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The purpose of this section is to prove that the definite integral [math]\int_a^b f[/math], defined on page 169 in terms of upper and lower sums, can be equivalently defined as the limit of Riemann sums. The fact that these two approaches to the integral are the same is stated without proof in Theorem (2.1), page 414, and we shall now supply the details of the argument. The “if” and the “only if” directions of the proof wil1 be treated separately. Let [math]f[/math] be a real-valued function which is bounded on the closed interval [math][a, b][/math]. This implies, according to our definition of boundedness, that [math][a, b][/math] is contained in the domain of [math]f[/math]. Let [math]\sigma = \{x_0, . . ., x_n \}[/math] be a partition of [math][a, b][/math] such that

[[math]] a = x_0 \leq x_1 \leq \cdots \leq \lt x_n = b. [[/math]]

If an arbitrary number [math]x_i^*[/math] is chosen in the ith subinterval [math][x_{i-1}, x_i][/math], then the sum

[[math]] R_\sigma = \sum_{i=1}^n f(x_i^*)(x_i - x_{i-1}) [[/math]]

is a Riemann sum for [math]f[/math] relative to [math]\sigma[/math]. The fineness of a partition [math]\sigma[/math] is measured by its mesh, which is denoted by [math]\|\sigma\|[/math] and defined by

[[math]] \|\sigma\| = \mbox{maximum}_{1 \leq i \leq n} \{(x_i - x_{i-1}) \}. [[/math]]

The first of the two theorems is:

THEOREM 1.

If [math]f[/math] is bounded on [math][a, b][/math] and if [math]\lim_{\| \sigma \| \rightarrow 0} R_\sigma = L[/math], then [math]f[/math] is integrable over [math][a, b][/math] and [math]\int_a^b f = L[/math].

Show Proof

We assume that [math]a \lt b[/math], since otherwise [math]L = 0 = \int_a^a f[/math] and the result is trivial. It is a consequence of the definition of integrability that the conclusion of Theorem 1 is implied by the following proposition: For any positive number [math]\epsilon [/math], there exists a partition [math]\sigma[/math] of [math][a, b][/math] such that, where [math]U_\sigma [/math] and [math]L_\sigma[/math] are, respectively, the upper and lower sums for [math]f[/math] relative to [math]\sigma[/math], then [math]|U_\sigma - L | \lt \epsilon[/math] and [math]|L - L_\sigma| \lt \epsilon[/math]. It is this that we shall prove. We first prove that, if [math]U_\sigma[/math] is the upper sum for [math]f[/math] relative to any partition [math]\sigma[/math] of [math][a, b][/math], then there exists a Riemann sum [math]R_\sigma^{(1)}[/math] for [math]f[/math] relative to [math]\sigma[/math] such that [math]| U_\sigma - R_\sigma^{(1)} |[/math] is arbitrarily small. Let [math]\sigma = \{ x_0, . . ., x_n \}[/math] be the partition with the usual proviso that

[[math]] a = x_0 \leq x_1 \leq \cdots \leq x_n = b, [[/math]]

and let [math]\epsilon[/math] be an arbitrary positive number. For each [math]i = 1, . . ., n[/math], set [math]M_i[/math] equal to the least upper bound of the values of [math]f[/math] in the subinterval [math][x_{i-1}, x_i][/math]. Then there exists a number [math]x_i^*[/math] in [math][x_{i-1}, x_i][/math] such that

[[math]] 0 \leq M_i - f(x_i^*) \leq \frac{\epsilon}{2(b - a)} . [[/math]]

Hence

[[math]] 0 \leq M_i (x_i - x_{i-1}) - f(x_i^*)(x_i - x_{i-1}) \leq \frac{\epsilon}{2(b - a)} (x_i - x_{i-1}) , [[/math]]

and so

[[math]] 0 \leq \sum_{i=1}^n M_i (x_i - x_{i-1}) - \sum_{i=1}^n f(x_i^*)(x_i - x_{i-1}) \leq \frac{\epsilon}{2(b - a)} \sum_{i=1}^n (x_i - x_{i-1}) [[/math]]

However,

[[math]] \begin{eqnarray*} \sum_{i=1}^n M_i (x_i - x_{i-1}) = U_\sigma , \\ \frac{\epsilon}{2 (b - a)} \sum_{i=1}^n (x_i - x_{i-1}) = \frac{\epsilon}{2(b - a)} (b - a) = \frac{\epsilon}{2} . \end{eqnarray*} [[/math]]

