Sigma-Algebras & Measures

Let $P = \{0 = x_0 < x_1 < \cdots < x_n = 1\}$ be any partition of $[0, 1]$. On every subinterval $[x_{i-1}, x_i]$ — no matter how small — we can find both a rational point and an irrational point. So $\sup_{x \in [x_{i-1}, x_i]} \mathbf{1}_\mathbb{Q}(x) = 1, \quad \inf_{x \in [x_{i-1}, x_i]} \mathbf{1}_\mathbb{Q}(x) = 0.$

Therefore the upper Darboux sum is $U(\mathbf{1}_\mathbb{Q}, P) = \sum_i 1 \cdot (x_i - x_{i-1}) = 1$ and the lower sum is $L(\mathbf{1}_\mathbb{Q}, P) = 0$, regardless of the partition. The Riemann integral exists only when $\sup_P L = \inf_P U$, but here the sup is $0$ and the inf is $1$. The Riemann integral does not exist.

Let $\Omega$ be a non-empty set. A sigma-algebra on $\Omega$ is a collection $\mathcal{F} \subseteq \mathcal{P}(\Omega)$ of subsets of $\Omega$ satisfying:

1. $\Omega \in \mathcal{F}$.
2. Closure under complement. If $A \in \mathcal{F}$, then $A^c = \Omega \setminus A \in \mathcal{F}$.
3. Closure under countable union. If $A_1, A_2, A_3, \ldots$ is a countable sequence in $\mathcal{F}$, then $\bigcup_{n=1}^\infty A_n \in \mathcal{F}$.

The pair $(\Omega, \mathcal{F})$ is called a measurable space. Members of $\mathcal{F}$ are called measurable sets.

A few easy consequences fall out. Since $\Omega \in \mathcal{F}$ and $\mathcal{F}$ is closed under complement, $\emptyset = \Omega^c \in \mathcal{F}$. Closure under countable union plus complement gives closure under countable intersection (de Morgan). And since the union of a finite number of sets is a special case of a countable union (pad with $\emptyset$), $\mathcal{F}$ is also closed under finite unions and finite intersections.

For any set $\Omega$, the collection $\{\emptyset, \Omega\}$ is a sigma-algebra — the smallest one possible. Closure under complement is immediate ($\emptyset^c = \Omega$ and vice versa), and any countable union of these two sets is again one of these two sets. This sigma-algebra “knows” only whether a point is in $\Omega$ or not — it carries no further information.

The full power set $\mathcal{P}(\Omega)$ is a sigma-algebra — the largest one possible. It contains every subset of $\Omega$ and is trivially closed under all set operations. When $\Omega$ is countable, $\mathcal{P}(\Omega)$ is the only sigma-algebra we ever use. But when $\Omega = \mathbb{R}$, we will see that $\mathcal{P}(\mathbb{R})$ is too big — it contains pathological sets that no consistent length function can measure.

Take $\Omega = \{1, 2, 3, 4\}$ and consider the partition $\{1, 2\}$ and $\{3, 4\}$. The sigma-algebra generated by this partition is $\mathcal{F} = \{\emptyset, \{1, 2\}, \{3, 4\}, \{1, 2, 3, 4\}\}.$

This collection is closed under complement (the complement of $\{1, 2\}$ is $\{3, 4\}$) and under union (any union of the four sets is again in the four sets). Note that $\{1\}$ is not in $\mathcal{F}$ — this sigma-algebra cannot distinguish $1$ from $2$. It encodes precisely the information “which side of the partition does this point belong to?” and nothing more.

The previous example hints at an interpretation that becomes essential in probability theory: a sigma-algebra is an abstract description of what information is available. A coarser sigma-algebra (fewer sets) corresponds to less information; a finer sigma-algebra (more sets) corresponds to more information. In the example above, $\mathcal{F}$ knows only “is this point in $\{1, 2\}$ or in $\{3, 4\}$?” but not “is this point exactly $1$?” In probability, a filtration — a nested sequence of sigma-algebras $\mathcal{F}_1 \subseteq \mathcal{F}_2 \subseteq \cdots$ — models information being revealed over time. This is the foundation of stochastic processes and the source of every “conditional expectation given $\mathcal{F}_t$” in mathematical finance and reinforcement learning.

Let $\mathcal{C} \subseteq \mathcal{P}(\Omega)$ be any collection of subsets of $\Omega$. The sigma-algebra generated by $\mathcal{C}$, written $\sigma(\mathcal{C})$, is the smallest sigma-algebra on $\Omega$ that contains $\mathcal{C}$. Equivalently, it is the intersection of all sigma-algebras containing $\mathcal{C}$: $\sigma(\mathcal{C}) = \bigcap \{\mathcal{F} : \mathcal{F} \text{ is a sigma-algebra on } \Omega \text{ with } \mathcal{C} \subseteq \mathcal{F}\}.$

This intersection is non-empty (the power set $\mathcal{P}(\Omega)$ always works) and is itself a sigma-algebra (the intersection of any family of sigma-algebras is a sigma-algebra — check the axioms). So $\sigma(\mathcal{C})$ is well-defined.

