] Since \(\epsilon\) is an arbitrary positive number, this implies .

If \(\phi\) is nondecreasing (rather than increasing), it may happen that \(x_{j-1}=x_j\) for some values of \(j\); however, this is no real complication, since it simply means that some terms in \(S(P)\) vanish.

By applying to \(-f\), we infer that \[\begin{eqnarray} \underline{\int_a^b} f(x)\,dx\ar=\underline{\int_c^d} f(\phi(t))\phi'(t)\,dt,\label{eq:3.3.32}\\ [2\jot] \arraytext{since}\nonumber\\ \overline{\int_a^b}(-f)(x)\,dx\ar=-\underline{\int_a^b}f(x)\,dx\nonumber\\ \arraytext{and}\nonumber\\ \overline{\int_c^d}(-f(\phi(t)\phi'(t))\,dt\ar=-\underline{\int_c^d} f(\phi(t))\phi'(t)\,dt.\nonumber \end{eqnarray} \nonumber \]

Now suppose that \(f\) is integrable on \([a,b]\). Then \[ \underline{\int_a^b} f(x)\,dx=\overline{\int_a^b}f(x)\,dx= \int_a^bf(x)\,dx, \nonumber \] by Theorem~. From this, , and , \[ \underline{\int^d_c}f(\phi(t))\phi'(t)\,dt= \overline{\int^d_c}f(\phi(t))\phi'(t)\,dt= \int^b_a f(x)\,dx. \nonumber \] This and Theorem~ (applied to \(f(\phi(t))\phi'(t)\)) imply that \(f(\phi(t))\phi'(t)\) is integrable on \([c,d]\) and \[\begin{equation} \label{eq:3.3.33} \int_a^b f(x)\,dx=\int_c^d f(\phi(t))\phi'(t)\,dt. \end{equation} \nonumber \] A similar argument shows that if \(f(\phi(t))\phi'(t)\) is integrable on \([c,d]\), then \(f\) is integrable on \([a,b]\), and holds.

\begin{exerciselist}

Prove Theorem~. 4pt

Prove Theorem~.

4pt

Can \(|f|\) be integrable on \([a,b]\) if \(f\) is not? 4pt

Complete the proof of Theorem~.

4pt

Prove: If \(f\) is integrable on \([a,b]\) and \(|f(x)|\ge\rho>0\) for \(a\le x\le b\), then \(1/f\) is integrable on \([a,b]\)

Suppose that \(f\) is integrable on \([a,b]\) and define \[ f^+(x)=\left\{\casespace\begin{array}{ll} f(x)&\mbox{\quad if $f(x)\ge0,$}\\[2\jot] 0&\mbox{\quad if $f(x)<0$,}\end{array}\right. \mbox{ and\quad } f^-(x)=\left\{\casespace\begin{array}{ll} 0& \mbox{\quad if $f(x)\ge0$},\\[2\jot] f(x)&\mbox{\quad if $f(x)<0$.\quad} \end{array}\right. \nonumber \] Show that \(f^+\) and \(f^-\) are integrable on \([a,b]\), and \[ \int_a^b f(x)\,dx=\int_a^b f^+(x)\,dx+\int_a^b f^-(x)\,dx. \nonumber \]

Find the weighted average \(\overline{u}\) of \(u(x)\) over \([a,b]\) with respect to \(v\), and find a point \(c\) in \([a,b]\) such that \(u(c)=\overline{u}\).

Prove Theorem~.

Show that \[ \int_a^c f(x)\,dx=\int_a^b f(x)\,dx+\int_b^c f(x)\,dx \nonumber \] for all possible relative orderings of \(a\), \(b\), and \(c\), provided that \(f\) is integrable on a closed interval containing them.

