The Lebesgue Integral

With Lebesgue measure in hand (3A.2 Lebesgue Measure), the integral almost builds itself. Riemann cut the domain into pieces and asked how high the function is on each. Lebesgue cuts the range into pieces and asks how much of the domain the function spends at each height. The second question needs a measure of complicated sets such as $\{x : f(x) > t\}$, which is exactly what the last chapter provided.

The payoff is three theorems about limits, which are the reason this integral is used everywhere in analysis. Monotone convergence: for increasing sequences of non-negative functions, the limit of the integrals is the integral of the limit. Fatou's lemma: in general, mass can be lost in a limit, but never gained. Dominated convergence: if all the functions are bounded by one fixed integrable function, nothing is lost. Each comes with a clear account of what goes wrong without it, and the failures are the same three escapes of 2B.3 Compactness and 2A.11 The Riemann Integral: mass moving off to infinity, spreading out thin, or concentrating into a spike. That last escape, concentration, is what a singularity of Ricci flow looks like.

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

• define measurable functions and the Lebesgue integral, and compute integrals by approximating with simple functions;
• prove and use the monotone convergence theorem, Fatou's lemma and the dominated convergence theorem;
• identify which escape to infinity defeats a given limit, and find a dominating function when there is one;
• differentiate under the integral sign, with a stated domination hypothesis;
• explain why the Lebesgue integral extends the Riemann integral and why $L^1$ is the completion of the continuous functions.

3.1Two ways to count money

3.2Measurable functions and simple functions

Every continuous function is measurable ($\{f > t\}$ is open). So is every function that you can build from measurable ones by sums, products, compositions with continuous functions, suprema, infima and, crucially, pointwise limits of sequences: $\{\sup_nf_n > t\} = \bigcup_n\{f_n > t\}$, a countable union, and $\lim f_n = \lim\sup f_n = \inf_N\sup_{n\geq N}f_n$. This closure under limits is the first thing the Riemann theory lacked.

A simple function is a finite linear combination of indicator functions of measurable sets, $\phi = \sum_{i=1}^kc_i\,1_{E_i}$. Its integral is the obvious one:

which doesn't depend on how $\phi$ is written (Tao, Lemma 1.3.4). This is Lebesgue's pile of coins: values times the sizes of the sets where they occur.

3.3The integral of a non-negative function

Immediate properties: the integral is monotone ($f \leq g$ implies $\int f \leq \int g$), $\int c f = c\int f$ for $c \geq 0$, and $\int 1_E = m(E)$. Two more, used constantly:

• Markov's inequality. For $t > 0$, $m(\{f \geq t\}) \leq \frac1t\int f$. (Compare $f$ with the simple function $t\,1_{\{f\geq t\}} \leq f$.)
• Vanishing. $\int f = 0$ if and only if $f = 0$ almost everywhere. (If $\int f = 0$, Markov gives $m(\{f \geq \frac1n\}) = 0$ for every $n$, and $\{f > 0\}$ is the countable union of these.)

So functions that agree almost everywhere have the same integral, and null sets are invisible to integration (3A.2 Lebesgue Measure).

3.4Monotone convergence

The factor $\lambda < 1$ is an "epsilon of room" (Tao's phrase): we can't expect $f_n \geq \phi$ eventually, only $f_n \geq \lambda\phi$, and that is enough. Two consequences follow at once:

• Linearity. $\int(f + g) = \int f + \int g$ for non-negative measurable $f, g$: it holds for simple functions, and approximating $f$ and $g$ by increasing simple functions, monotone convergence passes it to the limit.
• Series. For non-negative measurable $g_k$, $\int\sum_kg_k = \sum_k\int g_k$. The partial sums increase.

The second statement is an interchange of a limit and an integral with no hypothesis except non-negativity. Compare 2B.5 Uniform Convergence and Arzelà–Ascoli, where uniform convergence was needed.

3.5Fatou's lemma

Without monotonicity, the integral of the limit can be strictly smaller than the limit of the integrals. Fatou's lemma says it can never be larger.

