5.1 Partitions and the Riemann Sums
Throughout this chapter we work on a closed bounded interval \(I = [a,b]\) and, unless otherwise stated, all functions are real valued and bounded on \([a,b]\).
Definition 5.1.1. A partition \(P\) of \([a,b]\) is a finite set \(P = \{x_0, x_1, \dots , x_n\}\) such that \[a = x_0 < x_1 < \dots < x_{n - 1} < x_n = b .\] The points \(x_0, x_1, \dots , x_n\) are called the points of subdivision of \(P\). We write \(\Delta x_k = x_k - x_{k - 1}\); note that \(\sum ^n_{k = 1} \Delta x_k = b- a\).
Definition 5.1.2. A partition \(P^*\) of \([a,b]\) is a refinement of \(P\) if \(P\subseteq P^*\), that is, if every point of subdivision of \(P\) is also a point of subdivision of \(P^*\).
If \(P_1\) and \(P_2\) are partitions of \([a,b]\), a partition that refines both is called a common refinement of \(P_1\) and \(P_2\). Every pair of partitions has one: \(P_1 \cup P_2\).
The norm of a partition \(P\) is the length of its longest subinterval, \[\|P\| = \max _{1\leq k\leq n} (x_k - x_{k - 1}) .\] If \(P\subseteq P^*\) then \(\|P^*\|\leq \|P\|\), because each subinterval of \(P^*\) lies inside a subinterval of \(P\).
The set of all partitions of \([a,b]\) is denoted by \(\mathcal {P}[a,b]\).
Definition 5.1.3. Let \(P = \{x_0, x_1, \dots , x_n\}\) be a partition of \([a,b]\) and let \(t_k\) be a point of \([x_{k - 1}, x_k]\) for each \(k\). The sum \[S(P,f) = \sum ^n_{k = 1} f(t_k) \Delta x_k\] is called a Riemann sum of \(f\) for \(P\).
To define the Riemann integral we first define the upper and lower sums of \(f\) for a partition \(P\).
Definition 5.1.4. Let \(P\) be a partition of \([a,b]\), and for \(k = 1,\dots ,n\) let \begin {align*} M_k(f) &= \sup \{f(x): x\in [x_{k -1}, x_k]\},\\ m_k(f) &= \inf \{f(x): x\in [x_{k -1}, x_k]\}. \end {align*}
Then
- (i)
- \(U(P,f) = \sum ^n_{k =1}M_k(f)\Delta x_k\) is called the upper sum of \(f\) for \(P\);
- (ii)
- \(L(P,f) = \sum ^n_{k =1}m_k(f)\Delta x_k\) is called the lower sum of \(f\) for \(P\).
Since \(m_k(f) \leq f(t_k)\leq M_k(f)\) for every \(t_k \in [x_{k - 1}, x_k]\), every Riemann sum lies between them: \[L(P,f) \leq S(P,f)\leq U(P,f).\]
Note. When \(f\) is continuous and non-negative on \([a,b]\), \(U(P,f)\) is the total area of the rectangles that stand on the subintervals and reach up to the highest point of the graph over each, and \(L(P,f)\) is the total area of the rectangles that reach up to the lowest point.
Theorem 5.1.5. Let \(f\) be a bounded function on \([a,b]\).
- (i)
- If \(P^*\) is a refinement of a partition \(P\), then \(U(P^*, f)\leq U(P,f)\) and \(L(P^*,f)\geq L(P,f)\).
- (ii)
- For any two partitions \(P_1\) and \(P_2\), \(L(P_1, f)\leq U(P_2,f)\).
Proof. (i) Since \(P^*\) is obtained from \(P\) by adding finitely many points one at a time, it suffices to treat the case where \(P^*\) has exactly one more point than \(P\), say \(c\), lying in the \(j^{\text {th}}\) subinterval \((x_{j-1},x_j)\). Let \(M'\) and \(M''\) be the suprema of \(f\) on \([x_{j - 1}, c]\) and \([c,x_j]\). Then \(M'\leq M_j(f)\) and \(M''\leq M_j(f)\), so \begin {align*} U(P^*, f) - U(P,f) &= M'(c - x_{j - 1}) + M''(x_j - c)\\ &\qquad - M_j(f)(x_j - x_{j-1}) \leq 0 . \end {align*}
The inequality for lower sums is proved in the same way, with infima.
