The Riemann integral is one of the great ideas of calculus. It gives a precise meaning to the area under a curve, integrates every continuous function on a closed interval, and supports the fundamental theorem of calculus. For many computational problems, it is exactly the right tool.
So why develop another theory of integration?
The short answer is that analysis asks us to do more than integrate one well-behaved function at a time. We want to integrate functions with many discontinuities, take limits of sequences of functions, assign sizes to complicated sets, and so on. In these settings, the Riemann integral may show some difficulties.
This chapter motivates the need for a different theory of integration that resolve those difficulties. The point is not that Riemann integration is wrong; it is that its design is not broad enough for the questions we want to ask.
Construction of the Riemann integral
Suppose \(a, b \in \mathbb R\) with \(a < b\). A partition of the interval \([a,b]\) is a finite sequence of points \((x_i)_{i=0}^n\) such that \[
a = x_0 < x_1 < \cdots < x_n = b.
\]
We can think of a closed interval \([a,b]\) as union of closed subintervals: \[
[a, b] = [x_0, x_1] \cup [x_1, x_2] \cup \cdots \cup [x_{n-1}, x_n].
\]
Let \(f: [a,b] \to \mathbb R\) be a bounded function, and \(P = (x_i)_{i=0}^n\) a partition of \([a,b]\). The lower Riemann sum of \(L(f, P, [a,b])\) and the upper Riemann sum \(U(f, P, [a,b])\) are defined by \[
L(f,P,[a,b]) = \sum_{k=1}^n (x_k - x_{k-1})\inf_{[x_{k-1}, x_k]} f,
\] and \[
U(f,P,[a,b]) = \sum_{k=1}^n (x_k - x_{k-1})\sup_{[x_{k-1}, x_k]} f.
\]
Intuitively, the lower Riemann sum should be a lower bound for the area under the graph of \(f\), while the upper Riemann sum should be an upper bound.
Define \(f:[0,1]\to\mathbb{R}\) by \(f(x)=x^2\). Let \(P_n\) denote the partition \(0, \frac 1n, \frac 2n, \ldots, \frac{n-1}{n}, 1\) of \([0,1]\). For the partition \(P_n\), we have \(x_j - x_{j-1} = \frac 1n\) for each \(j=1,2,\ldots,n\). Thus \[
L(x^2, P_n, [0,1]) = \frac 1n \sum_{j=1}^n \frac{(j-1)^2}{n^2} = \frac{2n^2 - 3n + 1}{6n^2}
\] and \[
U(x^2, P_n, [0,1]) = \frac 1n \sum_{j=1}^n \frac{j^2}{n^2} = \frac{2n^2 + 3n + 1}{6n^2}.
\]
The next result shows that the lower and upper Riemann sums are compatible with refinement of partitions.
Let \(f: [a,b] \to \mathbb R\) be a bounded function, and let \(P\) and \(P'\) be partitions of \([a,b]\) such that \(P'\) is a refinement of \(P\), i.e. the list of points in \(P\) is a subset of the list of points in \(P'\). Then \[
L(f,P,[a,b]) \le L(f,P',[a,b]) \le U(f,P',[a,b]) \le U(f,P,[a,b]).
\]
To prove the first inequality, let \(P = (x_j)_{j=0}^n\) and \(P' = (x'_k)_{k=0}^{n'}\) be partitions of \([a,b]\) such that \(P'\) is a refinement of \(P\). Note that for each \(j = 1,2,\ldots,n\), there exists \(k \in \{0, ..., n'-1\}\) and a positive integer \(m\) such that \(x_{j-1} = x'_k < x'_{k+1} < \cdots < x'_{k+m} = x_j\). We have
\[\begin{align*}
(x_j - x_{j-1}) \inf_{[x_{j-1}, x_j]} f &= \sum_{i=1}^m (x'_{k+i} - x'_{k+i-1}) \inf_{[x_{j-1}, x_j]} f
\\ &\le \sum_{i=1}^m (x'_{k+i} - x'_{k+i-1}) \inf_{[x'_{k+i-1}, x'_{k+i}]} f.
\end{align*}\]
The inequality above implies that \(L(f, P, [a,b]) \le L(f, P', [a,b])\). The last inequalities can be proved similarly.
The middle inequality follows directly from the observation: \[
\inf_{[x_{k-1}, x_k]} f \le \sup_{[x_{k-1}, x_k]} f.
\]
Suppose \(f: [a,b] \to \mathbb R\) is a bounded function. Then for any two partitions \(P\) and \(P'\) of \([a,b]\), we have \[
L(f,P,[a,b]) \le U(f,P',[a,b]).
\]
Let \(P''\) be a common refinement of \(P\) and \(P'\). Then by the previous proposition, we have \[
L(f,P,[a,b]) \le L(f,P'',[a,b]) \le U(f,P'',[a,b]) \le U(f,P',[a,b]).