Prove: If \(f\) is integrable on \([a,b]\) and \(a=a_0<a_1<\cdots<a_n=b\), then \[ \int_a^bf(x)\,dx=\int_{a_0}^{a_1}f(x)\,dx+\int_{a_1}^{a_2}f(x)\,dx +\cdots+\int_{a_{n-1}}^{a_n}f(x)\,dx. \nonumber \]

Suppose that \(f\) is continuous on \([a,b]\) and \(P=\{x_0,x_1, \dots,x_n\}\) is a partition of \([a,b]\). Show that there is a Riemann sum of \(f\) over \(P\) that {} \(\int_a^b f(x)\,dx\).

Suppose that \(f'\) exists and \(|f'(x)|\le M\) on \([a,b]\). Show that any Riemann sum \(\sigma\) of \(f\) over any partition \(P\) of \([a,b]\) satisfies \[ \left|\sigma-\int_a^b f(x)\,dx\right|\le M(b-a)\|P\|. \nonumber \]

Prove: If \(f\) is integrable and \(f(x)\ge0\) on \([a,b]\), then \(\int_a^b f(x)\,dx\ge0\), with strict inequality if \(f\) is continuous and positive at some point in \([a,b]\).

Complete the proof of Theorem~.

State theorems analogous to Theorems~ and for the function \[ G(x)=\int_x^c f(t)\,dt, \nonumber \] and show how your theorems can be obtained from them.

The symbol \(\int f(x)\,dx\) denotes an antiderivative of \(f\). A plausible analog of Theorem~ would state that if \(f\) and \(g\) have antiderivatives on \([a,b]\), then so does \(f+g\), which is true, and \[ \int(f+g)(x)\,dx=\int f(x)\,dx+\int g(x)\,dx. \eqno{\rm(A)} \nonumber \] However, this is not true in the usual sense.

(See Exercise~.) Formulate a valid interpretation of the relation \[ \int(cf)(x)\,dx=c\int f(x)\,dx\quad (c\ne0). \nonumber \] Is your interpretation valid if \(c=0\)?

In addition to the assumptions of Theorem~, suppose that \(f(a)=0\), \(f\not\equiv0\), and \(g(x)>0\) \((a<x<b)\). Show that there is only one point \(c\) in \([a,b]\) with the property stated in Theorem~.

Assuming that Theorem~ is true under the additional assumption that \(f\) is nonnegative on \([a,b]\), show that it is true without this assumption.

Assuming that the conclusion of Theorem~ is true if \(\phi\) is nondecreasing, show that it is true if \(\phi\) is nonincreasing.

Suppose \(g'\) is integrable and \(f\) is continuous on \([a,b]\). Show that \(\int_a^b f(x)\,dg(x)\) exists and equals \(\int_a^b f(x)g'(x)\,dx\).

Suppose \(f\) and \(g''\) are bounded and \(fg'\) is integrable on \([a,b]\). Show that \(\int_a^b f(x)\,dg(x)\) exists and equals \(\int_a^b f(x)g'(x)\,dx\).

\end{exerciselist}

So far we have confined our study of the integral to bounded functions on finite closed intervals. This was for good reasons:

In this section we extend the definition of integral to include cases where \(f\) is unbounded or the interval is unbounded, or both.

We say \(f\) is {} on an interval \(I\) if \(f\) is integrable on every finite closed subinterval of \(I\). For example, \[ f(x)=\sin x \nonumber \] is locally integrable on \((-\infty,\infty)\); \[ g(x)=\frac{1}{ x(x-1)} \nonumber \] is locally integrable on \((-\infty,0)\), \((0,1)\), and \((1,\infty)\); and \[ h(x)=\sqrt{x} \nonumber \] is locally integrable on \([0,\infty)\).