(ii) Let \(P = P_1\cup P_2\). By (i), \[L(P_1,f) \leq L(P,f) \leq U(P,f) \leq U(P_2,f) .\] □
Note. If \(M = \sup \{f(x): x\in [a,b]\}\) and \(m = \inf \{f(x): x\in [a,b]\}\), then \(m(b-a)\leq L(P_1,f) \leq U(P_2,f)\leq M(b-a)\) for any partitions \(P_1\) and \(P_2\). In particular the sets of upper sums and of lower sums are bounded.
Definition 5.1.6. Let \(f\) be bounded on \([a,b]\). The upper Riemann integral and the lower Riemann integral of \(f\) over \([a,b]\) are \begin {align*} \overline {\int ^b_a}f(x)\, dx &= \inf \{U(P,f): P\in \mathcal {P}[a,b]\},\\ \underline {\int ^b_a}f(x)\, dx &= \sup \{L(P,f): P\in \mathcal {P}[a,b]\}, \end {align*}
the infimum and supremum being taken over all partitions of \([a,b]\). We shall also write them simply as \(\overline {\int ^b_a} f\) and \(\underline {\int ^b_a} f\).
Theorem 5.1.7. If \(f\) is bounded on \([a,b]\), then \(\displaystyle \underline {\int ^b_a} f \leq \overline {\int ^b_a} f\).
Proof. By Theorem 5.1.5(ii), every lower sum is at most every upper sum. Fix a partition \(P_2\): then \(U(P_2,f)\) is an upper bound for all lower sums, so \(\underline {\int ^b_a} f\leq U(P_2,f)\). This holds for every \(P_2\), so \(\underline {\int ^b_a} f\) is a lower bound for all upper sums, and therefore \(\underline {\int ^b_a} f\leq \overline {\int ^b_a} f\).
The inequality can be strict, as the next example shows. □
Example 5.1.8. Let \(f\) be defined on \([0,1]\) by \(f(x) = 1\) if \(x\) is rational and \(f(x) = 0\) if \(x\) is irrational. Every subinterval contains both rational and irrational numbers, so for every partition \(P\) of \([0,1]\) we have \(M_k(f) = 1\) and \(m_k(f) = 0\). Therefore \(U(P,f) = 1\) and \(L(P,f) = 0\) for all \(P\), and \[\overline {\int ^1_0} f = 1 ,\qquad \underline {\int ^1_0} f = 0.\]
Remark. If \(a< c < b\), then \[\overline {\int ^b_a} f = \overline {\int ^c_a} f + \overline {\int ^b_c} f \qquad \text {and}\qquad \underline {\int ^b_a} f = \underline {\int ^c_a} f + \underline {\int ^b_c} f .\] However, some properties of upper and lower integrals differ from those of integrals: in general only \[\overline {\int ^b_a} (f + g) \leq \overline {\int ^b_a} f + \overline {\int ^b_a} g\qquad \text {and}\qquad \underline {\int ^b_a} (f + g) \geq \underline {\int ^b_a} f + \underline {\int ^b_a} g\] hold, and the inequalities may be strict (take \(f\) as in the example and \(g = 1-f\)).
Definition 5.1.9. A bounded function \(f\) on \([a,b]\) is said to be Riemann integrable on \([a,b]\) if \[\underline {\int ^b_a} f = \overline {\int ^b_a} f .\] The common value is called the Riemann integral of \(f\) over \([a,b]\) and is denoted by \(\int ^b_a f\) or \(\int ^b_a f(x)\,dx\).
Example 5.1.10. Every constant function \(f(x) = C\) is Riemann integrable on \([a,b]\), and \(\int ^b_a C\,dx = C(b-a)\).
Solution. For any partition \(P\), \(M_k(f) = m_k(f) = C\) for every \(k\), so \[U(P,f) = \sum ^n_{k = 1} C \Delta x_k = C(b - a) = L(P,f).\] Hence both the upper and the lower integral equal \(C(b-a)\).