The Borel sigma-algebra on $\mathbb{R}$, denoted $\mathcal{B}(\mathbb{R})$, is the sigma-algebra generated by the collection of all open intervals: $\mathcal{B}(\mathbb{R}) = \sigma\left(\{(a, b) : a < b \in \mathbb{R}\}\right).$

Members of $\mathcal{B}(\mathbb{R})$ are called Borel sets. The collection $\mathcal{B}(\mathbb{R})$ contains every open set (each is a countable union of open intervals), every closed set (complements of open sets), every countable union of closed sets ($F_\sigma$ sets), every countable intersection of open sets ($G_\delta$ sets), and so on through a transfinite hierarchy. In short: every “reasonable” subset of $\mathbb{R}$ that you would write down without using the axiom of choice is Borel.

Equivalently, $\mathcal{B}(\mathbb{R})$ can be generated by any of the following: all open sets, all closed sets, all half-open intervals $(a, b]$, all rays $(-\infty, b]$, all rationals as singletons. The choice of generator doesn’t matter — the resulting sigma-algebra is the same.

The choice of countable (rather than finite) closure in the sigma-algebra axioms is the central technical decision of measure theory. With only finite closure, you get an algebra of sets — adequate for classical probability of dice and cards, but too weak for analysis. Countable closure is what makes limits behave correctly. For example, the set $\mathbb{Q} \cap [0, 1]$ is the countable union of singletons $\{q\}$ for $q$ rational; in the Borel sigma-algebra it is automatically measurable, and its measure is the countable sum $\sum_q \lambda(\{q\}) = 0$. With only finite additivity, this argument fails — the rationals would slip through the cracks. Countable additivity is exactly the right strength to make a.e. arguments, dominated convergence, and Fubini’s theorem all work.

Let $(\Omega, \mathcal{F})$ be a measurable space. A measure on $(\Omega, \mathcal{F})$ is a function $\mu: \mathcal{F} \to [0, \infty]$ satisfying:

1. Non-negativity. $\mu(A) \geq 0$ for all $A \in \mathcal{F}$.
2. Empty set has measure zero. $\mu(\emptyset) = 0$.
3. Countable additivity. For any countable sequence of pairwise disjoint sets $A_1, A_2, \ldots \in \mathcal{F}$, $\mu\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu(A_n).$

The triple $(\Omega, \mathcal{F}, \mu)$ is called a measure space. When $\mu(\Omega) = 1$, the triple is a probability space and $\mu$ is a probability measure — usually denoted $P$ or $\mathbb{P}$.

Note that measures take values in the extended non-negative reals $[0, \infty]$ — sets of infinite measure are allowed, and their arithmetic obeys the conventions $a + \infty = \infty$ for $a \geq 0$ and $0 \cdot \infty = 0$ (the latter is a convention, not a theorem, but it makes integrals of functions on null sets behave correctly).

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space. Then for all measurable sets $A, B, A_n \in \mathcal{F}$:

1. Monotonicity. If $A \subseteq B$ then $\mu(A) \leq \mu(B)$.
2. Subadditivity. $\mu\left(\bigcup_{n=1}^\infty A_n\right) \leq \sum_{n=1}^\infty \mu(A_n)$ (no disjointness required).
3. Continuity from below. If $A_1 \subseteq A_2 \subseteq \cdots$ is an increasing sequence with union $A = \bigcup_n A_n$, then $\mu(A_n) \to \mu(A)$ as $n \to \infty$.
4. Continuity from above. If $A_1 \supseteq A_2 \supseteq \cdots$ is a decreasing sequence with intersection $A = \bigcap_n A_n$, and $\mu(A_1) < \infty$, then $\mu(A_n) \to \mu(A)$ as $n \to \infty$.

Monotonicity (1). Write $B = A \sqcup (B \setminus A)$ as a disjoint union. Both $A$ and $B \setminus A = B \cap A^c$ are measurable. By countable additivity (with all but two terms equal to $\emptyset$), $\mu(B) = \mu(A) + \mu(B \setminus A) \geq \mu(A),$ since $\mu(B \setminus A) \geq 0$.

Continuity from below (3). Define a disjointified sequence: $B_1 = A_1, \quad B_2 = A_2 \setminus A_1, \quad B_n = A_n \setminus A_{n-1} \text{ for } n \geq 2.$ Each $B_n$ is measurable (difference of measurable sets), the $B_n$ are pairwise disjoint, and $\bigcup_{k=1}^n B_k = A_n, \qquad \bigcup_{k=1}^\infty B_k = \bigcup_{k=1}^\infty A_k = A.$

By countable additivity applied to the disjoint $B_k$, $\mu(A) = \sum_{k=1}^\infty \mu(B_k) = \lim_{n \to \infty} \sum_{k=1}^n \mu(B_k) = \lim_{n \to \infty} \mu(A_n),$ where the second equality is the definition of an infinite series and the third equality is finite additivity ($A_n$ is the disjoint union of $B_1, \ldots, B_n$).

