Everybody knows how long $[a, b)$ is. It's $b - a$. The whole job of Lebesgue measure is to take that one number and stretch it to as many sets as possible without breaking countable additivity. This page does it by hand, following Halmos. No functionals, no Urysohn: just lengths, covers, and one clever test.
The plan has three moves, and each section below is one of them.
Lebesgue Measure, the page before this one, gets Lebesgue measure out of the Riesz Representation Theorem instead: feed it the Riemann integral and a measure comes out. This page is the other road. At the very end, a uniqueness theorem says both roads lead to the same measure on the Borel sets.
The construction works for any measure on any ring, not just for length, so the theorems below are stated in that generality. Length on $\mathbb{R}$ is the example that runs through all of it.
A σ-algebra has to contain the whole space and every complement. Intervals are nowhere near that: the complement of $[0, 1)$ is two rays, and length has nothing to say about a ray yet. So we start with a weaker kind of collection, one that only asks for unions and differences.
A ring with no multiplication anywhere in sight. The algebraists would like a word.
Let $X$ be a set. A nonempty collection $\mathcal{R} \subseteq \mathcal{P}(X)$ is a ring if $E \cup F \in \mathcal{R}$ and $E \setminus F \in \mathcal{R}$ for all $E, F \in \mathcal{R}$.
Since $\mathcal{R}$ has some set $E$ in it, it has $E \setminus E = \emptyset$. It's closed under intersections too, because $E \cap F = E \setminus (E \setminus F)$, and then by induction under finite unions and intersections. What a ring does not have to contain is $X$. That's the whole gap between a ring and an algebra: a ring that contains $X$ has every complement $X \setminus E$, so it is an algebra.
The name isn't a coincidence. With symmetric difference as addition and intersection as multiplication, a ring of sets is a ring in the algebra sense: a commutative ring where every element satisfies E · E = E (a Boolean ring), with ∅ as zero. Older books (Halmos among them) define rings of sets this way. Don't mix the two up in conversation, though: in this course "ring" means a ring of sets.
The Pro Tip's claim has a checkable half: the two pairs of operations generate each other.
Show that a nonempty collection $\mathcal{R} \subseteq \mathcal{P}(X)$ is a ring if and only if it is closed under symmetric differences $E \triangle F$ and intersections $E \cap F$.
Proof. ($\implies$) If $\mathcal{R}$ is a ring, then $E \triangle F = (E \setminus F) \cup (F \setminus E) \in \mathcal{R}$ and $E \cap F = E \setminus (E \setminus F) \in \mathcal{R}$.
($\impliedby$) Suppose $\mathcal{R}$ is closed under $\triangle$ and $\cap$. A point lies in $(E \triangle F) \triangle (E \cap F)$ when it lies in exactly one of $E \triangle F$ and $E \cap F$; these two sets are disjoint, so that means it lies in one of them, that is, in $E \cup F$. So $E \cup F = (E \triangle F) \triangle (E \cap F) \in \mathcal{R}$. And since $E \cap F \subseteq E$, $E \triangle (E \cap F) = E \setminus (E \cap F) = E \setminus F \in \mathcal{R}$. $\blacksquare$
Now the example that runs the page. Write $\mathcal{I} = \{[a, b) : a, b \in \mathbb{R}, \ a \le b\}$ for the half-open intervals (with $[a, a) = \emptyset$).
Is $\mathcal{I}$ a ring? Is the collection of finite unions of sets in $\mathcal{I}$ a ring?
$\mathcal{I}$ is not: $[0, 3) \setminus [1, 2) = [0, 1) \cup [2, 3)$ is not an interval, and neither is $[0, 1) \cup [2, 3)$.
Finite unions do form a ring. Unions are clear: a union of two finite unions is a finite union. For differences, the key fact is that a difference of two half-open intervals is at most two of them: $$[a, b) \setminus [c, d) = [a, \min\{b, c\}) \cup [\max\{a, d\}, b),$$ where either piece may be empty. That one formula is why the page uses half-open intervals: closed or open ones would leave stray endpoints behind. The exercise below finishes the argument.
Write $\mathcal{R}(\mathcal{I})$ for this ring of finite unions; the notation is justified by the proposition after the exercise. The construction will lean on a slightly stronger fact: every set in $\mathcal{R}(\mathcal{I})$ is a finite union of disjoint intervals.
