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Basic Analysis I & II: Introduction to Real Analysis, Volumes I & II

Jiří Lebl

Section 5.2 Properties of the integral

Note: 2 lectures, integrability of functions with discontinuities can safely be skipped

Subsection 5.2.1 Additivity

Adding a bunch of things in two parts and then adding those two parts should be the same as adding everything all at once. The corresponding property for integrals is called the additive property of the integral. First, we prove the additivity property for the lower and upper Darboux integrals.

Proof.

If we have partitions \(P_1 = \{ x_0,x_1,\ldots,x_k \}\) of \([a,b]\) and \(P_2 = \{ x_k, x_{k+1}, \ldots, x_n \}\) of \([b,c]\text{,}\) then the set \(P \coloneqq P_1 \cup P_2 = \{ x_0, x_1, \ldots, x_n \}\) is a partition of \([a,c]\text{.}\) We find
\begin{equation*} L(P,f) = \sum_{i=1}^n m_i \Delta x_i = \sum_{i=1}^k m_i \Delta x_i + \sum_{i=k+1}^n m_i \Delta x_i = L(P_1,f) + L(P_2,f) . \end{equation*}
When we take the supremum of the right-hand side over all \(P_1\) and \(P_2\text{,}\) we are taking a supremum of the left-hand side over all partitions \(P\) of \([a,c]\) that contain \(b\text{.}\) If \(Q\) is a partition of \([a,c]\) and \(P = Q \cup \{ b \}\text{,}\) then \(P\) is a refinement of \(Q\) and so \(L(Q,f) \leq L(P,f)\text{.}\) Therefore, taking a supremum only over the \(P\) that contain \(b\) is sufficient to find the supremum of \(L(P,f)\) over all partitions \(P\text{,}\) see Exercise 1.1.9. Finally, recall Exercise 1.2.9 to compute
\begin{equation*} \begin{split} \underline{\int_a^c} f & = \sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,c] \bigr\} \\ & = \sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,c], b \in P \bigr\} \\ & = \sup \, \bigl\{ L(P_1,f) + L(P_2,f) : P_1 \text{ a partition of } [a,b], P_2 \text{ a partition of } [b,c] \bigr\} \\ & = \sup \, \bigl\{ L(P_1,f) : P_1 \text{ a partition of } [a,b] \bigr\} + \sup \, \bigl\{ L(P_2,f) : P_2 \text{ a partition of } [b,c] \bigr\} \\ &= \underline{\int_a^b} f + \underline{\int_b^c} f . \end{split} \end{equation*}
Similarly, for \(P\text{,}\) \(P_1\text{,}\) and \(P_2\) as above, we obtain
\begin{equation*} U(P,f) = \sum_{i=1}^n M_i \Delta x_i = \sum_{i=1}^k M_i \Delta x_i + \sum_{i=k+1}^n M_i \Delta x_i = U(P_1,f) + U(P_2,f) . \end{equation*}
We wish to take the infimum on the right over all \(P_1\) and \(P_2\text{,}\) and so we are taking the infimum over all partitions \(P\) of \([a,c]\) that contain \(b\text{.}\) If \(Q\) is a partition of \([a,c]\) and \(P = Q \cup \{ b \}\text{,}\) then \(P\) is a refinement of \(Q\) and so \(U(Q,f) \geq U(P,f)\text{.}\) Therefore, taking an infimum only over the \(P\) that contain \(b\) is sufficient to find the infimum of \(U(P,f)\) for all \(P\text{.}\) We obtain
\begin{equation*} \overline{\int_a^c} f = \overline{\int_a^b} f + \overline{\int_b^c} f . \qedhere \end{equation*}

Proof.