The limit in always exists if \([a,b)\) is finite and \(f\) is locally integrable and bounded on \([a,b)\). In this case, Definitions~ and assign the same value to \(\int_a^b f(x)\,dx\) no matter how \(f(b)\) is defined (Exercise~). However, the limit may also exist in cases where \(b=\infty\) or \(b<\infty\) and \(f\) is unbounded as \(x\) approaches \(b\) from the left. In these cases, Definition~ assigns a value to an integral that does not exist in the sense of Definition~, and \(\int_a^b f(x)\, dx\) is said to be an {} that {} to the limit in . We also say in this case that \(f\) is {} \([a,b)\) and that \(\int_a^b f(x)\,dx\) {}. If the limit in does not exist (finite), we say that the improper integral \(\int_a^b f(x)\,dx\) {}, and \(f\) is {} \([a,b)\). In particular, if \(\lim_{c\to b-}\int_a^c f(x)\,dx=\pm\infty\), we say that \(\int_a^b f(x)\,dx\) {} \(\pm\infty\), and we write \[ \int_a^b f(x)\,dx=\infty\mbox{\quad or\quad}\int_a^b f(x)\,dx=- \infty, \nonumber \] whichever the case may be.

Similar comments apply to the next two definitions.

The existence and value of \(\int_a^b f(x)\,dx\) according to Definition~ do not depend on the particular choice of \(\alpha\) in \((a,b)\) (Exercise~).

When we wish to distinguish between improper integrals and integrals in the sense of Definition~, we will call the latter {}.

In stating and proving theorems on improper integrals, we will consider integrals of the kind introduced in Definition~. Similar results apply to the integrals of Definitions~ and . We leave it to you to formulate and use them in the examples and exercises as the need arises.

If \(a<c<b\), then \[\begin{eqnarray*} \int_a^c (c_1f_1+c_2f_2+\cdots+c_nf_n)(x)\,dx\ar=c_1\int_a^c f_1(x)\,dx +c_2\int_a^c f_2(x)\,dx\\ \ar{}+\cdots+c_n\int_a^c f_n(x)\,dx, \end{eqnarray*} \nonumber \] by Theorem~. Letting \(c\to b-\) yields the stated result.

The theory of improper integrals of nonnegative functions is particularly simple.

2pt Since \(F\) is nondecreasing on \([a,b)\), Theorem~

implies the conclusion.

We often write 1pt \[ \int_a^b f(x)\,dx<\infty \nonumber \] 2pt

to indicate that an improper integral of a nonnegative function converges. Theorem~ justifies this convention, since it asserts that a divergent integral of this kind can only diverge to \(\infty\). Similarly, if \(f\) is nonpositive and \(\int_a^b f(x)\,dx\) converges, we write 1pt \[ \int_a^b f(x)\,dx>-\infty \nonumber \] 2pt

because a divergent integral of this kind can only diverge to \(-\infty\). (To see this, apply Theorem~ to \(-f\).) These conventions do not apply to improper integrals of functions that assume both positive and negative values in \((a,b)\), since they may diverge without diverging to \(\pm\infty\).

4pt

Assumption implies that \[ \int_a^x f(t)\,dt\le\int_a^x g(t)\,dt,\quad a\le x<b \nonumber \] (Theorem~), so \[ \sup_{a\le x<b}\,\int_a^x f(t)\,dt\le\sup_{a\le x\le b}\,\int_a^x g(t)\,dt. \nonumber \] If \(\int_a^b g(x)\,dx<\infty\), the right side of this inequality is finite by Theorem~, so the left side is also. This implies that \(\int_a^b f(x)\,dx<\infty\), again by Theorem~.

The proof is by contradiction. If \(\int_a^bg(x)\,dx<\infty\), then

implies that \(\int_a^bf(x)\,dx<\infty\), contradicting the assumption that \(\int_a^bf(x)\,dx=\infty\).

The comparison test is particularly useful if the integrand of the improper integral is complicated but can be compared with a function that is easy to integrate.

