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Five basic properties of the definite integral are listed at the beginning of Section 4 of Chapter 4. Of these, two are proved in the text and one is left as an exercise. The remaining two will be proved here. Let [math]f[/math] be a function which is bounded on a closed interval [math][a, b][/math]. This implies that [math][a, b][/math] is contained in the domain of [math]f[/math] and that there exists a positive number [math]B[/math] such that [math]|f(x)| \lt B[/math] for all [math]x[/math] in [math][a, b][/math]. We recall that, for every partition [math]\sigma[/math] of [math][a, b][/math], there are defined the upper and lower sums for [math]f[/math] relative to [math]\sigma[/math], which are denoted by [math]U_\sigma[/math] and [math]L_\sigma[/math], respectively. Moreover, it has been shown (see page 168) that

[[math]] L_\sigma \leq L_{\sigma \cup \tau} \leq U_{\sigma \cup \tau} \leq U_\tau, ( 1 ) [[/math]]

for any two partitions [math]\sigma[/math] and [math]\tau[/math] of [math][a, b][/math]. The function [math]f[/math] is defined to be integrable over [math][a, b][/math] if there exists one and only one number, denoted [math]\int_a^b f[/math], with the property that

[[math]] L_\sigma \leq \int_a^b f \leq U_\tau , [[/math]]

for any two partitions [math]\sigma[/math] and [math]\tau[/math] of [math][a, b][/math]. It is an immediate consequence of this definition and the inequalities (1) that [math]f[/math] is integrable over [math][a, b][/math] if and only if, for any positive number [math]\epsilon[/math], there exists a partition [math]\sigma[/math] of [math][a, b][/math] such that [math]U_\sigma - L_\sigma \lt \epsilon[/math]. A similar corollary, which we shall also usebin the subsequent proofs, is the statement that [math]f[/math] is integrable over [math][a, b][/math] and [math]\int_a^b f = J[/math] if and only if, for every positive number [math]\epsilon[/math], there exists a partition [math]\sigma[/math] of [math][a, b][/math] such that [math]|U_\sigma - J| \lt \epsilon[/math] and [math]|J - L_\sigma| \lt \epsilon[/math].

The first property of the definite integral, which we shall establish in this section, is presented in the following theorem:

THEOREM 1

The function [math]f[/math] is integrable over the intervals [math][a, b][/math] and [math][b, c][/math] if and only if it is integrable over their union [math][a, c][/math]. Furthermore,

[[math]] \int_a^b f + \int_b^c f = \int_a^c f. [[/math]]

Show Proof

We first assume that [math]f[/math] is integrable over [math][a, b][/math] and over [math][b, c][/math]. Let [math]\epsilon[/math] be an arbitrary positive number. Then there exists a partition [math]\sigma_1[/math] of [math][a, b][/math], and a partition [math]\sigma_2[/math] of [math][b, c][/math] such that the following inequalities hold:

[[math]] \begin{eqnarray*} \Big| U_{\sigma_1} - \int_a^b f \Big| \lt \frac{\epsilon}{2} , \;\;\; \Big| \int_a^b - L_{\sigma_1} f \Big| \lt \frac{\epsilon}{2} , \\ \Big| U_{\sigma_2} - \int_b^c f \Big| \lt \frac{\epsilon}{2} , \;\;\; \Big| \int_b^c - L_{\sigma_2} f \Big| \lt \frac{\epsilon}{2} . \end{eqnarray*} [[/math]]

It follows from these that

[[math]] \begin{eqnarray*} \Big| (U_{\sigma_1} + U_{\sigma_2}) - \Big( \int_a^b f + \int_b^c f \Big) \Big| \lt \epsilon ,\\ \Big| \Big(\int_a^b f + \int_b^c f \Big) - \Big(L_{\sigma_1} + L_{\sigma_2} \Big) \Big| \lt \epsilon . \end{eqnarray*} [[/math]]

Let us set [math]{\sigma_1} \cup {\sigma_2} = \sigma[/math]. This union is a partition of [math][a, c][/math], and it is obvious that

[[math]] \begin{eqnarray*} U_{\sigma_1} + U_{\sigma_2} = U_\sigma, \\ L_{\sigma_1} + L_{\sigma_2} = L_\sigma . \end{eqnarray*} [[/math]]

Hence

[[math]] \begin{eqnarray*} \Big| U_\sigma - \Big(\int_a^b f + \int_b^c f \Big) \Big| \leq \epsilon, \\ \Big| \Big(\int_a^b f + \int_b^c f \Big) - L_\sigma \Big| \leq \epsilon . \end{eqnarray*} [[/math]]

These inequalities imply that [math]f[/math] is integrable over [math][a, c][/math] and also that

[[math]] \int_a^c f = \int_a^b f + \int_b^c f . [[/math]]

It remains to prove that, if [math]f[/math] is integrable over [math][a, c][/math], then it is integrable over [math][a, b][/math] and over [math][b, c][/math]. We choose an arbitrary positive number [math]\epsilon[/math]. Since [math]f[/math] is integrable over [math][a, c][/math], there exists a partition [math]\sigma[/math] of [math][a, c][/math] such that [math]U_\sigma - L_\sigma \lt \epsilon[/math]. Let us form a refinement of the partition [math]\sigma[/math] by adjoining the number [math]b[/math]. That is, we set

[[math]] \sigma' = \sigma \cup \{ b \}. [[/math]]

(It is, of course, possible that [math]\sigma[/math] already contains [math]b[/math], in which case [math]\sigma' = \sigma[/math].) Then