Example 5.1.11. Let \(f(x) = x^2\) on \([0,1]\), and for each \(n\) let \(P_n = \left \{0, \frac {1}{n}, \frac {2}{n}, \dots , \frac {n}{n}\right \}\). Compute \(\lim _{n\rightarrow \infty } U(P_n,f)\) and \(\lim _{n \rightarrow \infty } L(P_n,f)\), and deduce \(\int _0^1x^2\,dx\).
Solution. The subintervals of \(P_n\) are \(\left [\frac {k-1}{n},\frac {k}{n}\right ]\), each of length \(\frac {1}{n}\). Since \(x^2\) is increasing on \([0,1]\), \[M_k(f) = \left (\frac {k}{n}\right )^2,\qquad m_k (f) = \left (\frac {k - 1}{n}\right )^2 .\] Using \(1^2 + 2^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)}{6}\), \[U(P_n,f) = \frac {1}{n^3}\sum ^n_{k = 1}k^2 = \frac {(n + 1)(2n + 1)}{6n^2}\longrightarrow \frac {2}{6} = \frac {1}{3},\] \[L(P_n,f) = \frac {1}{n^3}\sum ^n_{k = 1}(k-1)^2 = \frac {(n - 1)(2n - 1)}{6n^2} \longrightarrow \frac {1}{3} .\] Since \(L(P_n, f) \leq \underline {\int ^1_0} f \leq \overline {\int ^1_0} f \leq U(P_n, f)\) and both bounds tend to \(\frac {1}{3}\), \(f\) is integrable and \(\int ^1_0 x^2\, dx = \frac {1}{3}\).
Theorem 5.1.12 (Riemann’s Criterion). A bounded function \(f\) on \([a,b]\) is Riemann integrable if and only if for every \(\varepsilon >0\) there is a partition \(P\) of \([a,b]\) such that \(U(P, f) - L(P,f) < \varepsilon \).
Proof. We have already seen that \begin {equation} \label {eq:lowup} \underline {\int ^b_a} f \leq \overline {\int ^b_a} f . \end {equation} Suppose that for every \(\varepsilon > 0\) there is a partition \(P\) with \(U(P,f) - L(P, f) < \varepsilon \). Since \(\overline {\int ^b_a} f \leq U(P,f)\) and \(\underline {\int ^b_a} f \geq L(P,f)\), \[\overline {\int ^b_a}f - \underline {\int ^b_a}f \leq U(P,f) - L(P,f) < \varepsilon .\] Since \(\varepsilon > 0\) is arbitrary, \begin {equation} \label {eq:uplow} \overline {\int ^b_a}f \leq \underline {\int ^b_a}f , \end {equation} and (5.1) and (5.2) together show that \(f\) is Riemann integrable.
Conversely, suppose that \(f\) is Riemann integrable, with integral \(I\), and let \(\varepsilon > 0\). By the definitions of infimum and supremum there are partitions \(P_1\) and \(P_2\) with \[U(P_1,f) < I + \frac {\varepsilon }{2} \qquad \text {and}\qquad L(P_2,f) > I - \frac {\varepsilon }{2}.\] Let \(P = P_1\cup P_2\). By Theorem 5.1.5, \(U(P,f)\leq U(P_1,f)\) and \(L(P,f)\geq L(P_2,f)\), so \[U(P,f) - L(P,f) < \left (I + \frac {\varepsilon }{2}\right ) - \left (I - \frac {\varepsilon }{2}\right ) = \varepsilon .\] □
Theorem 5.1.13. Every continuous function on \([a,b]\) is Riemann integrable. More precisely, for every \(\varepsilon >0\) there is a \(\delta >0\) such that \(U(P,f) - L(P,f)<\varepsilon \) for every partition \(P\) with \(\|P\|<\delta \).
Proof. Let \(f\) be continuous on \([a,b]\) and \(\varepsilon > 0\). Since \(f\) is uniformly continuous on \([a,b]\), there is a \(\delta > 0\) such that \(|f(x) - f(y)| < \frac {\varepsilon }{b - a}\) whenever \(x, y\in [a,b]\) and \(|x - y| < \delta \).