Subadditivity (2). Disjointify as in (3): let $B_1 = A_1$ and $B_n = A_n \setminus (A_1 \cup \cdots \cup A_{n-1})$ for $n \geq 2$. Then $B_n \subseteq A_n$ (so $\mu(B_n) \leq \mu(A_n)$ by monotonicity), the $B_n$ are pairwise disjoint, and $\bigcup B_n = \bigcup A_n$. Countable additivity gives $\mu\left(\bigcup_{n} A_n\right) = \mu\left(\bigcup_{n} B_n\right) = \sum_{n} \mu(B_n) \leq \sum_{n} \mu(A_n).$

Continuity from above (4). This reduces to (3) by taking complements relative to $A_1$. The hypothesis $\mu(A_1) < \infty$ is essential — see Royden §17.4 for the cautionary counter-example without finiteness.

Let $\Omega = \mathbb{N}$ and $\mathcal{F} = \mathcal{P}(\mathbb{N})$ (every subset is measurable, since $\mathbb{N}$ is countable). Define $\mu(A) = |A|, \quad \text{the number of elements of } A,$ where $|A| = \infty$ if $A$ is infinite. This is the counting measure. It satisfies all three axioms: $\mu(\emptyset) = 0$, $\mu(A) \geq 0$, and the cardinality of a countable disjoint union is the sum of cardinalities. The counting measure is the foundation of discrete probability — every "$P(X = k)$" is integration against the counting measure.

Fix a point $a \in \Omega$. The Dirac measure at $a$ is $\delta_a(A) = \begin{cases} 1 & \text{if } a \in A, \\ 0 & \text{otherwise.} \end{cases}$

This is a probability measure ($\delta_a(\Omega) = 1$, since $a \in \Omega$). Countable additivity holds because $a$ belongs to at most one set in any disjoint collection. Dirac measures are the building blocks of point masses in probability and the simplest non-trivial example of a singular measure with respect to Lebesgue measure on $\mathbb{R}$.

On the measurable space $([0, 1], \mathcal{B}([0, 1]))$, there exists a unique measure $\lambda$ such that $\lambda([a, b]) = b - a$ for every interval $[a, b] \subseteq [0, 1]$. This is the Lebesgue measure on the unit interval, and constructing it rigorously is the work of Section 4 (we will need Carathéodory’s extension theorem). For now, observe that $\lambda([0, 1]) = 1$, so $([0, 1], \mathcal{B}([0, 1]), \lambda)$ is a probability space — the uniform distribution on the unit interval.

A measure $\mu$ on $(\Omega, \mathcal{F})$ is finite if $\mu(\Omega) < \infty$ and sigma-finite if $\Omega$ can be written as a countable union $\Omega = \bigcup_n \Omega_n$ with $\mu(\Omega_n) < \infty$ for every $n$. Probability measures are finite. Lebesgue measure on $\mathbb{R}$ is not finite ($\lambda(\mathbb{R}) = \infty$) but is sigma-finite — write $\mathbb{R} = \bigcup_n [-n, n]$. Most theorems we care about (Fubini, Radon-Nikodym) assume sigma-finiteness; the non-sigma-finite case is where measure theory gets pathological. Throughout this topic, every measure is sigma-finite unless we explicitly say otherwise.

For any subset $A \subseteq \mathbb{R}$, the Lebesgue outer measure of $A$ is $\lambda^*(A) \;=\; \inf\left\{\sum_{k=1}^\infty (b_k - a_k) \;:\; A \subseteq \bigcup_{k=1}^\infty (a_k, b_k)\right\}.$

The infimum is taken over all countable covers of $A$ by open intervals, and the value is the total length of the cover. Outer measure is defined for every subset of $\mathbb{R}$ — but as we will see, $\lambda^*$ alone is not countably additive on the full power set, so we need to restrict.

Some basic properties drop out immediately: $\lambda^*(\emptyset) = 0$ (use the empty cover), $\lambda^*$ is monotone ($A \subseteq B \Rightarrow \lambda^*(A) \leq \lambda^*(B)$), $\lambda^*$ is translation-invariant ($\lambda^*(A + t) = \lambda^*(A)$ — translating a cover gives a cover of the translate with the same total length), and $\lambda^*$ is countably subadditive ($\lambda^*(\bigcup_n A_n) \leq \sum_n \lambda^*(A_n)$). What we don’t yet have is countable additivity on disjoint sets — and that requires choosing the right sigma-algebra to restrict to.