Let $\mathcal{D}$ be the collection of finite disjoint unions $P_1 \cup \cdots \cup P_n$ of sets $P_i \in \mathcal{I}$. Show that $\mathcal{D}$ is a ring, and conclude that every finite union of sets in $\mathcal{I}$ is a finite disjoint union of sets in $\mathcal{I}$.
Proof. Two facts about intervals: $[a, b) \cap [c, d) = [\max\{a, c\}, \min\{b, d\})$ is in $\mathcal{I}$ (read it as $\emptyset$ when $\max \ge \min$), and $[a, b) \setminus [c, d)$ is a union of at most two disjoint sets of $\mathcal{I}$ (the formula in the example).
Intersections. If $E = \bigsqcup_i P_i$ and $F = \bigsqcup_l S_l$ are in $\mathcal{D}$, then $E \cap F = \bigcup_{i, l} (P_i \cap S_l)$. The pieces $P_i \cap S_l$ are in $\mathcal{I}$ and pairwise disjoint, so $E \cap F \in \mathcal{D}$.
Differences. $E \setminus F = \bigcup_i \bigcap_l (P_i \setminus S_l)$. Each $P_i \setminus S_l$ is in $\mathcal{D}$, so by the intersection step each $\bigcap_l (P_i \setminus S_l)$ is in $\mathcal{D}$, and it lies inside $P_i$. The $P_i$ are disjoint, so putting the pieces for different $i$ together is still a disjoint union: $E \setminus F \in \mathcal{D}$.
Unions. $E \cup F = (E \setminus F) \cup F$, a union of two disjoint sets of $\mathcal{D}$, so it is in $\mathcal{D}$.
So $\mathcal{D}$ is a ring containing $\mathcal{I}$. A ring is closed under finite unions, so every finite union of sets in $\mathcal{I}$ lies in $\mathcal{D}$. $\blacksquare$
Like σ-algebras, rings can be generated: start with any collection and close it up.
On σ-Algebras & Measurable Functions, the σ-algebra generated by $\mathcal{F}$ was the intersection of every σ-algebra containing $\mathcal{F}$. The same trick works for any kind of collection defined by closure rules, and we'll use it three times on this page.
For every collection $\mathcal{E} \subseteq \mathcal{P}(X)$ there is a unique ring $\mathcal{R}(\mathcal{E})$, the ring generated by $\mathcal{E}$, such that $\mathcal{E} \subseteq \mathcal{R}(\mathcal{E})$ and $\mathcal{R}(\mathcal{E}) \subseteq \mathcal{R}$ for every ring $\mathcal{R}$ with $\mathcal{E} \subseteq \mathcal{R}$.
Proof. Let $\Omega$ be the set of all rings $\mathcal{R}$ with $\mathcal{E} \subseteq \mathcal{R}$. It is not empty, since $\mathcal{P}(X)$ is one. Let $\mathcal{R}(\mathcal{E}) = \bigcap_{\mathcal{R} \in \Omega} \mathcal{R}$. It contains $\emptyset$ (every ring does), so it is nonempty. If $E, F \in \mathcal{R}(\mathcal{E})$, then $E$ and $F$ lie in every $\mathcal{R} \in \Omega$, each of which is a ring, so $E \cup F$ and $E \setminus F$ lie in every $\mathcal{R} \in \Omega$, hence in $\mathcal{R}(\mathcal{E})$. So $\mathcal{R}(\mathcal{E})$ is a ring. It contains $\mathcal{E}$ because each $\mathcal{R} \in \Omega$ does, and it is inside every $\mathcal{R} \in \Omega$ by construction.
If $\mathcal{R}_1$ and $\mathcal{R}_2$ both have the two properties, each is inside the other, so they are equal. $\blacksquare$
The intersection proves the ring exists, but it doesn't show you what's in it. For finitely many sets you can just see it: cut along every edge to get the atoms, and every union of atoms is in the ring.
That's why $\mathcal{R}(\mathcal{I})$ is the ring of finite unions of intervals: the finite unions form a ring containing $\mathcal{I}$ (the example), and any ring containing $\mathcal{I}$ contains them. Here's the same atom count on a finite set, where you can watch what's missing.
Let $X = \{1, 2, 3, 4\}$ and $\mathcal{E} = \{\{1, 2\}, \{2, 3\}\}$. List $\mathcal{R}(\mathcal{E})$. Is it an algebra on $X$?