If \(f\) is any function (not necessarily nonnegative) locally integrable on \([a,b)\), then \[ \int_a^c f(x)\,dx=\int_a^{a_1} f(x)\,dx+\int_{a_1}^c f(x)\,dx \nonumber \] if \(a_1\) and \(c\) are in \([a,b)\). Since \(\int_a^{a_1}f(x)\,dx\) is a proper integral, on letting \(c\to b-\) we conclude that if either of the improper integrals \(\int_a^b f(x)\,dx\) and \(\int_{a_1}^b f(x)\,dx\) converges then so does the other, and in this case \[ \int_a^b f(x)\,dx=\int_a^{a_1} f(x)\,dx+\int_{a_1}^b f(x)\,dx. \nonumber \]

This means that any theorem implying convergence or divergence of an improper integral \(\int_a^b f(x)\,dx\) in the sense of Definition~ remains valid if its hypotheses are satisfied on a subinterval \([a_1,b)\) of \([a,b)\) rather than on all of \([a,b)\). For example, Theorem~ remains valid if is replaced by \[ 0\le f(x)\le g(x),\quad a_1\le x<b, \nonumber \] where \(a_1\) is any point in \([a,b)\).

From this, you can see that if \(f(x)\ge0\) on some subinterval \([a_1,b)\) of \([a,b)\), but not necessarily for all \(x\) in \([a,b)\), we can still use the convention introduced earlier for positive functions; that is, we can write \(\int_a^bf(x)\,dx<\infty\) if the improper integral converges or \(\int_a^bf(x)\,dx=\infty\) if it diverges.

From , there is a point \(a_2\) in \([a_1,b)\) such that \[ 0<\frac{M}{2}<\frac{f(x)}{ g(x)}<\frac{3M}{2},\quad a_2\le x<b, \nonumber \] and therefore \[\begin{equation} \label{eq:3.4.4} \frac{M}{2} g(x)<f(x)<\frac{3M}{2} g(x),\quad a_2\le x<b. \end{equation} \nonumber \] Theorem~ and the first inequality in imply that \[ \int_{a_2}^b g(x)\,dx<\infty\mbox{\quad if \quad} \int_{a_2}^b f(x)\,dx<\infty. \nonumber \]

Theorem~ and the second inequality in imply that \[ \int_{a_2}^b f(x)\,dx<\infty \mbox{\quad if \quad} \int_{a_2}^b g(x)\,dx<\infty. \nonumber \] Therefore, \(\int_{a_2}^b f(x)\,dx\) and \(\int_{a_2}^b g(x)\,dx\) converge or diverge together, and in the latter case they must diverge to \(\infty\), since their integrands are nonnegative (Theorem~).

If \(M=\infty\), there is a point \(a_2\) in \([a_1,b)\) such that \[ f(x)\ge g(x),\quad a_2\le x\le b, \nonumber \] so Theorem~

implies that \(\int_a^bf(x)\,dx=\infty\).

If \(M=0\), there is a point \(a_2\) in \([a_1,b)\) such that \[ f(x)\le g(x),\quad a_2\le x\le b, \nonumber \] so Theorem~

implies that \(\int_a^bf(x)\,dx<\infty\).

The hypotheses of Theorem~ and do not imply that \(\int_a^bf(x)\,dx\) and \(\int_a^bg(x)\,dx\) necessarily converge or diverge together. For example, if \(b=\infty\), then \(f(x)=1/x\) and \(g(x)=1/x^2\) satisfy the hypotheses of Theorem~ , while \(f(x)=1/x^2\) and \(g(x)=1/x\) satisfy the hypotheses of Theorem~

. However, \(\int_1^\infty 1/x\,dx=\infty\), while \(\int_1^\infty 1/x^2\,dx<\infty\).

1em

{}

1em

1em

If \[ g(x)=|f(x)|-f(x), \nonumber \]

then \[ 0\le g(x)\le2|f(x)| \nonumber \] and \(\int_a^b g(x)\,dx<\infty\), because of Theorem~ and the absolute integrability of \(f\). Since \[ f=|f|-g, \nonumber \] Theorem~ implies that \(\int_a^b f(x)\,dx\) converges.