A set $A \subseteq \mathbb{R}$ is Carathéodory-measurable (with respect to $\lambda^*$) if for every test set $E \subseteq \mathbb{R}$, $\lambda^*(E) \;=\; \lambda^*(E \cap A) + \lambda^*(E \cap A^c).$

Let $\mathcal{L}(\mathbb{R})$ denote the collection of all such measurable sets. Then:

1. $\mathcal{L}(\mathbb{R})$ is a sigma-algebra containing every Borel set: $\mathcal{B}(\mathbb{R}) \subseteq \mathcal{L}(\mathbb{R})$.
2. The restriction $\lambda = \lambda^*|_{\mathcal{L}(\mathbb{R})}$ is a countably additive measure on $\mathcal{L}(\mathbb{R})$.
3. $\lambda([a, b]) = b - a$ for every interval, recovering the geometric notion of length.

This restriction $\lambda$ is the Lebesgue measure on $\mathbb{R}$, and $\mathcal{L}(\mathbb{R})$ is the Lebesgue sigma-algebra.

The full proof is long — about ten pages in Royden — and we will not reproduce it here. The structural ingredients are:

• $\mathcal{L}(\mathbb{R})$ contains $\emptyset$ (trivially) and is closed under complement (the Carathéodory criterion is symmetric in $A$ and $A^c$).
• Closure under finite union follows from a careful inclusion-exclusion argument on the criterion.
• Closure under countable union uses both subadditivity of $\lambda^*$ and continuity arguments to upgrade finite unions to countable.
• The fact that $\lambda^*$ is countably additive on $\mathcal{L}(\mathbb{R})$ — the punchline — uses subadditivity in one direction and the Carathéodory criterion applied to disjoint sets in the other.
• The inclusion $\mathcal{B}(\mathbb{R}) \subseteq \mathcal{L}(\mathbb{R})$ comes from showing every interval is Carathéodory-measurable, then using closure under countable operations to extend to all Borel sets.

For the complete argument, see Royden §2.4 or Folland §1.4. The takeaway for this topic is that $\mathcal{L}(\mathbb{R})$ is a strictly larger sigma-algebra than $\mathcal{B}(\mathbb{R})$ — every Borel set is Lebesgue-measurable, but there exist Lebesgue-measurable sets that are not Borel (their existence requires the axiom of choice). Both are smaller than $\mathcal{P}(\mathbb{R})$, which contains the non-measurable Vitali set we will construct in Section 8.

A set $A \subseteq \mathbb{R}$ is Lebesgue-measurable if it satisfies the Carathéodory criterion: for every $E \subseteq \mathbb{R}$, $\lambda^*(E) = \lambda^*(E \cap A) + \lambda^*(E \cap A^c).$ The collection of all such sets is the Lebesgue sigma-algebra $\mathcal{L}(\mathbb{R})$, and $\lambda = \lambda^*|_{\mathcal{L}(\mathbb{R})}$ is the Lebesgue measure on $\mathbb{R}$.

For every $A \in \mathcal{L}(\mathbb{R})$ and every $t \in \mathbb{R}$, the translate $A + t = \{a + t : a \in A\}$ is also Lebesgue-measurable, and $\lambda(A + t) = \lambda(A).$

We split the proof into two parts: first that $\lambda^*$ is translation-invariant on every subset of $\mathbb{R}$, then that $\mathcal{L}(\mathbb{R})$ is closed under translation.

Step 1: $\lambda^*(A + t) = \lambda^*(A)$ for every $A \subseteq \mathbb{R}$.

Let $\{(a_k, b_k)\}_{k=1}^\infty$ be any countable open cover of $A$. Then $\{(a_k + t, b_k + t)\}_{k=1}^\infty$ is a countable open cover of $A + t$ with the same total length: $\sum_{k=1}^\infty ((b_k + t) - (a_k + t)) = \sum_{k=1}^\infty (b_k - a_k).$

Taking the infimum over all covers of $A$ on the left and noting that every cover of $A$ produces a cover of $A + t$ of equal length (and vice versa, by translating by $-t$), we get $\lambda^*(A + t) = \lambda^*(A)$. So $\lambda^*$ is translation-invariant on the full power set.

Step 2: If $A \in \mathcal{L}(\mathbb{R})$, then $A + t \in \mathcal{L}(\mathbb{R})$.