\]
The lower and upper Riemann integrals are defined as follows.
The lower Riemann integral of \(f\) over \([a,b]\) is \[
L(f, [a,b]) = \sup_P L(f,P,[a,b]),
\] and the upper Riemann integral of \(f\) over \([a,b]\) is \[
U(f, [a,b]) = \inf_P U(f,P,[a,b]).
\] where the supremum and infimum are taken over all partitions \(P\) of \([a,b]\).
A similar inequality can be achieved for the lower and upper Riemann integrals with a similar proof.
Let \(f: [a,b] \to \mathbb R\) be a bounded function. Then \[
L(f, [a,b]) \le U(f, [a,b]).
\]
We would want the lower and upper Riemann integrals to be equal. We define the notion of Riemann integrability and the Riemann integral as follows.
A bounded function on a closed bounded interval is called Riemann integrable if its lower and upper Riemann integrals are equal. If \(f: [a,b] \to \mathbb R\) is Riemann integrable, then the common value of the lower and upper Riemann integrals is called the Riemann integral of \(f\) over \([a,b]\), denoted by \[
\int_a^b f = L(f, [a,b]) = U(f, [a,b]).
\] Alternatively, we can also write \(\int_a^b f(x) dx\), or use any dummy variable in place of \(x\).
Define \(f:[0,1]\to\mathbb{R}\) by \(f(x)=x^2\). Then \[
U(f, [0,1]) \le \inf_{n\in\mathbb N}\frac{2n^2 + 3n + 1}{6n^2} = \frac 13 \le \sup_{n\in\mathbb N}\frac{2n^2 - 3n + 1}{6n^2} \le L(f, [0,1]).
\] This shows that \(U(f, [0,1]) \le \frac 13 \le L(f, [0,1])\). Combine with the fact that the lower Riemann integral is less than or equal to the upper Riemann integral, this shows that \(L(f, [0,1]) = U(f, [0,1]) = \frac 13\). Therefore, \(f\) is Riemann integrable on \([0,1]\) and \(\int_0^1 x^2 dx = \frac 13\).
A key result regarding Riemann integrability is the following.
Every continuous real-valued function on each closed bounded interval is Riemann integrable.
The proof uses a fundamental property of continuous functions on closed bounded intervals, which is uniform continuity, that is, for every \(\epsilon > 0\), there exists \(\delta > 0\) such that for any \(x, y \in [a,b]\), if \(|x - y| < \delta\), then \(|f(x) - f(y)| < \epsilon\).
Suppose \(a, b \in \mathbb R\) with \(a < b\), and let \(f: [a,b] \to \mathbb R\) be a continuous function. Then \(f\) is uniformly continuous on \([a,b]\). Let \(\epsilon > 0\). By uniform continuity, there exists \(\delta > 0\) such that for any \(x, y \in [a,b]\), if \(|x - y| < \delta\), then \(|f(x) - f(y)| < \frac{\epsilon}{b-a}\).
Let \(P = (x_i)_{i=0}^n\) be a partition of \([a,b]\) such that \(\max_{1\le i\le n}(x_i - x_{i-1}) < \delta\). Then for each \(k = 1,2,\ldots,n\), we have \[
\sup_{[x_{k-1}, x_k]} f - \inf_{[x_{k-1}, x_k]} f < \frac{\epsilon}{b-a}.
\] Thus \[\begin{align*}
U(f, [a,b]) - L(f, [a,b]) &\le U(f,P,[a,b]) - L(f,P,[a,b]) \\
&= \sum_{k=1}^n (x_k - x_{k-1})\left(\sup_{[x_{k-1}, x_k]} f - \inf_{[x_{k-1}, x_k]} f\right) \\
&< \sum_{k=1}^n (x_k - x_{k-1})\frac{\epsilon}{b-a} = (b-a)\frac{\epsilon}{b-a} = \epsilon.
\end{align*}\] Since \(\epsilon > 0\) is arbitrary, we have \(U(f, [a,b]) - L(f, [a,b]) = 0\). Therefore, \(f\) is Riemann integrable on \([a,b]\).
Finally, this result gives bounds for a Riemann integral.
Let \(f: [a,b] \to \mathbb R\) be a Riemann integrable function. Then \[
(b-a)\inf_{[a,b]} f \le \int_a^b f \le (b-a)\sup_{[a,b]} f.
\]
Let \(P\) be the trivial partition \(x_0 = a, x_1 = b\). Then \[\begin{align*}
(b-a)\inf_{[a,b]} f &= L(f,P,[a,b]) \le L(f, [a,b]) = \int_a^b f
\\ &= U(f, [a,b]) \le U(f,P,[a,b]) = (b-a)\sup_{[a,b]} f.