-.15em We say that \(f\) is {} at \(b\negthickspace-(=\infty\mbox{\quad if\quad} b=\infty)\) if \(f\) is defined on \([a,b)\) and does not change sign on some subinterval \([a_1,b)\) of \([a,b)\). If \(f\) changes sign on every such subinterval, \(f\) is {} at \(b-\). For a function that is locally integrable on \([a,b)\) and nonoscillatory at \(b-\), convergence and absolute convergence of \(\int_a^b f(x)\,dx\) amount to the same thing (Exercise~), so absolute convergence is not an interesting concept in connection with such functions. However, an oscillatory function may be integrable, but not absolutely integrable, on \([a,b)\), as the next example shows. We then say that \(f\) is {} integrable on \([a,b)\), and that \(\int_a^b f(x)\,dx\) converges {}.

The method used in Example~ is a special case of the following test for convergence of improper integrals.

The continuous function \(fg\) is locally integrable on \([a,b)\). Integration by parts yields \[\begin{equation}\label{eq:3.4.10} \int_a^c f(x)g(x)\,dx=F(c)g(c)-\int_a^c F(x)g'(x)\,dx,\quad a\le c<b. \end{equation} \nonumber \] Theorem~ implies that the integral on the right converges absolutely as \(c\to b-\), since \(\int_a^b |g'(x)|\,dx<\infty\) by assumption, and \[ |F(x)g'(x)|\le M|g'(x)|, \nonumber \] where \(M\) is an upper bound for \(|F|\) on \([a,b)\). Moreover, and the boundedness of \(F\) imply that \(\lim_{c\to b-}F(c)g(c)=0\). Letting \(c\to b-\) in yields \[ \int_a^b f(x)g(x)\,dx=-\int_a^b F(x)g'(x)\,dx, \nonumber \] where the integral on the right converges absolutely.

Dirichlet’s test is useful only if \(f\) is oscillatory at \(b-\), since it can be shown that if \(f\) is nonoscillatory at \(b-\) and \(F\) is bounded on \([a,b)\), then \(\int_a^b |f(x)g(x)|\,dx<\infty\) if only \(g\) is locally integrable and bounded on \([a,b)\) (Exercise~).

The method used in Example~ is a special case of the following test for divergence of improper integrals.

The proof is by contradiction. Let \(f=uv\) and \(g=1/v\), and suppose that \(\int_a^bu(x)v(x)\,dx\) converges. Then \(f\) has the bounded antiderivative \(F(x)=\int_a^xu(t)v(t)\,dt\) on \([a,b)\), \(\lim_{x\to\infty}g(x)=0\) and \(g'=-v'/v^2\) is absolutely integrable on \([a,b)\). Therefore, Theorem~ implies that \(\int_a^b u(x)\,dx\) converges, a contradiction.

If Dirichlet’s test shows that \(\int_a^b f(x)g(x)\,dx\) converges, there remains the question of whether it converges absolutely or conditionally. The next theorem sometimes answers this question. Its proof can be modeled after the method of Example~ (Exercise~). The idea of an infinite sequence, which we will discuss in Section~4.1, enters into the statement of this theorem. We assume that you recall the concept sufficiently well from calculus to understand the meaning of the theorem.

-.4em The next theorem enables us to investigate an improper integral by transforming it into another whose convergence or divergence is known. It follows from Theorem~ and Definitions~, , and . We omit the proof.

In Section~3.2 we found necessary and sufficient conditions for existence of the proper Riemann integral, and in Section~3.3 we used them to study the properties of the integral. However, it is awkward to apply these conditions to a specific function and determine whether it is integrable, since they require computations of upper and lower sums and upper and lower integrals, which may be difficult. The main result of this section is an integrability criterion due to that does not require computation, but has to do with how badly discontinuous a function may be and still be integrable.

We emphasize that we are again considering proper integrals of bounded functions on finite intervals.