Let \(P\) be any partition with \(\|P\| < \delta \). On each closed subinterval \([x_{k-1},x_k]\) the continuous function \(f\) attains its maximum and minimum, say \(M_k = f(x_k')\) and \(m_k = f(x_k'')\). Since \(|x_k' - x_k''|\leq \Delta x_k < \delta \), we get \(M_k - m_k < \frac {\varepsilon }{b - a}\), and so \[U(P,f) - L(P,f) = \sum ^n_{k = 1}(M_k - m_k)\Delta x_k < \frac {\varepsilon }{b - a}\sum ^n_{k = 1}\Delta x_k = \varepsilon .\] By Riemann’s Criterion, \(f\) is Riemann integrable. □
Proof. Suppose \(f\) is increasing. If \(f(a) = f(b)\), then \(f\) is constant and integrable. Otherwise \(f(a)<f(b)\). Let \(\varepsilon > 0\), and let \(P\) be any partition with \(\|P\| < \frac {\varepsilon }{f(b) - f(a)}\). Since \(f\) is increasing, \(M_k(f) =f(x_k)\) and \(m_k(f) = f(x_{k-1})\) on \([x_{k - 1}, x_k]\). Hence \begin {align*} U(P,f) - L(P,f) &= \sum ^n_{k = 1} \bigl (f(x_k) - f(x_{k - 1})\bigr ) \Delta x_k\\ &< \frac {\varepsilon }{f(b) - f(a)} \sum ^n_{k = 1} \bigl (f(x_k) - f(x_{k - 1})\bigr ) = \varepsilon , \end {align*}
since the last sum telescopes to \(f(b)-f(a)\). By Riemann’s Criterion, \(f\) is integrable. If \(f\) is decreasing, apply this to \(-f\) and use Theorem 5.2.1 below, or repeat the argument with \(M_k(f) = f(x_{k-1})\) and \(m_k(f) = f(x_k)\). □
Example 5.1.15. Let \(f\) be continuous on \([0,1]\), let \(\sigma _n = \left \{0, \frac {1}{n}, \frac {2}{n}, \dots , \frac {n}{n}\right \}\), and let \(x^*_k\) be any point of \(\left [\frac {k - 1}{n}, \frac {k}{n}\right ]\).
- (i)
- Prove that \(L(\sigma _n, f) \leq \frac {1}{n} \sum _{k =1}^n f(x^*_k) \leq U(\sigma _n,f)\).
- (ii)
- Hence show that \(\lim _{n\rightarrow \infty }\frac {1}{n} \sum ^n_{k = 1} f(x^*_k) = \int ^1_0f(x)\, dx\).
Solution. (i) Here \(\Delta x_k = \frac {1}{n}\), so \(\frac {1}{n}\sum _{k=1}^n f(x_k^*)\) is a Riemann sum for \(\sigma _n\), and it lies between \(L(\sigma _n,f)\) and \(U(\sigma _n,f)\) because \(m_k(f)\leq f(x^*_k)\leq M_k(f)\).
(ii) Let \(I = \int _0^1 f\), which exists by Theorem 5.1.13. We also have \(L(\sigma _n,f)\leq I\leq U(\sigma _n,f)\). Let \(\varepsilon >0\) and choose \(\delta \) as in Theorem 5.1.13. For \(n>\frac {1}{\delta }\), \(\|\sigma _n\| = \frac {1}{n}<\delta \), so \(U(\sigma _n,f) - L(\sigma _n,f)<\varepsilon \). Both \(I\) and the Riemann sum lie in the interval \([L(\sigma _n,f), U(\sigma _n,f)]\), whose length is less than \(\varepsilon \), so \[\left |\frac {1}{n}\sum _{k=1}^n f(x^*_k) - I\right |<\varepsilon \qquad \text {for } n>\tfrac {1}{\delta }.\]
Example 5.1.16. Let \(f(x) = \frac {\sin x}{x}\) for \(0<x\leq 1\) and \(f(0) = 1\). Prove that \(f\) is Riemann integrable on \([0,1]\).
Solution. The function \(\frac {\sin x}{x}\) is continuous for \(x\neq 0\), and its limit at \(0\) is \(1 = f(0)\). So \(f\) is continuous on \([0,1]\), and Theorem 5.1.13 shows that it is Riemann integrable.