\end{align*}\]
When does this fail?
There are several occasions where Riemann integration fails. In particular, it fails to integrate functions with many discontinuities, it does not handle unbounded functions, and it does not work well with limits.
We provide some examples to illustrate these failures.
Define \(f: [0,1] \to \mathbb R\) by \[
f(x) = \begin{cases}
1, & \text{if } x \in \mathbb Q, \\
0, & \text{if } x \in \mathbb R \setminus \mathbb Q.
\end{cases}
\] If \([a, b] \subseteq [0,1]\) with \(a<b\), then \[
\inf_{[a,b]} f = 0 \quad \text{and} \quad \sup_{[a,b]} f = 1.
\] Thus for any partition \(P\) of \([0,1]\), we have \(L(f,P,[0,1]) = 0\) and \(U(f,P,[0,1]) = 1\). Therefore, \(L(f, [0,1]) = 0\) and \(U(f, [0,1]) = 1\), which shows that \(f\) is not Riemann integrable on \([0,1]\).
Trying to apply the Riemann integral to unbounded functions also fails.
Define \(f: [0,1] \to \mathbb R\) by \[
f(x) = \begin{cases}
\frac{1}{\sqrt x}, & \text{if } x \in (0,1], \\
0, & \text{if } x = 0.
\end{cases}
\] Then for any partition \(P\) of \([0,1]\), we would have \(\sup_{[x_0,x_1]} f = \infty\), and thus \(U(f,P,[0,1]) = \infty\). However, we should consider the area under the graph to be 2, since \[
\lim_{a \downarrow 0} \int_a^1 \frac{1}{\sqrt x} dx = \lim_{a \downarrow 0} \left(2 - 2\sqrt a\right) = 2.
\]
Calculus deals with issues by defining the improper Riemann integral as the limit of Riemann integrals over subdomains and then taking limits. This approach is not always satisfactory.
Let \(r_1,r_2,\dots\) be a sequence that includes each rational number in \((0,1)\) exactly once. Define \(f_k: [0,1] \to \mathbb R\) by \[
f_k(x) = \begin{cases}
\frac{1}{\sqrt{x-r_k}}, & \text{if } x > r_k, \\
0, & \text{if } x \le r_k.
\end{cases}
\]
Define \(f: [0,1] \to [0, \infty]\) by \[
f(x) = \sum_{k=1}^\infty \frac{f_k(x)}{2^k}.
\] Because every nonempty open interval in \((0,1)\) contains a rational number, the function \(f\) is unbounded on every such subinterval. Thus the Riemann integral of \(f\) is undefined on every subinterval of \([0,1]\) with more than one element.
However, the area under the graph of each \(f_k\) is less than 2, and the formula defining \(f\) shows that we should expect the area under the graph of \(f\) to be less than 2 instead of being undefined.
Furthermore, the pointwise limit of a sequence of Riemann integrable bounded functions need not be Riemann integrable.
Let \(r_1,r_2,\dots\) be a sequence that includes each rational number in \((0,1)\) exactly once. Define \(f_k: [0,1] \to \mathbb R\) by \[
f_k(x) = \begin{cases}
1, & \text{if } x \in \{r_1, r_2,...,r_k\}, \\
0, & \text{otherwise.}
\end{cases}
\]
Then each \(f_k\) is Riemann integrable and \(\int_0^1 f_k = 0\). Define \(f: [0,1]\to \mathbb R\) by \[
f(x) = \begin{cases}
1, & \text{if } x \text{ is rational}, \\
0, & \text{otherwise.}
\end{cases}
\] Clearly \[
\lim_{k\to\infty} f_k(x) = f(x) \qquad \text{ for each } x \in[0,1].
\] However, \(f\) is not Riemann integrable.
The previous example points out a drawback in that limit and Riemann integral cannot be interchanged. Indeed, the conditions for interchangability of limit and Riemann integral can be a bit restrictive, in that it requires the limiting function \(f\) to also be Riemann integrable.
Let \(a, b \in \mathbb R\) with \(a<b\). Let \(f_1,f_2,\dots\) be a sequence of Riemann integrable functions on \([a,b]\) such that \[
|f_k(x)| \le M \qquad \forall k \in \mathbb N, x\in[a,b]
\] for some \(M \in\mathbb R\). Suppose \(\lim_{k\to\infty}f_k(x)\) exists for each \(x \in [a,b]\). Denote \[
f(x) = \lim_{k\to\infty} f_k(x)
\] If \(f\) is Riemann integrable on \([a,b]\), then \[
\int_a^b f = \lim_{k\to\infty}\int_a^b f_k
\]
The proof for the above result is more intricate than the other proofs we have on Riemann integration, we do not provide one here. We will give a clean proof of a stronger result using the tools of measure theory that will be developed in the next chapters.
Where does this leave us?