The Riemann Integral

Derivatives measure rates. Integrals measure totals: area under a curve, distance from speed, work from force, mass from density. This last chapter of Book 2A defines the integral as Riemann did in 1854, by squeezing the area between sums of rectangles from below and above. It proves that continuous functions can be integrated, and proves the fundamental theorem of calculus, which says that integration and differentiation undo each other.

Two consequences of the fundamental theorem matter more than any other for this guidebook. Integration by parts is the single most-used computation from here to Perelman: every energy estimate for the heat equation and every monotonicity formula for Ricci flow is an integration by parts. And the chapter ends by showing where Riemann's integral breaks down. It does not cope with limits, and that failure is the starting point of Course 3.

By the end of this chapter you will be able to:

• define the Riemann integral by upper and lower sums, and prove a function is or isn't integrable;
• prove that continuous and monotone functions are integrable;
• prove and use both halves of the fundamental theorem of calculus;
• integrate by parts and change variables, and use integration by parts to prove that an energy decreases;
• explain, with examples, why the Riemann integral is not good enough for analysis with limits.

11.1Area by rectangles

The idea is ancient. Archimedes, in the third century BCE, found the area of a parabolic segment by filling it with triangles and summing a geometric series (2A.7 Series). What Riemann supplied in 1854, in the same habilitation thesis as his rearrangement theorem, was a precise definition that works for a large class of functions at once.

Work on a closed interval $[a, b]$ with a bounded function $f$. A partition $P$ of $[a, b]$ cuts it into finitely many subintervals $[x_{k-1}, x_k]$ with $a = x_0 < x_1 < \cdots < x_n = b$. On each piece, the smallest and largest heights of $f$ are $m_k = \inf_{[x_{k-1}, x_k]} f$ and $M_k = \sup_{[x_{k-1}, x_k]} f$. Rectangles of those heights give an area that is certainly too small and one that is certainly too big:

Refining a partition (adding points) can only raise $L$ and lower $U$, and every lower sum is at most every upper sum (Exercise 11.10). So the following definition makes sense.

The two integrals exist because of the least upper bound property (2A.4 The Real Numbers), which keeps appearing. The working criterion is a single ε statement: $f$ is integrable if and only if for every $\varepsilon > 0$ there is a partition $P$ with $U(f, P) - L(f, P) \leq \varepsilon$. In words, the thin strips in Figure 11.1 can be given total area as small as you like.

11.1.1Which functions are integrable

Here is the payoff of uniform continuity: one $\delta$ for the whole interval means one mesh size controls every strip at once. With mere continuity, the mesh needed would vary from place to place, and the argument would fail.

Monotone functions are integrable too, even with jumps: for $n$ equal pieces of an increasing function, $U - L = \frac{b - a}{n}\big(f(b) - f(a)\big)$, because the strips telescope (Exercise 11.11). And a bounded function with finitely many discontinuities is integrable, by putting the discontinuities in tiny intervals.

But not every bounded function is integrable.

Riemann's integral cannot assign an area to this function. That would be a curiosity, except that the function arises naturally as a limit of perfectly good integrable functions, as the last section of this chapter shows.

11.1.2Basic properties

For integrable $f$ and $g$ on $[a, b]$:

• Linearity: $\int (f + g) = \int f + \int g$ and $\int cf = c\int f$.
• Monotonicity: if $f \leq g$ then $\int f \leq \int g$. In particular $\big|\int f\big| \leq \int |f|$ (and $|f|$ is integrable).
• Additivity: for $a < c < b$, $\int_a^b f = \int_a^c f + \int_c^b f$.
• Products and compositions: $fg$ is integrable, and so is $\phi \circ f$ for continuous $\phi$.

Each is proved by comparing upper and lower sums. The second property gives the estimate used constantly: if $|f| \leq M$ on $[a, b]$, then $\big|\int_a^b f\big| \leq M(b - a)$.

11.2The fundamental theorem of calculus

Part 1 says that integration produces antiderivatives, and part 2 says that antiderivatives compute integrals. That the slope problem and the area problem are inverse to each other was the great discovery of Newton and Leibniz in the 1660s–1680s. The proofs above, using only the definitions, show how much the 19th-century foundations of 2A.4 The Real Numbers and 2A.9 Continuous Functions were needed to make it rigorous.

11.2.1Integration by parts

The formula moves a derivative from one factor to the other at the cost of a boundary term. When the boundary term vanishes, as it does for periodic functions or for functions that vanish at the ends, the derivative simply moves across with a change of sign. That one move drives a remarkable amount of mathematics. Here is the first example.