Suppose \(f \in \sR\bigl([a,c]\bigr)\text{,}\) then it is bounded and \(\overline{\int_a^c} f = \underline{\int_a^c} f = \int_a^c f\text{.}\) The lemma gives
\begin{equation*} \int_a^c f = \underline{\int_a^c} f = \underline{\int_a^b} f + \underline{\int_b^c} f \leq \overline{\int_a^b} f + \overline{\int_b^c} f = \overline{\int_a^c} f = \int_a^c f . \end{equation*}
Thus the inequality is an equality,
\begin{equation*} \underline{\int_a^b} f + \underline{\int_b^c} f = \overline{\int_a^b} f + \overline{\int_b^c} f . \end{equation*}
As we also know \(\underline{\int_a^b} f \leq \overline{\int_a^b} f\) and \(\underline{\int_b^c} f \leq \overline{\int_b^c} f\text{,}\) we conclude
\begin{equation*} \underline{\int_a^b} f = \overline{\int_a^b} f \qquad \text{and} \qquad \underline{\int_b^c} f = \overline{\int_b^c} f . \end{equation*}
Thus \(f\) is Riemann integrable on \([a,b]\) and \([b,c]\) and the desired formula holds.
Now assume \(f\) is Riemann integrable on \([a,b]\) and on \([b,c]\text{.}\) Again it is bounded, and the lemma gives
\begin{equation*} \underline{\int_a^c} f = \underline{\int_a^b} f + \underline{\int_b^c} f = \int_a^b f + \int_b^c f = \overline{\int_a^b} f + \overline{\int_b^c} f = \overline{\int_a^c} f . \end{equation*}
Therefore, \(f\) is Riemann integrable on \([a,c]\text{,}\) and the integral is computed as indicated.
An easy consequence of the additivity is the following corollary. We leave the details to the reader as an exercise.

Subsection 5.2.2 Linearity and monotonicity

A sum is a linear function of the summands. So is the integral.

Proof.

Let us prove the first item for \(\alpha \geq 0\text{.}\) Let \(P\) be a partition of \([a,b]\text{,}\) and \(m_i \coloneqq \inf \bigl\{ f(x) : x \in [x_{i-1},x_i] \bigr\}\) as usual. As \(\alpha \geq 0\text{,}\) the multiplication by \(\alpha\) moves past the infimum,
\begin{equation*} \inf \bigl\{ \alpha f(x) : x \in [x_{i-1},x_i] \bigr\} = \alpha \inf \bigl\{ f(x) : x \in [x_{i-1},x_i] \bigr\} = \alpha m_i . \end{equation*}
Therefore,
\begin{equation*} L(P,\alpha f) = \sum_{i=1}^n \alpha m_i \Delta x_i = \alpha \sum_{i=1}^n m_i \Delta x_i = \alpha L(P,f). \end{equation*}
Similarly,
\begin{equation*} U(P,\alpha f) = \alpha U(P,f) . \end{equation*}
Again, as \(\alpha \geq 0\text{,}\) we may move multiplication by \(\alpha\) past the supremum. Hence,
\begin{equation*} \begin{split} \underline{\int_a^b} \alpha f(x)\,dx & = \sup \, \bigl\{ L(P,\alpha f) : P \text{ a partition of } [a,b] \bigr\} \\ & = \sup \, \bigl\{ \alpha L(P,f) : P \text{ a partition of } [a,b] \bigr\} \\ & = \alpha \, \sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,b] \bigr\} \\ & = \alpha \underline{\int_a^b} f(x)\,dx . \end{split} \end{equation*}
Similarly, we show
\begin{equation*} \overline{\int_a^b} \alpha f(x)\,dx = \alpha \overline{\int_a^b} f(x)\,dx . \end{equation*}
The conclusion now follows for \(\alpha \geq 0\text{.}\)
To finish the proof of the first item (for \(\alpha < 0\)), we need to show that \(-f\) is Riemann integrable and \(\int_a^b - f(x)\,dx = - \int_a^b f(x)\,dx\text{.}\) The proof of this fact is left as Exercise 5.2.1.
The proof of the second item is left as Exercise 5.2.2. It is not difficult, but it is not as trivial as it may appear at first glance.
The second item in the proposition does not hold with equality for the Darboux integrals, but we do obtain inequalities. The proof of the following proposition is Exercise 5.2.16. It follows for upper and lower sums on a fixed partition by Exercise 1.3.7, that is, supremum of a sum is less than or equal to the sum of suprema and similarly for infima.
Adding up smaller numbers should give us a smaller result. That is true for an integral as well.