We must show that $A + t$ satisfies the Carathéodory criterion: for every $E \subseteq \mathbb{R}$, $\lambda^*(E) = \lambda^*(E \cap (A + t)) + \lambda^*(E \cap (A + t)^c).$

Apply Step 1 to translate $E$ by $-t$: $\lambda^*(E) = \lambda^*(E - t).$

Now $E - t$ can be split using the measurability of $A$ (which we are given): $\lambda^*(E - t) = \lambda^*((E - t) \cap A) + \lambda^*((E - t) \cap A^c).$

We translate each piece on the right side back by $+t$. Note that $(E - t) \cap A$ translated by $+t$ gives $E \cap (A + t)$, and $(E - t) \cap A^c$ translated by $+t$ gives $E \cap (A + t)^c$. By Step 1 again, translation does not change outer measure: $\lambda^*((E - t) \cap A) = \lambda^*(E \cap (A + t)),$ $\lambda^*((E - t) \cap A^c) = \lambda^*(E \cap (A + t)^c).$

Substituting back: $\lambda^*(E) = \lambda^*(E \cap (A + t)) + \lambda^*(E \cap (A + t)^c).$

This is exactly the Carathéodory criterion for $A + t$, so $A + t \in \mathcal{L}(\mathbb{R})$. Step 1 then gives $\lambda(A + t) = \lambda^*(A + t) = \lambda^*(A) = \lambda(A)$.

Let $[a, b] \subset \mathbb{R}$ with $a < b$. The interval is open-cover-able by $(a - \varepsilon, b + \varepsilon)$ for any $\varepsilon > 0$, so $\lambda^*([a, b]) \leq (b - a) + 2\varepsilon$. Taking $\varepsilon \to 0$ gives $\lambda^*([a, b]) \leq b - a$. The reverse inequality $\lambda^*([a, b]) \geq b - a$ is the non-trivial direction — it requires showing that no countable open cover of $[a, b]$ can have total length less than $b - a$. The argument uses the Heine-Borel theorem (compactness of $[a, b]$, from Topic 3) to extract a finite sub-cover, then a clean overlapping-intervals argument to bound the total length below by $b - a$. Combined with the fact that $[a, b]$ is Borel and hence Lebesgue-measurable, we get $\lambda([a, b]) = b - a$.

The rationals in $[0, 1]$ form a countable set: $\mathbb{Q} \cap [0, 1] = \{q_1, q_2, q_3, \ldots\}$. For any $\varepsilon > 0$, cover the $n$-th rational $q_n$ by the open interval $(q_n - \varepsilon/2^{n+1}, q_n + \varepsilon/2^{n+1})$, which has length $\varepsilon/2^n$. The total length of this cover is $\sum_{n=1}^\infty \frac{\varepsilon}{2^n} = \varepsilon.$

So $\lambda^*(\mathbb{Q} \cap [0, 1]) \leq \varepsilon$ for every $\varepsilon > 0$, hence $\lambda^*(\mathbb{Q} \cap [0, 1]) = 0$. Since $\mathbb{Q} \cap [0, 1]$ is Borel (a countable union of singletons), it is Lebesgue-measurable, so $\lambda(\mathbb{Q} \cap [0, 1]) = 0$.

This is the precise statement of “the rationals are negligible.” Combined with $\lambda([0, 1]) = 1$, we get that the irrationals in $[0, 1]$ have measure $1$ — almost every point in the unit interval is irrational. The Dirichlet function $\mathbf{1}_\mathbb{Q}$ is therefore zero almost everywhere with respect to Lebesgue measure, and its Lebesgue integral is $0$ — exactly what our intuition wanted in Section 1.

A bounded function $f: [a, b] \to \mathbb{R}$ is Riemann-integrable if and only if its set of discontinuities has Lebesgue measure zero (Lebesgue’s criterion for Riemann integrability). For every Riemann-integrable $f$, the Lebesgue integral exists and equals the Riemann integral. So Lebesgue integration is a strict extension of Riemann integration — it agrees with the old theory wherever the old theory applies, and assigns sensible values in many cases where the old theory fails (like the Dirichlet function).

Define a sequence of sets recursively. Start with $C_0 = [0, 1]$. To form $C_{n+1}$, remove the open middle third of every interval in $C_n$: $C_1 = [0, 1/3] \cup [2/3, 1],$ $C_2 = [0, 1/9] \cup [2/9, 3/9] \cup [6/9, 7/9] \cup [8/9, 1],$ and so on. The Cantor set is the intersection $C = \bigcap_{n=0}^\infty C_n.$

After $n$ iterations, $C_n$ is a disjoint union of $2^n$ intervals each of length $(1/3)^n$, so the total length of $C_n$ is $(2/3)^n$. As $n \to \infty$, this length tends to zero — but the intersection $C$ is nevertheless non-empty and, as we will prove, uncountable.

The Cantor set $C$ is a closed subset of $[0, 1]$ that is:

1. Uncountable — there is a bijection $C \leftrightarrow \{0, 1\}^\mathbb{N}$, so $|C| = 2^{\aleph_0} =$ continuum.
2. Of Lebesgue measure zero — $\lambda(C) = 0$.

Measure zero. Since each $C_n$ has Lebesgue measure $(2/3)^n$ and $C \subseteq C_n$ for every $n$, monotonicity of $\lambda$ gives $0 \leq \lambda(C) \leq \lambda(C_n) = (2/3)^n.$

Since $(2/3)^n \to 0$ as $n \to \infty$, the squeeze gives $\lambda(C) = 0$.