The ring contains $\{1, 2\} \setminus \{2, 3\} = \{1\}$, $\{1, 2\} \cap \{2, 3\} = \{2\}$, and $\{2, 3\} \setminus \{1, 2\} = \{3\}$. These are the atoms: every point of $\{1, 2, 3\}$ sits in exactly one. Every union of atoms is in the ring, and the unions of atoms already form a ring (unions and differences of unions of atoms are unions of atoms), so $$\mathcal{R}(\mathcal{E}) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\},$$ $2^3 = 8$ sets. The point $4$ is in no generator, so it is in no set of the ring: $X \notin \mathcal{R}(\mathcal{E})$, and the ring is not an algebra.
Build it yourself, then break the pattern on purpose.
So the ring generated by finitely many sets has $2^k$ sets for $k$ atoms, and only the points the generators touch ever show up. Count a few.
Loading practice engine...
A ring is closed under finite unions. Measures care about countable ones, so the σ-version comes next.
Let $X$ be a set. A ring $\mathcal{S} \subseteq \mathcal{P}(X)$ is a σ-ring if $\bigcup_{j=1}^{\infty} E_j \in \mathcal{S}$ whenever $E_1, E_2, \ldots \in \mathcal{S}$.
A σ-ring is closed under countable intersections as well: $\bigcap_j E_j = E_1 \setminus \bigcup_j (E_1 \setminus E_j)$. And the intersection trick from the proposition works word for word for σ-rings, so every collection generates one.
For $\mathcal{E} \subseteq \mathcal{P}(X)$, the σ-ring generated by $\mathcal{E}$, written $\mathcal{S}(\mathcal{E})$, is the smallest σ-ring containing $\mathcal{E}$: the intersection of all σ-rings containing $\mathcal{E}$.
Two examples on $\mathbb{R}$ show the two ways a collection can fall short.
(a) Show that the finite subsets of $\mathbb{R}$ form a ring but not a σ-ring. (b) Show that the countable subsets of $\mathbb{R}$ form a σ-ring but not a σ-algebra.
(a) A union or difference of two finite sets is finite, so it's a ring. But $\bigcup_n \{n\} = \mathbb{N}$ is a countable union of finite sets that is not finite.
(b) Unions and differences of countable sets are countable, and so is a countable union of countable sets. So it's a σ-ring. It isn't a σ-algebra because $\mathbb{R}$ itself is not countable, so it isn't in the collection.
In (b), the only thing missing for a σ-algebra was $\mathbb{R}$. That's always the story.
If $\mathcal{S}$ is a σ-ring on $X$ and $X \in \mathcal{S}$, then $\mathcal{S}$ is a σ-algebra on $X$.
Proof. $X \in \mathcal{S}$ by assumption, and $\mathcal{S}$ is closed under countable unions because it is a σ-ring. For $E \in \mathcal{S}$, $E^c = X \setminus E$ is a difference of two sets of $\mathcal{S}$, so $E^c \in \mathcal{S}$. $\blacksquare$
So there are four places a collection can land: not a ring at all, a ring that loses a countable union, a σ-ring that's only missing the whole space, or a σ-algebra. Sort some.
Notice the pattern in the bad bins: open sets, closed sets, and intervals all fail on a difference, and that's the operation people forget to check. Your turn.
Loading practice engine...
σ-rings are hard to check directly: you have to handle every countable union at once. Some collections are much easier to check against a weaker rule, closure under monotone limits only. That rule is exactly what continuity of measures (from below and from above) is good at. So it gets a name.
A nonempty collection $\mathcal{K} \subseteq \mathcal{P}(X)$ is a monotone class if $\bigcup_n E_n \in \mathcal{K}$ whenever $E_1 \subseteq E_2 \subseteq \cdots$ are in $\mathcal{K}$, and $\bigcap_n E_n \in \mathcal{K}$ whenever $E_1 \supseteq E_2 \supseteq \cdots$ are in $\mathcal{K}$.
Every σ-ring is a monotone class, since it is closed under all countable unions and intersections. An intersection of monotone classes containing $\mathcal{E}$ is again one, so here's the generated version.