Proof.

Let \(P = \{ x_0, x_1, \ldots, x_n \}\) be a partition of \([a,b]\text{.}\) Then let
\begin{equation*} m_i \coloneqq \inf \, \bigl\{ f(x) : x \in [x_{i-1},x_i] \bigr\} \qquad \text{and} \qquad \widetilde{m}_i \coloneqq \inf \, \bigl\{ g(x) : x \in [x_{i-1},x_i] \bigr\} . \end{equation*}
As \(f(x) \leq g(x)\text{,}\) then \(m_i \leq \widetilde{m}_i\text{.}\) Therefore,
\begin{equation*} L(P,f) = \sum_{i=1}^n m_i \Delta x_i \leq \sum_{i=1}^n \widetilde{m}_i \Delta x_i = L(P,g) . \end{equation*}
We take the supremum over all \(P\) (see Proposition 1.3.7) to obtain
\begin{equation*} \underline{\int_a^b} f \leq \underline{\int_a^b} g . \end{equation*}
Similarly, we obtain the same conclusion for the upper integrals. Finally, if \(f\) and \(g\) are Riemann integrable all the integrals are equal, and the conclusion follows.

Subsection 5.2.3 Continuous functions

Let us show that continuous functions are Riemann integrable. We can even allow some discontinuities. We start with a function continuous on the whole closed interval \([a,b]\text{.}\)

Proof.

As \(f\) is continuous on a closed bounded interval, it is bounded and uniformly continuous. Given \(\epsilon > 0\text{,}\) find a \(\delta > 0\) such that \(\abs{x-y} < \delta\) implies \(\abs{f(x)-f(y)} < \frac{\epsilon}{b-a}\text{.}\)
Let \(P = \{ x_0, x_1, \ldots, x_n \}\) be a partition of \([a,b]\) such that \(\Delta x_i < \delta\) for all \(i = 1,2, \ldots, n\text{.}\) For example, take \(n\) such that \(\frac{b-a}{n} < \delta\text{,}\) and let \(x_i \coloneqq \frac{i}{n}(b-a) + a\text{.}\) Then for all \(x, y \in [x_{i-1},x_i]\text{,}\) we have \(\abs{x-y} \leq \Delta x_i < \delta\text{,}\) and so
\begin{equation*} f(x)-f(y) \leq \abs{f(x)-f(y)} < \frac{\epsilon}{b-a} . \end{equation*}
As \(f\) is continuous on \([x_{i-1},x_i]\text{,}\) it attains a maximum and a minimum on this interval. Let \(x\) be a point where \(f\) attains the maximum and \(y\) be a point where \(f\) attains the minimum. Then \(f(x) = M_i\) and \(f(y) = m_i\) in the notation from the definition of the integral. Therefore,
\begin{equation*} M_i-m_i = f(x)-f(y) < \frac{\epsilon}{b-a} . \end{equation*}
And so
\begin{equation*} \begin{split} \overline{\int_a^b} f - \underline{\int_a^b} f & \leq U(P,f) - L(P,f) \\ & = \left( \sum_{i=1}^n M_i \Delta x_i \right) - \left( \sum_{i=1}^n m_i \Delta x_i \right) \\ & = \sum_{i=1}^n (M_i-m_i) \Delta x_i \\ & < \frac{\epsilon}{b-a} \sum_{i=1}^n \Delta x_i \\ & = \frac{\epsilon}{b-a} (b-a) = \epsilon . \end{split} \end{equation*}
As \(\epsilon > 0\) was arbitrary,
\begin{equation*} \overline{\int_a^b} f = \underline{\int_a^b} f , \end{equation*}
and \(f\) is Riemann integrable on \([a,b]\text{.}\)
The second lemma says that we need the function to only be “Riemann integrable inside the interval,” as long as it is bounded. It also tells us how to compute the integral.

Proof.