Uncountability. We use the ternary expansion characterization of $C$. Every $x \in [0, 1]$ has a base-3 (ternary) expansion $x = 0.d_1 d_2 d_3 \ldots_{(3)} = \sum_{k=1}^\infty \frac{d_k}{3^k}, \quad d_k \in \{0, 1, 2\}.$

The ternary expansion is unique except at terminating expansions, where the same number has two representations (analogous to $0.999\ldots = 1$ in base 10). For terminating expansions we adopt the convention that ends in all $2$‘s when possible — this resolves the ambiguity.

The middle third $(1/3, 2/3)$ removed at step $1$ is precisely the set of $x$ whose ternary expansion has $d_1 = 1$ (and $d_1 = 1$ cannot be replaced by ending in $0\overline{2}$ for any of these — the boundary points $1/3$ and $2/3$ themselves have alternative expansions with $d_1 \in \{0, 2\}$). So $C_1$ consists of points with $d_1 \in \{0, 2\}$. By induction, $C_n$ consists of points whose first $n$ ternary digits are all in $\{0, 2\}$. Therefore $C = \{x \in [0, 1] : x \text{ has a ternary expansion with all digits } d_k \in \{0, 2\}\}.$

The map $\phi: C \to \{0, 1\}^\mathbb{N}$ defined by sending $0.d_1 d_2 \ldots_{(3)}$ to the binary sequence $(d_1/2, d_2/2, d_3/2, \ldots)$ is a bijection. (It is well-defined and injective by the uniqueness of the chosen ternary expansion; it is surjective because every binary sequence corresponds to a valid ternary string in $\{0, 2\}^\mathbb{N}$.)

By Cantor’s diagonal argument (Topic 3), $\{0, 1\}^\mathbb{N}$ is uncountable, with cardinality $2^{\aleph_0}$ — the cardinality of the continuum. Therefore $|C| = 2^{\aleph_0}$, and $C$ is uncountable.

The standard middle-thirds Cantor set has measure zero because we remove a constant fraction $1/3$ at every step. If instead we remove a shrinking fraction at each step, we get a Cantor-like set with positive measure. Specifically, at step $k \geq 1$, remove a centered interval of length $1/4^k$ from each of the $2^{k-1}$ remaining intervals. The total length removed is $\sum_{k=1}^\infty 2^{k-1} \cdot \frac{1}{4^k} \;=\; \sum_{k=1}^\infty \frac{1}{2^{k+1}} \;=\; \frac{1}{2}.$

So the limiting set, the Smith-Volterra-Cantor set (or “fat Cantor set”), has Lebesgue measure $1 - 1/2 = 1/2$. It is closed, nowhere dense, and uncountable — exactly like the standard Cantor set — but it has positive measure. This shows that “Cantor-like” does not imply “measure zero”: uncountability and measure zero are independent properties, and the standard middle-thirds set is a particular case where they happen to coincide.

$C$ is closed (it is the intersection of closed sets $C_n$) and bounded (a subset of $[0, 1]$), so by Heine-Borel from Topic 3, $C$ is compact. It is also totally disconnected: between any two distinct points $x, y \in C$ there is a deleted middle interval at some stage of the construction, so the connected components of $C$ are single points. A non-empty compact totally disconnected metric space without isolated points is called a Cantor space, and the Cantor set is the prototype. (Every two such spaces are homeomorphic — the Cantor set is, up to homeomorphism, the Cantor space.)

Let $(\Omega, \mathcal{F})$ be a measurable space. A function $f: \Omega \to \mathbb{R}$ is $\mathcal{F}$-measurable (or just measurable when the sigma-algebra is clear from context) if for every Borel set $B \in \mathcal{B}(\mathbb{R})$, $f^{-1}(B) = \{\omega \in \Omega : f(\omega) \in B\} \in \mathcal{F}.$

Equivalently — and this is usually how you check it in practice — $f$ is measurable if and only if $f^{-1}((-\infty, t]) \in \mathcal{F}$ for every $t \in \mathbb{R}$. The “if” direction follows from the fact that the half-rays $\{(-\infty, t] : t \in \mathbb{R}\}$ generate $\mathcal{B}(\mathbb{R})$, and preimages commute with countable set operations.

If $f: \mathbb{R} \to \mathbb{R}$ is continuous and $U \subseteq \mathbb{R}$ is open, then $f^{-1}(U)$ is open, hence Borel, hence Lebesgue-measurable. Since the open sets generate $\mathcal{B}(\mathbb{R})$ and preimages preserve countable set operations, $f^{-1}(B) \in \mathcal{B}(\mathbb{R})$ for every Borel $B$. So every continuous function is Borel-measurable. This is why the integral of a continuous function on a compact interval — the bread and butter of single-variable calculus — is just a special case of the Lebesgue integral.