Moreover, [math]\sum_{i=1}^n f(x_i^*)(x_i - x_{i-1})[/math] is a Riemann sum for [math]f[/math] relative to [math]\sigma[/math], which we denote by [math]R_\sigma^{(1)}[/math]. Thus the preceding inequalities become

[[math]] 0 \leq U_\sigma - R_\sigma^{(1)} \leq \frac{\epsilon}{2}, ( 1 ) [[/math]]

which proves the assertion at the beginning of the paragraph. In an entirely analogous manner, we can prove that, if [math]L_\sigma[/math] is the lower sum relative to an arbitrary partition [math]\sigma[/math] of [math][a, b][/math] and if [math]\epsilon[/math] is any positive number, then there exists a Riemann sum [math]R_\sigma^{(2)}[/math] such that

[[math]] 0 \leq R_\sigma^{(2)} - L_\sigma \leq \frac{\epsilon}{2} . ( 2 ) [[/math]]

We are now ready to use the premise of Theorem 1 --- the fact that [math]\lim_{\|\sigma\| \rightarrow 0} R_\sigma = L[/math]. Let [math]\epsilon[/math] be an arbitrary positive number. Then there exists a positive number [math]\delta[/math] such that, if [math]\sigma[/math] is any partition of [math][a, b][/math] with mesh less than 3, then

[[math]] |R_\sigma - L| \lt \frac{\epsilon}{2}, [[/math]]

for every Riemann sum [math]R_\sigma[/math]. Accordingly, let [math]\sigma[/math] be a partition of [math][a, b][/math] with [math]\| \sigma \| \lt \delta[/math], and let [math]U_\sigma[/math] and [math]L_\sigma[/math] be, respectively, the upper and lower sums for [math]f[/math] relative to this partition. It follows from the preceding two paragraphs that there exist Riemann sums [math]R_\sigma^{(1)}[/math] and [math]R_\sigma^{(2)}[/math], for [math]f[/math] relative to [math]\sigma[/math] such that

[[math]] \begin{eqnarray*} |U_\sigma - R_\sigma^{(1)}| &\leq& \frac{\epsilon}{2} , \\ |R_\sigma^{(2)} - L_\sigma| &\leq& \frac{\epsilon}{2} , \end{eqnarray*} [[/math]]

[see inequalities (1) and (2)]. Since the mesh of [math]\sigma[/math] is less than [math]\delta[/math], we have

[[math]] \begin{eqnarray*} |R_\sigma^{(1)} - L_\sigma| & \lt & \frac{\epsilon}{2} ,\\ |L_\sigma - R_\sigma^{(2)} | = |R_\sigma^{(2)} - L_\sigma| & \lt & \frac{\epsilon}{2} . \end{eqnarray*} [[/math]]

Hence

[[math]] \begin{eqnarray*} |U_\sigma - L| &=& |(U_\sigma - R_\sigma) + (R_\sigma - L)| \\ &\leq& |U_\sigma - R_\sigma + I R_\sigma - L | \\ & \lt & \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon, \end{eqnarray*} [[/math]]

and, similarly,

[[math]] \begin{eqnarray*} |L - L_\sigma| &=& |(L - R_\sigma^{(2)}) + (R_\sigma ^{(2)} - L_\sigma)\\ &\leq& |L - R_\sigma ^{(2)}| + |R_\sigma ^{(2)} - L |\\ & \lt & \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{eqnarray*} [[/math]]

Thus both [math]|U_\sigma - L|[/math] and [math]|L - L_\sigma|[/math] are less than [math]\epsilon[/math], and the proof of Theorem 1 is complete.

■

The converse proposition is the following:

THEOREM 2

If [math]f[/math] is integrable over [math][a, b][/math], then [math]\lim_{\|\sigma \| \rightarrow 0} R_\sigma = \int_a^b f.[/math]

Show Proof

We assume from the outset that [math]a \lt b[/math]. Let [math]\epsilon[/math] be an arbitrary positive number. Since [math]f[/math] is integrable, there exist partitions of [math][a, b][/math] with upper and lower sums arbitrarily close to [math]\int_a^b f[/math]. By taking, if necessary, the common refinement [math]\sigma \cup \tau[/math] of two partitions [math]\sigma[/math] and [math]\tau[/math] (see the inequalities [math]L_\sigma \leq L_{\sigma \cup \tau} \leq U_{\sigma \cup \tau} \leq U_\tau[/math] on page 168), we may choose a partition [math]\sigma_0 = \{ x_0, . . ., x_n \}[/math] of [math][a, b][/math] such that