Let \(M > 0\) be a real number such that \(\abs{f(x)} \leq M\text{.}\) As \((b-a) \geq (b_n-a_n)\text{,}\)
\begin{equation*} -M(b-a) \leq -M(b_n-a_n) \leq \int_{a_n}^{b_n} f \leq M(b_n-a_n) \leq M(b-a) . \end{equation*}
Therefore, the sequence of numbers \(\bigl\{ \int_{a_n}^{b_n} f \bigr\}_{n=1}^\infty\) is bounded and by Bolzano–Weierstrass has a convergent subsequence indexed by \(n_k\text{.}\) Let us call \(L\) the limit of the subsequence \(\bigl\{ \int_{a_{n_k}}^{b_{n_k}} f \bigr\}_{k=1}^\infty\text{.}\)
Lemma 5.2.1 says that the lower and upper integral are additive and the hypothesis says that \(f\) is integrable on \([a_{n_k},b_{n_k}]\text{.}\) Therefore
\begin{equation*} \underline{\int_a^b} f = \underline{\int_a^{a_{n_k}}} f + \int_{a_{n_k}}^{b_{n_k}} f + \underline{\int_{b_{n_k}}^b} f \geq -M(a_{n_k}-a) + \int_{a_{n_k}}^{b_{n_k}} f - M(b-b_{n_k}) . \end{equation*}
We take the limit as \(k\) goes to \(\infty\) on the right-hand side,
\begin{equation*} \underline{\int_a^b} f \geq -M\cdot 0 + L - M\cdot 0 = L . \end{equation*}
Next we use additivity of the upper integral,
\begin{equation*} \overline{\int_a^b} f = \overline{\int_a^{a_{n_k}}} f + \int_{a_{n_k}}^{b_{n_k}} f + \overline{\int_{b_{n_k}}^b} f \leq M(a_{n_k}-a) + \int_{a_{n_k}}^{b_{n_k}} f + M(b-b_{n_k}) . \end{equation*}
We take the same subsequence \(\{ \int_{a_{n_k}}^{b_{n_k}} f \}_{k=1}^\infty\) and take the limit to obtain
\begin{equation*} \overline{\int_a^b} f \leq M\cdot 0 + L + M\cdot 0 = L . \end{equation*}
Thus \(\overline{\int_a^b} f = \underline{\int_a^b} f = L\) and hence \(f\) is Riemann integrable and \(\int_a^b f = L\text{.}\) In particular, no matter what subsequence we chose, the \(L\) is the same number.
To prove the final statement of the lemma we use Proposition 2.3.7. We have shown that every convergent subsequence \(\bigl\{ \int_{a_{n_k}}^{b_{n_k}} f \bigr\}_{k=1}^\infty\) converges to \(L = \int_a^b f\text{.}\) Therefore, the sequence \(\bigl\{ \int_{a_n}^{b_n} f \bigr\}_{n=1}^\infty\) is convergent and converges to \(\int_a^b f\text{.}\)
We say a function \(f \colon [a,b] \to \R\) has finitely many discontinuities if there exists a finite set \(S = \{ x_1, x_2, \ldots, x_n \} \subset [a,b]\text{,}\) and \(f\) is continuous at all points of \([a,b] \setminus S\text{.}\)

Proof.

We divide the interval into finitely many intervals \([a_i,b_i]\) so that \(f\) is continuous on the interior \((a_i,b_i)\text{.}\) If \(f\) is continuous on \((a_i,b_i)\text{,}\) then it is continuous and hence integrable on \([c_i,d_i]\) whenever \(a_i < c_i < d_i < b_i\text{.}\) By Lemma 5.2.8, the restriction of \(f\) to \([a_i,b_i]\) is integrable. By additivity of the integral (and induction), \(f\) is integrable on the union of the intervals.

Subsection 5.2.4 More on integrable functions

Sometimes it is convenient (or necessary) to change certain values of a function and then integrate. The next result says that if we change the values at finitely many points, the integral does not change.

Proof.