For a fixed \(x\) in \((a,b)\), \(W_f(x-h,x+h)\) is a nonnegative and nondecreasing function of \(h\) for \(0<h<\min(x-a,b-x)\); therefore, \(w_f(x)\) exists and is nonnegative, by Theorem~. Similar arguments apply to \(w_f(a)\) and \(w_f(b)\).

Suppose that \(a<x_0<b\). First, suppose that \(w_f(x_0)=0\) and \(\epsilon>0\). Then \[ W_f[x_0-h,x_0+h]<\epsilon \nonumber \] for some \(h>0\), so \[ |f(x)-f(x')|<\epsilon\mbox{\quad if\quad} x_0-h\le x,x'\le x_0+h. \nonumber \] Letting \(x'=x_0\), we conclude that \[ |f(x)-f(x_0)|<\epsilon\mbox{\quad if\quad} |x-x_0|<h. \nonumber \] Therefore, \(f\) is continuous at \(x_0\).

Conversely, if \(f\) is continuous at \(x_0\) and \(\epsilon>0\), there is a \(\delta>0\) such that \[ |f(x)-f(x_0)|<\frac{\epsilon}{2}\mbox{\quad and\quad} |f(x')-f(x_0)|< \frac{\epsilon}{2} \nonumber \] if \(x_0-\delta\le x\), \(x'\le x_0+\delta\). From the triangle inequality, \[ |f(x)-f(x')|\le|f(x)-f(x_0)|+|f(x')-f(x_0)|<\epsilon, \nonumber \] so \[ W_f[x_0-h,x_0+h]\le\epsilon\mbox{\quad if\quad} h<\delta; \nonumber \] therefore, \(w_f(x_0)=0\). Similar arguments apply if \(x_0=a\) or \(x_0=b\).

We use the Heine–Borel theorem (Theorem~). If \(w_f(x)<\epsilon\), there is an \(h_x>0\) such that \[\begin{equation} \label{eq:3.5.1} |f(x')-f(x'')|<\epsilon \end{equation} \nonumber \]

if \[\begin{equation} \label{eq:3.5.2} x-2h_x<x',x''<x+2h_x\mbox{\quad and\quad} x',x'' \in [a,b]. \end{equation} \nonumber \] If \(I_x=(x-h_x,x+h_x)\), then the collection \[ {\mathcal H}=\set{I_x}{a\le x\le b} \nonumber \] is an open covering of \([a,b]\), so the Heine–Borel theorem implies that there are finitely many points \(x_1\), \(x_2\), , \(x_n\) in \([a,b]\) such that \(I_{x_1}\), \(I_{x_2}\), , \(I_{x_n}\) cover \([a,b]\). Let \[ h=\min_{1\le i\le n}h_{x_i} \nonumber \] and suppose that \([a_1,b_1]\subset [a,b]\) and \(b_1-a_1<h\). If \(x'\) and \(x''\) are in \([a_1,b_1]\), then \(x'\in I_{x_r}\) for some \(r\ (1\le r \le n)\), so \[ |x'-x_r|<h_{x_r}. \nonumber \] Therefore, \[\begin{eqnarray*} |x''-x_r|\ar\le |x''-x'|+|x'-x_r| <b_1-a_1+h_{x_r}\\ &<&h+h_{x_r} \le2h_{x_r}. \end{eqnarray*} \nonumber \] Thus, any two points \(x'\) and \(x''\) in \([a_1,b_1]\) satisfy with \(x=x_r\), so they also satisfy . Therefore, \(\epsilon\) is an upper bound for the set \[ \set{|f(x')-f(x'')|}{x',x''\in [a_1,b_1]}, \nonumber \] which has the supremum \(W_f[a_1,b_1]\). Hence, \(W_f[a_1,b_1]\le\epsilon\).

In the following, \(L(I)\) is the length of the interval \(I\).