- (i)
- The function \(f\) on \([0,1]\) with \(f(x) = 1\) for rational \(x\) and \(f(x) = 0\) for irrational \(x\) is bounded but not Riemann integrable, since its upper and lower integrals are \(1\) and \(0\).
- (ii)
- A discontinuous function may be integrable. Let \(f(x) = 1\) for \(x\in \left [0, \frac {1}{2}\right ]\) and \(f(x) = 0\) for \(x\in \left (\frac {1}{2}, 1\right ]\). Then \(f\) is discontinuous at \(\frac {1}{2}\). Let \(P\) be any partition of \([0,1]\), and let \(k\) be the index with \(x_{k-1}\leq \frac {1}{2}<x_k\). On the subintervals before the \(k^{\text {th}}\), \(M_i = m_i = 1\); on those after it, \(M_i = m_i = 0\); and on \([x_{k-1},x_k]\), which contains \(\frac {1}{2}\) and points beyond it, \(M_k = 1\) and \(m_k = 0\). Hence \[U(P,f) = x_k\qquad \text {and}\qquad L(P,f) = x_{k-1},\] so \(L(P,f)\leq \frac {1}{2}<U(P,f)\) and \(U(P,f) - L(P,f) = \Delta x_k\leq \|P\|\). Taking partitions with small norm, Riemann’s Criterion (Theorem 5.1.12) shows that \(f\) is integrable, and since \(\frac {1}{2}\) lies between every lower and every upper sum, \(\int ^1_0 f = \frac {1}{2}\), the length of the interval on which \(f = 1\).
Theorem 5.1.18. A bounded function \(f\) on \([a,b]\) is Riemann integrable if
- (i)
- it has only finitely many points of discontinuity in \([a,b]\), or
- (ii)
- its set of points of discontinuity has only finitely many limit points.
Proof. Let \(M = \sup f\) and \(m = \inf f\) on \([a,b]\). If \(M = m\), \(f\) is constant; so assume \(M>m\). Let \(\varepsilon >0\).
(i) Let \(\alpha _1<\alpha _2<\dots <\alpha _p\) be the points of discontinuity. Enclose them in non-overlapping intervals \([x_1', x''_1], \dots , [x_p',x_p'']\), each containing its point \(\alpha _r\) in its interior (or as an end point if \(\alpha _r = a\) or \(b\)), with total length \[\sum ^p_{r = 1}(x''_r - x'_r) < \frac {\varepsilon }{2(M - m)} .\] On each of the \(p + 1\) remaining closed intervals \([a,x_1'], [x''_1,x'_2], \dots , [x''_p,b]\) the function \(f\) is continuous, hence integrable, so by Riemann’s Criterion there are partitions \(P_1, \dots , P_{p+1}\) of them with \[U(P_r,f) - L(P_r,f) < \frac {\varepsilon }{2(p+1)}, \qquad r =1, 2, \dots , p + 1 .\] Let \(P\) be the partition of \([a,b]\) formed by all the points of \(P_1,\dots ,P_{p+1}\). Its subintervals are those of the \(P_r\) together with the intervals \([x'_r,x''_r]\), on which the oscillation \(M'_r - m'_r\) is at most \(M - m\). Hence \begin {align*} U(P,f) - L(P,f) &\leq (M-m)\sum ^p_{r=1}(x''_r-x'_r) + \sum ^{p + 1}_{r=1}\bigl (U(P_r,f) - L(P_r,f)\bigr )\\ &< \frac {\varepsilon }{2} + \frac {\varepsilon }{2} = \varepsilon , \end {align*}
and \(f\) is integrable by Riemann’s Criterion.
(ii) Let \(\alpha _1,\dots ,\alpha _p\) be the limit points of the set \(D\) of discontinuities, and enclose them as in (i) in intervals of total length less than \(\frac {\varepsilon }{2(M-m)}\). Each of the remaining closed intervals contains only finitely many points of \(D\): otherwise, by the Bolzano–Weierstrass Theorem, \(D\) would have a limit point in that closed interval, but all limit points of \(D\) lie in the enclosing intervals’ interiors. By part (i), \(f\) is integrable on each remaining interval, and the argument of (i) gives a partition \(P\) with \(U(P,f) - L(P,f) < \varepsilon \). □
Example 5.1.19. Using upper and lower sums, show that \(\int _0^b x\,dx = \frac {b^2}{2}\) for \(b>0\).