Read it as a statement about energy. If $\int f_n$ is an energy, Fatou says energy can be lost in a limit, never created. In the language of 2B.5 Uniform Convergence and Arzelà–Ascoli and 4A.6 Weak Convergence and the Direct Method, the integral is lower semicontinuous under pointwise convergence. That is exactly the property needed to show a minimum exists: take a sequence whose energies approach the infimum, extract a limit by some compactness theorem, and Fatou ensures the limit's energy is no larger than the infimum, so it is a minimiser. This direct method proves the existence of minimisers in 4A.6 Weak Convergence and the Direct Method and 6A.9 Calculus of Variations and Gradient Flows, and, in 12A.3 The 𝓦-Entropy, the existence of the minimiser behind Perelman's $\mu$-functional.

3.6Integrable functions and dominated convergence

A measurable $f : \mathbb{R}^d \to \mathbb{R}$ is absolutely integrable, written $f \in L^1(\mathbb{R}^d)$, if $\int|f| < \infty$. Then its integral is $\int f = \int f^+ - \int f^-$, where $f^+ = \max(f, 0)$ and $f^- = \max(-f, 0)$; for complex $f$, integrate the real and imaginary parts. The integral is linear on $L^1$, and $\big|\int f\big| \leq \int|f|$.

The proof is three lines because Fatou does the work, applied to the right non-negative functions. The hypothesis to check in practice is the dominating function $G$: one integrable function above all the $|f_n|$ at once. Bounded convergence on a set of finite measure is the special case $G = M\,1_E$.

3.6.1The Riemann integral is a special case

A bounded Riemann-integrable function $f$ on $[a, b]$ is Lebesgue integrable with the same integral. The lower and upper step functions of finer and finer partitions (2A.11 The Riemann Integral) increase and decrease to functions $\ell \leq f \leq u$; by monotone (or dominated) convergence their Lebesgue integrals converge to the lower and upper Riemann integrals, which are equal; so $\int(u - \ell) = 0$, $u = \ell = f$ almost everywhere, and $f$ is measurable with $\int f = \int_a^bf$ (Tao, Exercise 1.3.17). The Lebesgue integral loses nothing, and gains Dirichlet's function: $1_{\mathbb{Q}\cap[0,1]} = 0$ almost everywhere, so its integral is $0$.

3.7The three escapes

When the dominated convergence theorem doesn't apply, mass can disappear in a limit. Tao's §1.4.5 names three ways, all on the real line with Lebesgue measure.

1. Escape to horizontal infinity. $f_n = 1_{[n, n+1]}$: a block of height $1$ sliding to the right. $f_n \to 0$ pointwise, but $\int f_n = 1$.
2. Escape to width infinity. $f_n = \frac1n1_{[0, n]}$: a plateau of height $\frac1n$ spreading out. $f_n \to 0$ uniformly, but $\int f_n = 1$.
3. Escape to vertical infinity. $f_n = n\,1_{[1/n, 2/n]}$: a spike getting taller and thinner. $f_n \to 0$ pointwise, but $\int f_n = 1$.

In each case the mass $1$ disappears from the limit, and in each case there is no integrable $G$ above all the $f_n$: the supremum $\sup_nf_n$ is $1_{[1, \infty)}$ (roughly), $1_{[0,1]} + \sum_n\frac1n1_{(n-1, n]}$ (like $\frac1x$), and of size $\frac1x$ near $0$, none of them integrable. Fatou's inequality is strict in all three. These are the compactness failures of 2B.3 Compactness (escape, a missing point, oscillation into new directions) seen through integrals, and they return as the obstructions to compact Sobolev embeddings (4A.10 Sobolev Embeddings and Critical Exponents).

3.8Differentiating under the integral sign

Variations of functionals, the calculations at the heart of Ricci flow, require moving a derivative inside an integral. The dominated convergence theorem says exactly when that is allowed.

3.9$L^1$ completes the Riemann integral

The opening complaint of this book is now answered. Write $L^1(\mathbb{R}^d)$ for the integrable functions, with functions equal almost everywhere identified, and $\|f\|_1 = \int|f|$.