(Sketch of proof) Using additivity of the integral, split the interval \([a,b]\) into smaller intervals such that \(f(x) = g(x)\) holds for all \(x\) except at the endpoints (details are left to the reader).
Therefore, without loss of generality suppose \(f(x) = g(x)\) for all \(x \in (a,b)\text{.}\) The proof follows by Lemma 5.2.8, and is left as Exercise 5.2.3.
Finally, monotone (increasing or decreasing) functions are always Riemann integrable. The proof is left to the reader as part of Exercise 5.2.14.

Subsection 5.2.5 Exercises

Exercise 5.2.1.

Finish the proof of the first part of Proposition 5.2.4. Let \(f\) be in \(\sR\bigl([a,b]\bigr)\text{.}\) Prove that \(-f\) is in \(\sR\bigl([a,b]\bigr)\) and
\begin{equation*} \int_a^b - f(x) \,dx = - \int_a^b f(x) \,dx . \end{equation*}

Exercise 5.2.2.

Prove the second part of Proposition 5.2.4. Let \(f\) and \(g\) be in \(\sR\bigl([a,b]\bigr)\text{.}\) Prove, without using Proposition 5.2.5, that \(f+g\) is in \(\sR\bigl([a,b]\bigr)\) and
\begin{equation*} \int_a^b \bigl( f(x)+g(x) \bigr) \,dx = \int_a^b f(x) \,dx + \int_a^b g(x) \,dx . \end{equation*}
Hint: One way to do it is to use Proposition 5.1.7 to find a single partition \(P\) such that \(U(P,f)-L(P,f) < \nicefrac{\epsilon}{2}\) and \(U(P,g)-L(P,g) < \nicefrac{\epsilon}{2}\text{.}\)

Exercise 5.2.3.

Let \(f \colon [a,b] \to \R\) be Riemann integrable, and \(g \colon [a,b] \to \R\) be such that \(f(x) = g(x)\) for all \(x \in (a,b)\text{.}\) Prove that \(g\) is Riemann integrable and that
\begin{equation*} \int_a^b g = \int_a^b f. \end{equation*}

Exercise 5.2.4.

Prove the mean value theorem for integrals: If \(f \colon [a,b] \to \R\) is continuous, then there exists a \(c \in [a,b]\) such that \(\int_a^b f = f(c)(b-a)\text{.}\)

Exercise 5.2.5.

Let \(f \colon [a,b] \to \R\) be a continuous function such that \(f(x) \geq 0\) for all \(x \in [a,b]\) and \(\int_a^b f = 0\text{.}\) Prove that \(f(x) = 0\) for all \(x\text{.}\)

Exercise 5.2.6.

Let \(f \colon [a,b] \to \R\) be a continuous function and \(\int_a^b f = 0\text{.}\) Prove that there exists a \(c \in [a,b]\) such that \(f(c) = 0\text{.}\) (Compare with the previous exercise.)

Exercise 5.2.7.

Let \(f \colon [a,b] \to \R\) and \(g \colon [a,b] \to \R\) be continuous functions such that \(\int_a^b f = \int_a^b g\text{.}\) Show that there exists a \(c \in [a,b]\) such that \(f(c) = g(c)\text{.}\)

Exercise 5.2.8.

Let \(f \in \sR\bigl([a,b]\bigr)\text{.}\) Let \(\alpha, \beta, \gamma\) be arbitrary numbers in \([a,b]\) (not necessarily ordered in any way). Prove
\begin{equation*} \int_\alpha^\gamma f = \int_\alpha^\beta f + \int_\beta^\gamma f . \end{equation*}
Recall what \(\int_a^b f\) means if \(b \leq a\text{.}\)

Exercise 5.2.10.

Suppose \(f \colon [a,b] \to \R\) is bounded and has finitely many discontinuities. Show that as a function of \(x\) the expression \(\abs{f(x)}\) is bounded with finitely many discontinuities and is thus Riemann integrable. Then show
\begin{equation*} \abs{\int_a^b f(x)\,dx} \leq \int_a^b \abs{f(x)}\,dx . \end{equation*}

Exercise 5.2.11.