Solution. Let \(P_n\) divide \([0,b]\) into \(n\) equal parts, \(x_k = \frac {kb}{n}\). Since \(f(x) = x\) is increasing, \(M_k = \frac {kb}{n}\) and \(m_k = \frac {(k-1)b}{n}\), so with \(\Delta x_k = \frac {b}{n}\), \[U(P_n,f) = \frac {b^2}{n^2}\sum _{k=1}^nk = \frac {b^2}{n^2}\cdot \frac {n(n+1)}{2} = \frac {b^2}{2}\left (1+\frac {1}{n}\right ),\qquad L(P_n,f) = \frac {b^2}{2}\left (1-\frac {1}{n}\right ).\] Both tend to \(\frac {b^2}{2}\), and \(U(P_n,f) - L(P_n,f) = \frac {b^2}{n}\longrightarrow 0\). By Riemann’s Criterion \(f\) is integrable, and its integral lies between \(L(P_n,f)\) and \(U(P_n,f)\) for every \(n\), so it equals \(\frac {b^2}{2}\).
Example 5.1.20. Compute the midpoint Riemann sum of \(f(x) = \frac {1}{1+x^2}\) on \([0,1]\) with \(n = 4\), and compare it with \(\int _0^1\frac {dx}{1+x^2} = \frac {\pi }{4}\).
Solution. The midpoints of the four subintervals are \(\frac {1}{8}, \frac {3}{8}, \frac {5}{8}, \frac {7}{8}\), where \(f\) takes the values \[0.984615,\qquad 0.876712,\qquad 0.719101,\qquad 0.566372 .\] The Riemann sum is \(\frac {1}{4}(0.984615 + 0.876712 + 0.719101 + 0.566372) = 0.786700\), while \(\frac {\pi }{4} = 0.785398\). Four evaluations already give \(\pi \approx 4\times 0.7867 = 3.1468\). Since \(f\) is decreasing on \([0,1]\), the upper and lower sums for this partition are guaranteed bounds: \(L = \frac {1}{4}\bigl (f(\frac 14)+f(\frac 12)+f(\frac 34)+f(1)\bigr )\approx 0.7203\) and \(U = L + \frac {1}{4}\bigl (f(0) - f(1)\bigr )\approx 0.8453\), so \(0.7203<\frac {\pi }{4}<0.8453\).
Example 5.1.21 (Thomae’s function is integrable). Let \(f\) be Thomae’s function on \([0,1]\): \(f(x) = \frac {1}{q}\) if \(x = \frac {p}{q}\) in lowest terms, and \(f(x) = 0\) if \(x\) is irrational (with \(f(0) = f(1) = 1\)). Show that \(f\in R[0,1]\) and \(\int _0^1f = 0\), although \(f\) is discontinuous at every rational point.
Solution. Every subinterval contains irrational points, so \(m_k = 0\) and \(L(P,f) = 0\) for every partition. It remains to make \(U(P,f)\) small. Let \(\varepsilon >0\) and choose \(N\) with \(\frac {1}{N}<\frac {\varepsilon }{2}\). Only finitely many points of \([0,1]\), say \(K\) of them, have \(f(x)\geq \frac {1}{N}\) (the rationals with denominator at most \(N\)). Choose a partition \(P\) in which each of these \(K\) points lies in subintervals of total length less than \(\frac {\varepsilon }{2}\) (at most two subintervals per point). On those subintervals \(M_k\leq 1\); on all the others \(M_k<\frac {1}{N}\). Hence \[U(P,f) < 1\cdot \frac {\varepsilon }{2} + \frac {1}{N}\cdot 1<\varepsilon .\] By Riemann’s Criterion \(f\) is integrable, and \(0 = L(P,f)\leq \int _0^1f\leq U(P,f)<\varepsilon \) for every \(\varepsilon \), so \(\int _0^1 f = 0\). This function has infinitely many discontinuities, with every point of \([0,1]\) a limit point of them, so the last theorem does not apply; yet it is integrable.
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