• $L^1$ is complete: every Cauchy sequence in $\|\cdot\|_1$ converges in $L^1$ (the Riesz–Fischer theorem, proved for all $L^p$ in 3A.7 Lᵖ Spaces and Jensen’s Inequality).
• Continuous functions with compact support are dense in $L^1$: every integrable function is within $\varepsilon$ in $\|\cdot\|_1$ of one (Littlewood's second principle, Tao §1.3.5; proved by approximating indicators of sets of finite measure by indicators of finite unions of boxes, then smoothing the corners).

So $L^1$ is the completion of the continuous compactly supported functions under the distance $\int|f - g|$, exactly as $\mathbb{R}$ is the completion of $\mathbb{Q}$ (2A.4 The Real Numbers). The incomplete space of 2B.2 Completeness and Contraction, continuous functions with the area metric, has been completed, and the limits that were missing are now integrable functions.

3.10History

Lebesgue's thesis (1902) defined the integral and proved the bounded convergence theorem. Beppo Levi proved the monotone convergence theorem in 1906, and Pierre Fatou his lemma in the same year, in a thesis on Fourier series and analytic functions. Lebesgue proved the dominated convergence theorem in 1908. Frigyes Riesz and Ernst Fischer proved the completeness of square-integrable functions in 1907, which is what gave the new integral its first great success: Fourier series of square-integrable functions finally converged to something (2B.7 Fourier Series and the First Heat Equation). John Edensor Littlewood formulated his three principles in his 1944 Lectures on the Theory of Functions.

3.11Exercises

Solution

(a) $0$ and $1$ in each case. (b) $f_n = 1_{[0,1]}$ for even $n$ and $f_n = 0$ for odd $n$: then $\liminf f_n = 0$ and $\liminf\int f_n = 0$, so Fatou holds with equality, while $\limsup\int f_n = 1$. A sharper example: $f_n = 2\cdot1_{[0, 1/2]}$ for even $n$ and $2\cdot1_{(1/2, 1]}$ for odd $n$, with $\int f_n = 1$ always but $\liminf f_n = 0$. (c) $\liminf f_n = 0$ with integral $0$, but $\int f_n = -1$ for all $n$.

Solution

For $0 \leq x < 1$, $\frac{-\log(1-x)}{x} = \sum_{k\geq1}\frac{x^{k-1}}{k}$, a series of non-negative functions. Integrating term by term (monotone convergence) gives $\sum_k\frac{1}{k}\cdot\frac{1}{k}$.

Solution

(a) $|n\sin(x/n)| \leq x$, so the integrand is dominated by $\frac{1}{1+x^2}$, and it converges to $\frac{1}{1 + x^2}$; the limit is $\frac\pi2$. (b) No domination: $nx^{n-1}$ concentrates at $x = 1$ (vertical escape). Substitute: $\int_0^1\frac{d(x^n)}{1 + x}$, and the mass concentrates at $1$, so the limit is $\frac12$. (c) $(1 + x^2/n)^{-n}$ decreases in $n$ for each $x$ (for $n \geq 1$), so it is dominated by $\frac1{1 + x^2}$, and it converges to $e^{-x^2}$; the limit is $\sqrt\pi$ (3A.5 Product Measures and Change of Variables).

Solution

$|\partial_a e^{-ax^2}| = x^2e^{-ax^2} \leq x^2e^{-a_0x^2}$, integrable. So $-\int x^2e^{-ax^2} = \frac{d}{da}\sqrt\pi a^{-1/2} = -\frac{\sqrt\pi}2a^{-3/2}$. Differentiating again (dominated by $x^4e^{-a_0x^2}$) gives the second.

Solution

$|\partial_tf\,e^{-f}| \leq C(1 + |x|^2)e^{C - c|x|^2}$, which is integrable on $\mathbb{R}^d$ (it is a polynomial times a Gaussian, 3A.5 Product Measures and Change of Variables) and doesn't depend on $t \in [t_0, t_1]$. Apply Theorem 3.7.