Lebesgue Measure

3A.1 The Problem of Measure found the problem: Jordan measure approximates a set by finitely many boxes, so it can't see countable sets, and it can't even measure some open sets. Lebesgue's fix is a single change. Approximate from outside by countably many boxes. The resulting outer measure is defined for every set; on a very large class of sets, the Lebesgue measurable ones, it is countably additive. That class contains every open set, every closed set, every set of outer measure zero, and everything that can be built from these by countable unions, intersections and complements. In practice, every set you will meet.

Two ideas from this chapter are used constantly afterwards. The first is null sets and the phrase almost everywhere: a property that holds except on a set of measure zero is, for the purposes of integration, as good as true everywhere. The second is that measure and topology measure different things. The Cantor set is uncountable but has measure zero; a "fat" Cantor set contains no interval at all but has positive measure.

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

• compute Lebesgue outer measures of simple sets, and prove that countable sets are null;
• state the definition of Lebesgue measurability and list the operations under which measurable sets are closed;
• use countable additivity and its consequences, continuity from below and above;
• work with null sets and "almost everywhere", and explain why probability zero doesn't mean impossible;
• construct the Cantor set and a fat Cantor set and compute their measures.

2.1Probability zero is not impossible

2.2Lebesgue outer measure

The only difference from Jordan outer measure is "countably many boxes" in place of "finitely many". It changes everything at once.

The basic properties of $m^*$ hold for all sets.

The proof uses an ε/2ⁿ trick, splitting an error budget $\varepsilon$ into countably many pieces that add up to $\varepsilon$. It is the countable version of the ε/2 trick of 2A.5 Quantifiers and the Shape of a Proof, and it appears in nearly every proof of this book.

Next, outer measure gives boxes the right size. This is not obvious, and it is where compactness enters.

Compactness converted a countable cover into a finite one, and finite covers are what elementary measure understands. Without that step there would be nothing to prevent the $\varepsilon/2^n$ covers of the rationals from also "covering" an interval with total length $\varepsilon$, which is exactly what happens over $\mathbb{Q}$ (whose closed bounded subsets need not be compact).

Outer measure is not additive on all sets: Vitali's construction (3A.1 The Problem of Measure) produces disjoint sets $A$, $B$ with $m^*(A \cup B) < m^*(A) + m^*(B)$ (Tao, Exercise 1.2.26). So we restrict to sets where additivity does hold.

2.3Measurable sets

A measurable set is one that can be approximated from outside by open sets, with an error of small outer measure. This is Tao's definition, and it makes the following list, the working toolkit, natural to prove.

Architecture of the proof (Tao, Lemmas 1.2.11–1.2.15, worth reading in full). The steps are: (a) every open set is a countable union of almost disjoint closed cubes (dyadic cubes inside it), so open sets are measurable and their measure is the total volume of those cubes; (b) compact sets are measurable, by approximating their complements; then closed sets, as countable unions of compact ones; (c) null sets are measurable directly from the definition; (d) countable unions by the $\varepsilon/2^n$ trick; (e) complements, the hardest step, using (b) and (c); (f) countable additivity, first for compact sets, which are at positive distance from each other when disjoint, then for bounded sets by inner approximation, then in general.

The class of measurable sets is therefore a σ-algebra: it contains $\varnothing$ and is closed under complements and countable unions. The smallest σ-algebra containing the open sets is the class of Borel sets. Every Borel set is Lebesgue measurable, but not conversely: Lebesgue measurable sets are exactly the sets that differ from a Borel set by a null set, and there are more of them.

2.3.1Continuity of measure

Countable additivity has two immediate consequences that are used more often than additivity itself.

The finiteness hypothesis in (2) is needed: $E_n = [n, \infty)$ decrease to the empty set, but each has infinite measure. That is the "escape to horizontal infinity" of 3A.3 The Lebesgue Integral in its simplest form.

2.3.2Measurable sets are nearly simple

Lebesgue measurable sets can be wild, but they are always close to tame ones. Measurability says they are nearly open from outside. Equally (Tao, Exercises 1.2.7, 1.2.15 and 1.2.16):

• (Inner regularity) $m(E) = \sup\{m(K) : K \subseteq E \text{ compact}\}$;
• (Approximation by boxes) if $m(E) < \infty$, then for every $\varepsilon > 0$ there is an elementary set $A$ (a finite union of boxes) with $m^*(E \,\triangle\, A) \leq \varepsilon$, where $\triangle$ is the symmetric difference.