The indicator $\mathbf{1}_\mathbb{Q}: [0, 1] \to \{0, 1\}$ takes only two values, so its preimage of any Borel set $B$ is one of $\emptyset$, $\mathbb{Q} \cap [0, 1]$, $[0, 1] \setminus \mathbb{Q}$, or $[0, 1]$, depending on which of $0$ and $1$ lie in $B$. All four sets are Borel ($\mathbb{Q} \cap [0, 1]$ is a countable union of singletons), so $\mathbf{1}_\mathbb{Q}$ is Borel-measurable. Combined with our earlier observation that $\lambda(\mathbb{Q} \cap [0, 1]) = 0$, this is the foundation for showing $\int \mathbf{1}_\mathbb{Q} \, d\lambda = 0$ — once we have built the integral in Topic 26.

A measurable function $s: \Omega \to \mathbb{R}$ is simple if it takes only finitely many values. Equivalently, $s$ has a representation $s = \sum_{k=1}^n c_k \cdot \mathbf{1}_{A_k}$ where $c_1, \ldots, c_n$ are distinct real numbers and $A_1, \ldots, A_n$ is a finite measurable partition of $\Omega$ (each $A_k$ is measurable, the $A_k$ are pairwise disjoint, and $\Omega = \bigsqcup_k A_k$). The integral of a non-negative simple function with respect to a measure $\mu$ is defined to be $\int s \, d\mu = \sum_{k=1}^n c_k \cdot \mu(A_k).$

(With the convention $0 \cdot \infty = 0$, this is well-defined even when some $c_k = 0$ but $\mu(A_k) = \infty$.)

Let $f: \Omega \to [0, \infty]$ be a non-negative measurable function. Then there exists a sequence $(s_n)_{n=1}^\infty$ of non-negative simple measurable functions such that:

1. $s_1 \leq s_2 \leq s_3 \leq \cdots$ (the sequence is monotone increasing).
2. $s_n(\omega) \to f(\omega)$ for every $\omega \in \Omega$.
3. The convergence is uniform on every set where $f$ is bounded.

We construct $s_n$ explicitly via dyadic slicing of the range. For each $n \geq 1$, partition the interval $[0, n)$ into $n \cdot 2^n$ dyadic subintervals of length $2^{-n}$: $[0, 2^{-n}), \quad [2^{-n}, 2 \cdot 2^{-n}), \quad \ldots, \quad [(n \cdot 2^n - 1) \cdot 2^{-n}, n \cdot 2^n \cdot 2^{-n}) = [(n - 2^{-n}), n).$

Define $s_n(\omega) = \begin{cases} k \cdot 2^{-n} & \text{if } k \cdot 2^{-n} \leq f(\omega) < (k + 1) \cdot 2^{-n} \text{ for some } k = 0, 1, \ldots, n \cdot 2^n - 1, \\ n & \text{if } f(\omega) \geq n. \end{cases}$

Each $s_n$ is a finite sum of step values times indicators of preimage sets, so it is a simple function. Each preimage set $\{f \in [k \cdot 2^{-n}, (k+1) \cdot 2^{-n})\}$ is measurable (preimage of a Borel set under a measurable function), so $s_n$ is measurable.

Monotonicity ($s_n \leq s_{n+1}$). Suppose $s_n(\omega) = k \cdot 2^{-n}$ for some $k < n \cdot 2^n$. Then $f(\omega) \in [k \cdot 2^{-n}, (k+1) \cdot 2^{-n})$. The next-finer dyadic partition splits this interval in half: $f(\omega)$ lies in either $[2k \cdot 2^{-(n+1)}, (2k + 1) \cdot 2^{-(n+1)})$ or $[(2k + 1) \cdot 2^{-(n+1)}, (2k + 2) \cdot 2^{-(n+1)})$. In the first case $s_{n+1}(\omega) = 2k \cdot 2^{-(n+1)} = k \cdot 2^{-n} = s_n(\omega)$. In the second case $s_{n+1}(\omega) = (2k + 1) \cdot 2^{-(n+1)} > s_n(\omega)$. Either way $s_n(\omega) \leq s_{n+1}(\omega)$. The case where $s_n(\omega) = n$ (the “cap”) is handled similarly: increasing $n$ either keeps $s_n$ at the cap or moves it into a finer dyadic level above $n$.

Pointwise convergence ($s_n(\omega) \to f(\omega)$). Fix $\omega$. If $f(\omega) < \infty$, choose $n_0$ so large that $n_0 > f(\omega)$. For all $n \geq n_0$, $f(\omega) \in [k_n \cdot 2^{-n}, (k_n + 1) \cdot 2^{-n})$ for some $k_n$ in the dyadic partition, so $s_n(\omega) = k_n \cdot 2^{-n} \in [f(\omega) - 2^{-n}, f(\omega)].$

Hence $|f(\omega) - s_n(\omega)| < 2^{-n} \to 0$. If $f(\omega) = \infty$, then $f(\omega) > n$ for every $n$, so $s_n(\omega) = n \to \infty = f(\omega)$.