Why "energy"? If $u(x, t)$ is the temperature in a ring of metal, it obeys the heat equation $\partial_t u = \partial_x^2 u$. Differentiating $\int u^2$ under the integral sign (justified later, in 3A.3 The Lebesgue Integral) and applying the proposition at each time gives

The quantity $\int u^2$ can only decrease, and it stays constant only if $\partial_x u \equiv 0$, that is, when the temperature is uniform. This is the first monotonicity formula of the guidebook: a quantity that changes in one direction only, with a rate given by a square that vanishes exactly in the equilibrium state. It is proved by integration by parts. Fourier's solution of the heat equation (2B.7 Fourier Series and the First Heat Equation) shows the same decay frequency by frequency.

11.2.2Change of variables

The factor $\phi'(x)$ records how much the substitution stretches lengths. In several variables it becomes the Jacobian determinant (1A.9 Multiple Integrals and Change of Variables, 3A.5 Product Measures and Change of Variables), and on a manifold it becomes the way volumes transform under a change of coordinates (8A.8 Differential Forms and Stokes’ Theorem).

11.3Sums and integrals

Integrals and sums can be compared directly, which settles the question left open in 2A.7 Series about how fast the harmonic series grows.

For $f(x) = 1/x$, define $\ln x = \int_1^x \frac{dt}{t}$, which is the natural logarithm (2B.6 Power Series, Exponentials and Bump Functions shows it agrees with the inverse of the exponential). The integral test gives $\ln(n + 1) \leq H_n \leq 1 + \ln n$. The difference $H_n - \ln n$ is decreasing and bounded below by $0$, so by monotone convergence (2A.6 Sequences) it converges. Its limit is the Euler–Mascheroni constant $\gamma = 0.5772\ldots$, which explains the estimate $H_n \approx \ln n + 0.5772$ used in 2A.7 Series.

11.4Riemann–Stieltjes integrals, briefly

Tao's chapter includes a generalisation that is worth knowing exists, even if it can be skimmed now. Replace the length $x_k - x_{k-1}$ of each piece by the increase $\alpha(x_k) - \alpha(x_{k-1})$ of an increasing function $\alpha$. The resulting $\int_a^b f\,d\alpha$ weights different parts of the interval differently. If $\alpha$ has a jump, that point gets a positive weight of its own.

11.5Where Riemann's integral fails

Here is the problem that drives Course 3. Integrals and limits should be interchangeable in reasonable situations: if $f_n \to f$, we would like $\int f_n \to \int f$. Riemann's integral fails at this in two distinct ways.

The limit may not be integrable. List the rationals in $[0, 1]$ as $q_1, q_2, q_3, \dots$ (2A.8 Infinite Sets), and let $f_n$ be $1$ at $q_1, \dots, q_n$ and $0$ elsewhere. Each $f_n$ differs from $0$ at finitely many points, so it is integrable with $\int f_n = 0$. The sequence increases to Dirichlet's function at every point, and Dirichlet's function has no Riemann integral at all (Example 11.4).

The integrals may not converge to the integral of the limit. Let $f_n$ be a narrow triangular spike of height $n$ on $[0, \tfrac2n]$, peaking at $\tfrac1n$, and zero elsewhere (Figure 11.6). For each fixed $x$, eventually the spike has moved past $x$ (or $x = 0$, where every $f_n$ is $0$), so $f_n(x) \to 0$ for every $x$. But every spike has area $\tfrac12 \cdot \tfrac2n \cdot n = 1$. So

The area hasn't disappeared. It has concentrated into an ever thinner spike and escaped through the limit.

The second failure is not really the integral's fault. The spikes show something true about limits, which any integral must respect, and the theorems of 3A.3 The Lebesgue Integral say exactly when it can't happen: when the functions are dominated by a single integrable function. The first failure is the integral's fault, and it is fixed by building a better one.

11.6Exercises

Solution

Because $f$ is increasing, on the $k$-th piece $M_k = f(x_k)$ and $m_k = f(x_{k-1})$. The sum $\sum_k (f(x_k) - f(x_{k-1}))\frac{b-a}{n}$ telescopes to $\frac{b-a}{n}(f(b) - f(a))$, which tends to $0$.

Solution

With $F(y) = \int_0^y \sqrt{1 + t^3}\,dt$, the expression is $F(x^2)$, whose derivative is $F'(x^2)\cdot 2x = 2x\sqrt{1 + x^6}$.

Solution

(a) $E' = 2\int u\,u_{xx} = -2\int u_x^2 \leq 0$. (b) $D' = 2\int u_x\,u_{xt} = 2\int u_x (u_{xx})_x = -2\int u_{xx}^2 \leq 0$, integrating by parts once more. (c) $\frac{d}{dt}\int u = \int u_{xx} = [u_x]_0^1 = 0$ by periodicity.