For a nonempty $\mathcal{E} \subseteq \mathcal{P}(X)$, the monotone class generated by $\mathcal{E}$, written $\mathcal{K}(\mathcal{E})$, is the smallest monotone class containing $\mathcal{E}$: the intersection of all monotone classes containing $\mathcal{E}$.
Monotone classes can be very small, much smaller than the σ-ring they sit inside.
Let $\mathcal{E} = \{[0, 1), [1, 2)\}$. Find $\mathcal{K}(\mathcal{E})$ and compare it with $\mathcal{S}(\mathcal{E})$.
An increasing sequence drawn from a two-set collection of incomparable sets is constant, and so is a decreasing one. Its union or intersection is that same set, which is in $\mathcal{E}$. So $\mathcal{E}$ is already a monotone class, and $\mathcal{K}(\mathcal{E}) = \mathcal{E}$.
But $\mathcal{S}(\mathcal{E})$ contains $\emptyset$ and $[0, 2) = [0, 1) \cup [1, 2)$, which are not in $\mathcal{E}$. So $\mathcal{K}(\mathcal{E}) \subsetneq \mathcal{S}(\mathcal{E})$.
The trouble is that $\mathcal{E}$ isn't a ring. Start from a ring, and the monotone class it generates catches up with the σ-ring. The proof is the cleverest few lines on the page.
If $\mathcal{R}$ is a ring, then $\mathcal{K}(\mathcal{R})$ is a σ-ring.
Proof. Step 1: a monotone class that is a ring is a σ-ring. Let $\mathcal{K}$ be both, and let $E_1, E_2, \ldots \in \mathcal{K}$. Put $F_n = E_1 \cup \cdots \cup E_n$. Each $F_n \in \mathcal{K}$ because $\mathcal{K}$ is a ring, the $F_n$ increase, and $\bigcup_n F_n = \bigcup_j E_j$. So $\bigcup_j E_j \in \mathcal{K}$ because $\mathcal{K}$ is a monotone class.
Step 2: $\mathcal{K}(\mathcal{R})$ is a ring. For $F \subseteq X$, let $$\mathcal{L}(F) = \{E \subseteq X : E \setminus F, \ F \setminus E, \ E \cup F \in \mathcal{K}(\mathcal{R})\}.$$ The definition is symmetric: $E \in \mathcal{L}(F)$ exactly when $F \in \mathcal{L}(E)$.
Whenever $\mathcal{L}(F)$ is nonempty, it is a monotone class. If $E_1 \subseteq E_2 \subseteq \cdots$ lie in $\mathcal{L}(F)$ with union $E$, then $E \setminus F = \bigcup_j (E_j \setminus F)$ and $E \cup F = \bigcup_j (E_j \cup F)$ are increasing unions of sets in $\mathcal{K}(\mathcal{R})$, and $F \setminus E = \bigcap_j (F \setminus E_j)$ is a decreasing intersection of sets in $\mathcal{K}(\mathcal{R})$, so all three lie in $\mathcal{K}(\mathcal{R})$ and $E \in \mathcal{L}(F)$. Decreasing sequences work the same way, with unions and intersections trading places.
Now take $F \in \mathcal{R}$. For every $E \in \mathcal{R}$, the three sets $E \setminus F$, $F \setminus E$, $E \cup F$ are in $\mathcal{R} \subseteq \mathcal{K}(\mathcal{R})$, so $\mathcal{R} \subseteq \mathcal{L}(F)$. Since $\mathcal{L}(F)$ is a monotone class containing $\mathcal{R}$, $\mathcal{K}(\mathcal{R}) \subseteq \mathcal{L}(F)$.
So for $E \in \mathcal{K}(\mathcal{R})$ and $F \in \mathcal{R}$ we have $E \in \mathcal{L}(F)$, and by symmetry $F \in \mathcal{L}(E)$. That says $\mathcal{R} \subseteq \mathcal{L}(E)$ for each $E \in \mathcal{K}(\mathcal{R})$, and the same argument gives $\mathcal{K}(\mathcal{R}) \subseteq \mathcal{L}(E)$. Unpacked: for all $E, F \in \mathcal{K}(\mathcal{R})$, the sets $E \setminus F$ and $E \cup F$ are in $\mathcal{K}(\mathcal{R})$. So $\mathcal{K}(\mathcal{R})$ is a ring, and by Step 1 a σ-ring. $\blacksquare$
The move to remember is the bootstrapping: first fix $F$ in the small collection and let $E$ roam, then use symmetry to let $F$ roam too. It gives the theorem this whole section exists for.