Uniform convergence on bounded sets. If $|f| \leq M$ on a set $E \subseteq \Omega$, then for $n > M$ we have $f(\omega) < n$ for all $\omega \in E$, and the bound $|f(\omega) - s_n(\omega)| < 2^{-n}$ holds uniformly in $\omega \in E$. So $s_n \to f$ uniformly on $E$.

Theorem 5 is the constructive heart of Lebesgue integration. Notice what it does: it slices the range of $f$ into dyadic levels and uses the preimages of those levels as the building blocks for simple functions. This is the exact opposite of the Riemann strategy, which slices the domain into uniform subintervals. The Riemann approach fails on the Dirichlet function because no domain partition can separate rationals from irrationals; the Lebesgue approach succeeds because the range of $\mathbf{1}_\mathbb{Q}$ has only two values and their preimages ($\mathbb{Q} \cap [0, 1]$ and its complement) are both Borel-measurable. The “partition the range” insight is the single most important conceptual move in measure theory, and it is what makes the Lebesgue integral strictly more powerful than the Riemann integral.

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space. A set $N \in \mathcal{F}$ is a null set (or $\mu$-null set) if $\mu(N) = 0$. Examples on $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \lambda)$: every singleton $\{x\}$, every countable set (such as $\mathbb{Q}$), and the Cantor set $C$.

A property $P(\omega)$ is said to hold almost everywhere with respect to $\mu$ — written $\mu$-a.e. or simply a.e. when the measure is clear — if the set of $\omega$ for which $P(\omega)$ fails is contained in a null set: $\mu(\{\omega \in \Omega : P(\omega) \text{ does not hold}\}) = 0.$

In probability, where $\mu = P$ is a probability measure, “almost everywhere” is usually called almost surely, abbreviated a.s. or “with probability $1$.” When you read “the gradient flow converges to a critical point with probability 1,” the “with probability 1” is the measure-theoretic statement that the bad set has Lebesgue measure zero.

On $([0, 1], \mathcal{B}([0, 1]), \lambda)$ the rationals form a null set ($\lambda(\mathbb{Q} \cap [0, 1]) = 0$, by Example 9). So $\mathbf{1}_\mathbb{Q}(x) = 0 \text{ for almost every } x \in [0, 1],$ and the almost-everywhere equivalence class of $\mathbf{1}_\mathbb{Q}$ is the same as that of the constant zero function. Once we have the Lebesgue integral, this will give $\int_0^1 \mathbf{1}_\mathbb{Q} \, d\lambda = 0$ — exactly the answer our intuition demanded back in Section 1.

There exists a function $\phi: [0, 1] \to [0, 1]$, called the Cantor function or devil’s staircase, with the following remarkable properties:

1. $\phi$ is continuous.
2. $\phi$ is non-decreasing, with $\phi(0) = 0$ and $\phi(1) = 1$.
3. $\phi$ is differentiable almost everywhere with $\phi'(x) = 0$ a.e.

The construction is iterative: $\phi$ is constant on every “removed middle third” of the Cantor set construction (taking the value $1/2$ on $(1/3, 2/3)$, the values $1/4$ and $3/4$ on the next two removed middles, and so on). On the Cantor set itself, $\phi$ is defined by a base-$3$-to-base-$2$ digit substitution.

The “missing increase” paradox is striking: $\phi$ goes from $0$ to $1$ continuously, yet its derivative is $0$ almost everywhere. All of the “increase” happens on the Cantor set $C$, which has measure zero. The fundamental theorem of calculus fails for $\phi$: $\phi(1) - \phi(0) = 1 \;\neq\; 0 \;=\; \int_0^1 \phi'(x) \, dx.$

The FTC requires absolute continuity, and the Cantor function is the canonical example of a continuous, non-decreasing function that is not absolutely continuous. This is exactly the kind of pathology that motivates measure theory — the Riemann picture cannot detect the difference, but the measure-theoretic picture can.

The property “$\mu$-almost everywhere” depends entirely on which measure $\mu$ you have in mind. The set $\{1\} \subset \mathbb{R}$ has Lebesgue measure zero (so $\mathbf{1}_{\{1\}} = 0$ Lebesgue-a.e.), but it has counting measure $1$ (so $\mathbf{1}_{\{1\}} \neq 0$ counting-a.e.). When you read or write “a.e.” in measure theory, always know which measure you mean — switching measures changes which sets are negligible.

A measure $\mu$ on $(\Omega, \mathcal{F})$ is complete if every subset of every null set is itself measurable (and therefore also has measure zero). Equivalently, if $N \in \mathcal{F}$ with $\mu(N) = 0$ and $A \subseteq N$, then $A \in \mathcal{F}$. A measure is incomplete if there exists a null set $N$ with a non-measurable subset.