[[math]] \begin{eqnarray*} U_{\sigma_0} - \int_a^b f & \lt & \frac{\epsilon}{2} , \\ \int_a^b f - L_{\sigma_0} & \lt & \frac{\epsilon}{2} . \end{eqnarray*} [[/math]]

The assumption of integrability implies that the function [math]f[/math] is bounded on [math][a, b][/math]. Thus there exists a positive number [math]B[/math] such that [math]If(x)l \leq B[/math] for every [math]x[/math] in [math][a, b][/math]. We define

[[math]] \delta = \frac{\epsilon}{4Bn} . [[/math]]

Next, let [math]\sigma[/math] be any partition of [math][a, b][/math] with mesh less than [math]\delta[/math]. Consider the common refinement [math]\sigma \cup \sigma_0[/math]. Since

[[math]] L_{\sigma_0} \leq L_{\sigma \cup \sigma_0} \leq \int_a^b f \leq U_{\sigma \cup \sigma_0} \leq U_{\sigma_0} , [[/math]]

we have

[[math]] U_{\sigma \cup \sigma_0} - \int_a^b f \lt \frac{\epsilon}{2} , ( 3 ) [[/math]]

[[math]] \int_a^b f - L_{\sigma \cup \sigma_0} \lt \frac{\epsilon}{2} . ( 4 ) [[/math]]

The partition [math]\sigma \cup \sigma_0[/math] may be regarded as having been obtained from [math]\sigma[/math] by the addition of at most [math]n - 1[/math] new points of [math]\sigma_0[/math]. Hence at most [math]n - 1[/math] of the subintervals of [math]\sigma[/math] have been further partitioned by the inclusion of points of [math]\sigma_0[/math] in their interiors. Each of these further partitioned subintervals has length less than [math]b[/math]. It follows that, on each of them, the contribution to the difference [math]U_\sigma - U_{\sigma \cup \sigma_0}[/math] is less than the product [math]\delta (2B)[/math]. On those subintervals of [math]\sigma[/math] which have not been hit by points of [math]\sigma_0[/math] in their interiors, the corresponding terms of [math]U_\sigma[/math] and of [math]U_{\sigma \cup \sigma_0}[/math] are the same. We conclude that

[[math]] U_\sigma - U_{\sigma \cup \sigma_0} \lt (n - 1) \delta (2B), [[/math]]

and, similarly,

[[math]] L_{\sigma \cup \sigma_0} - L_\sigma \lt (n - 1) \delta (2B). [[/math]]

However,

[[math]] (n - 1) \delta (2B) = (n - 1)\frac{\epsilon}{4nB} (2B) \lt \frac{\epsilon}{2} . [[/math]]

Hence

[[math]] \begin{eqnarray*} U_\sigma - U_{\sigma \cup \sigma_0} & \lt & \frac{\epsilon}{2} ,\\ L_{\sigma \cup \sigma_0} - L_\sigma & \lt & \frac{\epsilon}{2} . \end{eqnarray*} [[/math]]

Combining these inequalities with (3) and (4), we conclude that

[[math]] U_\sigma - \int_a^b f \lt \epsilon, ( 5 ) [[/math]]

[[math]] \int_a^b f - L_\sigma \lt \epsilon. ( 6 ) [[/math]]

for every partition [math]\sigma[/math] with mesh less than [math]b[/math]. Finally, let [math]R_\sigma[/math] be an arbitrary Riernann sum for [math]f[/math] relative to a partition [math]\sigma[/math] of [math][a, b][/math] with mesh less than [math]\delta[/math]. We know that

[[math]] L_\sigma \leq R_\sigma \leq U_\sigma [[/math]]

(see page 413). These inequalities together with those in (5) and (6) immediately imply that

[[math]] |R_\sigma - \int_a^b f| \lt \epsilon . [[/math]]

Hence [math]\lim_{||\sigma|| \rightarrow 0} R_\sigma = \int_a^b f[/math] and the proof of Theorem 2 is complete.

■

The conjunction of Theorems 1 and 2 is equivalent to Theorem (2.1), page 414. We have therefore proved that the definite integral defined in terms of upper and lower sums is the same as the limit of Riemann sums.

General references

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