The second statement is Littlewood's first principle: every measurable set is nearly a finite union of intervals. Its companions, that every measurable function is nearly continuous and every convergent sequence of measurable functions is nearly uniformly convergent, come in 3A.3 The Lebesgue Integral and 3A.6 Modes of Convergence and Differentiation. Together they are the reason measure theory is usable: prove a statement for boxes, continuous functions or uniform convergence, then pass to the general case through a small error.

2.4Null sets and "almost everywhere"

A null set is a set of measure zero. Countable unions of null sets are null (subadditivity), and subsets of null sets are null (monotonicity) and measurable.

Examples beyond countable sets: a line segment in $\mathbb{R}^2$ (cover it by $N$ squares of side $\frac1N$); more generally the graph of a continuous function (3A.1 The Problem of Measure); any hyperplane, or any smooth hypersurface in $\mathbb{R}^d$; the boundary of a ball. A $(d-1)$-dimensional set is always negligible for $d$-dimensional volume.

Two functions are equal almost everywhere if they differ only on a null set. Dirichlet's function ($1$ on rationals, $0$ elsewhere) equals $0$ almost everywhere. In 3A.3 The Lebesgue Integral such functions have the same integral, and in 3A.7 Lᵖ Spaces and Jensen’s Inequality they are regarded as the same element of a function space. This is the identification that makes the area distance $\int|f - g|$ a genuine metric (2B.1 Metric Spaces).

2.5The Cantor set, thin and fat

So a set can be uncountable, the same cardinality as $[0, 1]$, and still have measure zero. Cardinality and measure are different notions of size. (A third notion, Hausdorff dimension from 3A.1 The Problem of Measure, gives $C$ dimension $\log 2/\log 3 \approx 0.63$, between a countable set and an interval.)

The fat Cantor set is topologically small (it has empty interior and is nowhere dense) and measure-theoretically large. Its indicator function $1_F$ is discontinuous at every point of $F$, a set of positive measure, and so it is not Riemann integrable (a bounded function is Riemann integrable exactly when its discontinuities form a null set, Lebesgue's criterion). Its complement $[0, 1] \setminus F$ is an open set that is not Jordan measurable. In the Lebesgue theory both are perfectly measurable, with measure $\tfrac12$.

2.6History

Émile Borel (1898) defined measure for sets built from intervals by countable unions and complements, the sets now named after him. Henri Lebesgue's thesis (1902) defined outer measure by countable covers and the measurable sets, and built the integral on them. Constantin Carathéodory gave an abstract criterion for measurability in 1914 (Tao, Exercise 1.2.17), which is the route to general measures (3A.4 Measures, Probability and Weights). Georg Cantor published his ternary set in 1883; Henry Smith (1875) and Vito Volterra (1881) had already constructed nowhere dense sets of positive measure, which is why fat Cantor sets are also called Smith–Volterra–Cantor sets. Borel's normal number theorem (1909) was one of the first results stated in terms of "almost all" numbers.

2.7Exercises

Solution

A finite subcover $I_{n_1}, \ldots, I_{n_k}$ exists. Choose one containing $0$, then one containing its right endpoint, and so on; the chain reaches past $1$, and the sum of the lengths of the chain is at least $1$ because consecutive intervals overlap. For $\mathbb{Q} \cap [0, 1]$ there is no compactness: the set is not closed, and the cover of Example 2.2 has no finite subcover.

Solution

For every $N$, $\bigcap_M\bigcup_{n\geq M}E_n \subseteq \bigcup_{n\geq N}E_n$, whose measure is at most $\sum_{n\geq N}m(E_n) \to 0$.

Solution

The set $E_q$ has measure at most $(q + 1)\frac{2}{q^3} \leq \frac4{q^2}$, and $\sum_q\frac4{q^2} < \infty$. By Borel–Cantelli almost every $x$ lies in only finitely many $E_q$, and for each $q$ only finitely many $p$ work.

Hint

For (b), $\tfrac12C$ consists of numbers with ternary digits $0$ and $1$ only. Every $z \in [0, 1]$ is $\tfrac12(x + y)$ with $x, y \in C$: split each ternary digit $d \in \{0, 1, 2\}$ of $z$ as $d = a + b$ with $a, b \in \{0, 1\}$.

Solution

$m(E) \leq m(E \cup N) \leq m(E) + m(N) = m(E)$, and $m(E \setminus N) \geq m(E) - m(N) = m(E)$ by additivity on $E = (E \setminus N) \cup (E \cap N)$. Integrals over a set are unchanged by adding or removing a null set (3A.3 The Lebesgue Integral), so the bad set contributes nothing. Local Lipschitz continuity is what guarantees that the derivatives, which exist almost everywhere, are bounded and the integration by parts formulas still hold.