(Hard)   Show that the Thomae or popcorn function (see Example 3.2.12) is Riemann integrable. Therefore, there exists a function discontinuous at all rational numbers (a dense set) that is Riemann integrable.
That is, define \(f \colon [0,1] \to \R\) by
\begin{equation*} f(x) \coloneqq \begin{cases} \nicefrac{1}{k} & \text{if } x=\nicefrac{m}{k} \text{ where } m,k \in \N \text{ and } m \text{ and } k \text{ have no common divisors,} \\ 0 & \text{if } x \text{ is irrational.} \end{cases} \end{equation*}
Show \(\int_0^1 f = 0\text{.}\)
If \(I \subset \R\) is a bounded interval, then the function
\begin{equation*} \varphi_I(x) \coloneqq \begin{cases} 1 & \text{if } x \in I, \\ 0 & \text{otherwise,} \end{cases} \end{equation*}
is called an elementary step function.

Exercise 5.2.12.

Let \(I\) be an arbitrary bounded interval (you should consider all types of intervals: closed, open, half-open) and \(a < b\text{,}\) then using only the definition of the integral show that the elementary step function \(\varphi_I\) is integrable on \([a,b]\text{,}\) and find the integral in terms of \(a\text{,}\) \(b\text{,}\) and the endpoints of \(I\text{.}\)
A function \(f\) is called a step function if it can be written as
\begin{equation*} f(x) = \sum_{k=1}^n \alpha_k \varphi_{I_k} (x) \end{equation*}
for some real numbers \(\alpha_1,\alpha_2, \ldots, \alpha_n\) and some bounded intervals \(I_1,I_2,\ldots,I_n\text{.}\)

Exercise 5.2.13.

Using Exercise 5.2.12, show that a step function (see above) is integrable on every interval \([a,b]\text{.}\) Furthermore, find the integral in terms of \(a\text{,}\) \(b\text{,}\) the endpoints of \(I_k\) and the \(\alpha_k\text{.}\)

Exercise 5.2.14.

Let \(f \colon [a,b] \to \R\) be a function.
  1. Show that if \(f\) is increasing, then it is Riemann integrable. Hint: Use a uniform partition; each subinterval of same length.
  2. Use part a) to show that if \(f\) is decreasing, then it is Riemann integrable.
  3. Suppose 1  \(h = f-g\) where \(f\) and \(g\) are increasing functions on \([a,b]\text{.}\) Show that \(h\) is Riemann integrable.

Exercise 5.2.15.

(Challenging)   Suppose \(f \in \sR\bigl([a,b]\bigr)\text{,}\) then the function that takes \(x\) to \(\abs{f(x)}\) is also Riemann integrable on \([a,b]\text{.}\) Then show the same inequality as Exercise 5.2.10.

Exercise 5.2.16.

Suppose \(f \colon [a,b] \to \R\) and \(g \colon [a,b] \to \R\) are bounded.
  1. Show \(\overline{\int_a^b} (f+g) \leq \overline{\int_a^b}f+\overline{\int_a^b}g\) and \(\underline{\int_a^b} (f+g) \geq \underline{\int_a^b}f+\underline{\int_a^b}g\text{.}\)
  2. Find example \(f\) and \(g\) where the inequality is strict. Hint: \(f\) and \(g\) should not be Riemann integrable.

Exercise 5.2.17.

Suppose \(f \colon [a,b] \to \R\) is continuous and \(g \colon \R \to \R\) is Lipschitz continuous. Define
\begin{equation*} h(x) \coloneqq \int_a^b g(t-x) f(t) \, dt . \end{equation*}
Prove that \(h\) is Lipschitz continuous.

Exercise 5.2.18.

Prove a version of the so-called Riemann–Lebesgue Lemma (one of several such named): Suppose \(f \colon [a,b] \to \R\) is continuous and define the sequence \(\{ x_n \}_{n=1}^\infty\) by
\begin{equation*} x_n \coloneqq \int_a^b f(t) \sin(nt) \, dt . \end{equation*}
Prove that \(\lim\limits_{n\to\infty} x_n = 0\text{.}\)
Such an \(h\) is said to be of bounded variation.
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