[[math]] L_\sigma \leq L_{\sigma'} \leq U_{\sigma'} \leq U_{\sigma'}, [[/math]]

from which it follows that [math]U_{\sigma'} - L_{\sigma'}, \lt \epsilon[/math]. But, since [math]b[/math] belongs to [math]\sigma'[/math], we can write [math]\sigma' = \sigma_1 \cup \sigma_2[/math], where [math]\sigma_1[/math] is a partition of [math][a, b][/math] and [math]\sigma_2[/math] is a partition of [math][b, c][/math]. Moreover,

[[math]] \begin{eqnarray*} U_{\sigma'} = U_{\sigma_1} + U_{\sigma_2},\\ L_{\sigma'} = L_{\sigma_1} + L_{\sigma_2} . \end{eqnarray*} [[/math]]

Hence

[[math]] (U_{\sigma_1} - L_{\sigma_1}) + (U_{\sigma_2} - L_{\sigma_2}) = U_{\sigma'} - L_{\sigma'} \lt \epsilon, [[/math]]

Since [math]U_{\sigma_1} - L_{\sigma_1}[/math] and [math]U_{\sigma_2} - L_{\sigma_2}[/math] are both nonnegative, it follows that

[[math]] \begin{eqnarray*} U_{\sigma_1} - L_{\sigma_1} \lt \epsilon,\\ U_{\sigma_2} - L_{\sigma_2} \lt \epsilon . \end{eqnarray*} [[/math]]

The first of these inequalities implies that [math]f[/math] is integrable over [math][a, b][/math], and the second that [math]f[/math] is integrable over [math][b, c][/math]. This completes the proof of Theorem 1.

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The second result to be proved is the following:

THEOREM 2

If [math]f[/math] and [math]g[/math] are integrable over [math][a, b][/math], then so is their sum and

[[math]] \int_a^b (f + g) = \int_a^b f + \int_a^b g . [[/math]]

Show Proof

Let [math]\epsilon[/math] be an arbitrary positive number. By taking, if necessary, the common refinement [math]\sigma_1 \cup \sigma_2[/math] of two partitions of [math][a, b][/math], we may select a partition [math]\sigma[/math] of [math][a, b][/math] such that

[[math]] \begin{array}{ll} \Big| U_\sigma^{(f)} - \int_a^b f \Big| \lt \frac{\epsilon}{2} , \;\;\;& \Big| \int_a^b f - L_\sigma^{(f)} \Big| \lt \frac{\epsilon}{2} , \\ \Big| U_\sigma^{(g)} - \int_a^b g \Big| \lt \frac{\epsilon}{2},\;\;\;& \Big| \int_a^b g - L_\sigma^{(g)} \Big| \lt \frac{\epsilon}{2} , \end{array} [[/math]]

where [math]U_\sigma^{(f)}[/math] and [math]L_\sigma^{(f)}[/math] are, respectively, the upper and lower sums for [math]f[/math] relative to [math]\sigma[/math], and [math]U_\sigma^{(g)}[/math] and [math]L_\sigma^{(g)}[/math] are the same for [math]g[/math]. We conclude from the above inequalities that

[[math]] \Big| ( U_\sigma^{(f)} + U_\sigma^{(g)} ) - \Big(\int_a^b f + \int_a^b g \Big) \Big| \lt \epsilon, ( 2 ) [[/math]]

[[math]] \Big| \Big(\int_a^b f + \int_a^b g \Big) - (L_\sigma^{(f)} + L_\sigma^{(g)}) \Big| \lt \epsilon . ( 3 ) [[/math]]

Let [math][x_{i-1}, x_i][/math] be the ith subinterval of the partition [math]\sigma[/math]. We denote by [math]M_i^{(f)}[/math] and [math]M_i^{(g)}[/math] the least upper bounds of the values of [math]f[/math] and of [math]g[/math], respectively, on [math][x_{i-1}, x_i][/math], and by [math]m_i^{(f)}[/math] and [math]m_i^{(g)}[/math] the analogous greatest lower bounds. Then

[[math]] m_i^{(f)} + m_i^{(g)} \leq f(x) + g(x) \leq M_i^{(f)} + M_i^{(g)}, [[/math]]

for every [math]x[/math] in [math][x_{i-1}, x_i][/math]. It follows that

[[math]] m_i^{(f)} + m_i^{(g)} \leq m_i^{(f+g)} \leq M_i^{(f+g)} \leq M_i^{(f)} + M_i^{(g)} , [[/math]]

where [math]m_i^{(f+g)}[/math] and [math]M_i^{(f+g)}[/math] are, respectively, the greatest lower bound and the least upper bound of the values of [math]f + g[/math] on [math][x_{i-1}, x_i][/math]. By multiplying each term in the preceding chain of inequalities by [math](x_i - x_{i-1})[/math] and then summing on [math]i[/math], we obtain

[[math]] L_\sigma^{(f)} + L_\sigma^{(g)} \leq L_\sigma^{(f+g)} \leq U_\sigma^{(f+g)} \leq U_\sigma^{(f)} + U_\sigma^{(g)}, (4 ) [[/math]]

where [math]U_\sigma^{(f+g)}[/math] and [math]L_\sigma^{(f+g)}[/math] are the upper and lower sums, respectively, for [math]f + g[/math] relative to [math]\sigma[/math]. The inequalities (2), (3), and (4) imply that

[[math]] \begin{eqnarray*} \Big| U_\sigma^{(f+g)} - \Big(\int_a^b f + \int_a^b g \Big) \Big| \lt \epsilon, \\ \Big| \Big(\int_a^b f + \int_a^b g \Big) - L_\sigma^{(f+g)} \Big| \lt \epsilon . \end{eqnarray*} [[/math]]

It follows from these two inequalities that the function [math]f + g[/math] is integrable over [math][a, b][/math] and that

[[math]] \int_a^b (f + g) = \int_a^b f + \int_a^b g. [[/math]]

This completes the proof of Theorem 2.

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General references

Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.