We first show that \(E_\rho\) is closed. Suppose that \(x_0\) is a limit point of \(E_\rho\). If \(h>0\), there is an \(\overline{x}\) from \(E_\rho\) in \((x_0-h,x_0+h)\). Since \([\overline{x}-h_1,\overline{x}+h_1] \subset [x_0-h,x_0+h]\) for sufficiently small \(h_1\) and \(W_f[\overline{x}-h_1,\overline{x}+h_1]\ge\rho\), it follows that \(W_f[x_0-h,x_0+h]\ge\rho\) for all \(h>0\). This implies that \(x_0\in E_\rho\), so \(E_\rho\) is closed (Corollary~).

Now we will show that the stated condition in necessary for integrability. Suppose that the condition is not satisfied; that is, there is a \(\rho>0\) and a \(\delta>0\) such that \[ \sum_{j=1}^p L(I_j)\ge\delta \nonumber \]

for every finite set \(\{I_1,I_2, \dots, I_p\}\) of open intervals covering \(E_\rho\). If \(P= \{x_0,x_1, \dots,x_n\}\) is a partition of \([a,b]\), then \[\begin{equation} \label{eq:3.5.4} S(P)-s(P)=\sum_{j\in A} (M_j-m_j)(x_j-x_{j-1})+\sum_{j\in B} (M_j-m_j)(x_j-x_{j-1}), \end{equation} \nonumber \] where \[ A=\set{j}{[x_{j-1},x_j]\cap E_\rho\ne\emptyset}\mbox{\quad and\quad} B=\set{j}{[x_{j-1},x_j]\cap E_\rho=\emptyset}\negthickspace. \nonumber \]

Since \(\bigcup_{j\in A} (x_{j-1},x_j)\) contains all points of \(E_\rho\) except any of \(x_0\), \(x_1\), , \(x_n\) that may be in \(E_\rho\), and each of these finitely many possible exceptions can be covered by an open interval of length as small as we please, our assumption on \(E_\rho\) implies that \[ \sum_{j\in A} (x_j-x_{j-1})\ge\delta. \nonumber \] Moreover, if \(j\in A\), then \[ M_j-m_j\ge\rho, \nonumber \] so implies that \[ S(P)-s(P)\ge\rho\sum_{j\in A} (x_j-x_{j-1})\ge\rho\delta. \nonumber \] Since this holds for every partition of \([a,b]\), \(f\) is not integrable on \([a,b]\), by Theorem~. This proves that the stated condition is necessary for integrability.

For sufficiency, let \(\rho\) and \(\delta\) be positive numbers and let \(I_1\), \(I_2\), , \(I_p\) be open intervals that cover \(E_\rho\) and satisfy . Let \[ \widetilde{I}_j=[a,b]\cap\overline{I}_j. \nonumber \] (\(\overline{I}_j=\mbox{closure of } I\).) After combining any of \(\widetilde{I}_1\), \(\widetilde{I}_2\), , \(\widetilde{I}_p\) that overlap, we obtain a set of pairwise disjoint closed subintervals \[ C_j=[\alpha_j,\beta_j],\quad 1\le j\le q\ (\le p), \nonumber \] of \([a,b]\) such that \[\begin{equation} \label{eq:3.5.5} a\le\alpha_1<\beta_1<\alpha_2<\beta_2\cdots< \alpha_{q-1}<\beta_{q-1}<\alpha_q<\beta_q\le b, \end{equation} \nonumber \] \[\begin{equation} \label{eq:3.5.6} \sum_{i=1}^q\, (\beta_i-\alpha_i)<\delta \end{equation} \nonumber \] and \[ w_f(x)<\rho,\quad\beta_j\le x\le\alpha_{j+1},\quad 1\le j\le q-1. \nonumber \] Also, \(w_f(x)<\rho\) for \(a\le x\le\alpha_1\) if \(a<\alpha_1\) and for \(\beta_q\le x\le b\) if \(\beta_q<b\).

Let \(P_0\) be the partition of \([a,b]\) with the partition points indicated in , and refine \(P_0\) by partitioning each subinterval \([\beta_j,\alpha_{j+1}]\) (as well as \([a,\alpha_1]\) if \(a<\alpha_1\) and \([\beta_q,b]\) if \(\beta_q<b\)) into subintervals on which the oscillation of \(f\) is not greater than \(\rho\). This is possible by Lemma~. In this way, after renaming the entire collection of partition points, we obtain a partition \(P=\{x_0,x_1, \dots,x_n\}\) of \([a,b]\) for which \(S(P)-s(P)\) can be written as in , with \[ \sum_{j\in A}\, (x_j-x_{j-1})=\sum_{i=1}^q\, (\beta_i-\alpha_i)<\delta \nonumber \] (see ) and \[ M_j-m_j\le\rho,\quad j\in B. \nonumber \] For this partition, \[ \sum_{j\in A}\, (M_j-m_j)(x_j-x_{j-1})\le2K\sum_{j\in A}\,(x_j-x_{j-1})<2K\delta, \nonumber \] where \(K\) is an upper bound for \(|f|\) on \([a,b]\) and \[ \sum_{j\in B}\, (M_j-m_j)(x_j-x_{j-1})\le\rho(b-a). \nonumber \] We have now shown that if \(\rho\) and \(\delta\) are arbitrary positive numbers, there is a partition \(P\) of \([a,b]\) such that \[\begin{equation} \label{eq:3.5.7} S(P)-s(P)<2K\delta+\rho(b-a). \end{equation} \nonumber \] If \(\epsilon>0\), let \[ \delta=\frac{\epsilon}{4K}\mbox{\quad and\quad}\rho=\frac{\epsilon}{ 2(b-a)}. \nonumber \] Then yields \[ S(P)-s(P)<\epsilon, \nonumber \] and Theorem~ implies that \(f\) is integrable on \([a,b]\).

We need the next definition to state Lebesgue’s integrability condition.

-3em

Note that any subset of a set of Lebesgue measure zero is also of Lebesgue measure zero. (Why?)

-2em

There are also nondenumerable sets of Lebesgue measure zero, but it is beyond the scope of this book to discuss examples.

The next theorem is the main result of this section.

From Theorem~, \[ S=\set{x\in [a,b]}{w_f(x)>0}\negthickspace. \nonumber \] Since \(w_f(x)>0\) if and only if \(w_f(x)\ge1/i\) for some positive integer \(i\), we can write \[\begin{equation} \label{eq:3.5.12} S=\bigcup^\infty_{i=1} S_i, \end{equation} \nonumber \] where \[ S_i=\set{x\in [a,b]}{w_f(x)\ge1/i}. \nonumber \]

Now suppose that \(f\) is integrable on \([a,b]\) and \(\epsilon>0\). From Lemma~, each \(S_i\) can be covered by a finite number of open intervals \(I_{i1}\), \(I_{i2}\), , \(I_{in}\) of total length less than \(\epsilon/2^i\). We simply renumber these intervals consecutively; thus, \[ I_1,I_2, \dots= I_{11}, \dots,I_{1n_1},I_{21}, \dots,I_{2n_2}, \dots, I_{i1}, \dots,I_{in_i}, \dots. \nonumber \] Now and hold because of and , and we have shown that the stated condition is necessary for integrability.

For sufficiency, suppose that the stated condition holds and \(\epsilon>0\). Then \(S\) can be covered by open intervals \(I_1,I_2, \dots\) that satisfy . If \(\rho>0\), then the set \[ E_\rho=\set{x\in [a,b]}{w_f(x)\ge\rho} \nonumber \] of Lemma~ is contained in \(S\) (Theorem~), and therefore \(E_\rho\) is covered by \(I_1,I_2, \dots\). Since \(E_\rho\) is closed (Lemma~) and bounded, the Heine–Borel theorem implies that \(E_\rho\) is covered by a finite number of intervals from \(I_1,I_2, \dots\). The sum of the lengths of the latter is less than \(\epsilon\), so Lemma~ implies that \(f\) is integrable on \([a,b]\).