If $\mathcal{R}$ is a ring, then $\mathcal{S}(\mathcal{R}) = \mathcal{K}(\mathcal{R})$.
Proof. By the theorem above, $\mathcal{K}(\mathcal{R})$ is a σ-ring containing $\mathcal{R}$, and $\mathcal{S}(\mathcal{R})$ is the smallest one, so $\mathcal{S}(\mathcal{R}) \subseteq \mathcal{K}(\mathcal{R})$. Conversely, $\mathcal{S}(\mathcal{R})$ is a σ-ring, hence a monotone class, containing $\mathcal{R}$, and $\mathcal{K}(\mathcal{R})$ is the smallest one, so $\mathcal{K}(\mathcal{R}) \subseteq \mathcal{S}(\mathcal{R})$. $\blacksquare$
Here's how it gets used. To prove that some property holds for every set in $\mathcal{S}(\mathcal{R})$, check it on the ring, then check that the sets with the property form a monotone class. Two measures agreeing on a ring is the classic case, and that's how this page ends.
Product Measures & Fubini runs the same play: to show a property holds for every set in the product σ-algebra, it checks finite unions of rectangles and then shows the good sets form a monotone class.
A measure on a ring is defined just like a measure on a σ-algebra, with one adjustment: a ring doesn't promise that a countable union of its sets is still in it, so countable additivity is only asked for when it is.
Let $\mathcal{R}$ be a ring. A measure on $\mathcal{R}$ is a function $\mu\colon \mathcal{R} \to [0, \infty]$ with $\mu(\emptyset) = 0$ and $\mu\left(\bigcup_{j=1}^{\infty} E_j\right) = \sum_{j=1}^{\infty} \mu(E_j)$ whenever $E_1, E_2, \ldots \in \mathcal{R}$ are pairwise disjoint and $\bigcup_j E_j \in \mathcal{R}$.
Padding with empty sets gives finite additivity, and then monotonicity comes for free: if $F \subseteq E$ in $\mathcal{R}$, then $\mu(E) = \mu(F) + \mu(E \setminus F) \ge \mu(F)$. The proofs on Measures of continuity from below and from above only ever use countable unions and intersections that stay inside the collection, so on a σ-ring they work word for word.
Now length. On the intervals themselves it's the obvious thing.
The length of $[a, b) \in \mathcal{I}$ is $\ell([a, b)) = b - a$.
So $\ell(\emptyset) = 0$. Finite additivity on $\mathcal{I}$ is believable from a picture. Countable additivity is the real content: $[0, 1)$ can be cut into infinitely many disjoint pieces, and nothing so far rules out the pieces' lengths adding up to less than $1$. The proof needs compactness, through this lemma about finite covers.
Prove: if $[a, b] \subseteq \bigcup_{j=1}^{n} (a_j, b_j)$ for finitely many open intervals, then $b - a \le \sum_{j=1}^{n} (b_j - a_j)$.
Proof. Induction on $n$. For $n = 1$: $[a, b] \subseteq (a_1, b_1)$ means $a_1 < a$ and $b < b_1$, so $b - a < b_1 - a_1$.
Suppose the claim holds for covers by $n - 1$ intervals, and let $[a, b] \subseteq \bigcup_{j=1}^{n} (a_j, b_j)$. Some interval contains $b$; renumber so that it is $(a_n, b_n)$, so $a_n < b < b_n$. If $a_n < a$, then $b - a < b_n - a_n$ and we are done. Otherwise $a \le a_n < b$, and no point of $[a, a_n]$ lies in $(a_n, b_n)$, so $[a, a_n] \subseteq \bigcup_{j=1}^{n-1} (a_j, b_j)$. By the induction hypothesis, $a_n - a \le \sum_{j=1}^{n-1} (b_j - a_j)$. Then $$b - a = (b - a_n) + (a_n - a) < (b_n - a_n) + \sum_{j=1}^{n-1} (b_j - a_j). \qquad \blacksquare$$
Now the main event. The trick is to trade the half-open pieces for slightly bigger open ones, paying $\varepsilon/2^j$ for the $j$th, so the total bill is $\varepsilon$, and then let compactness throw away all but finitely many.
ε/2 + ε/4 + ε/8 + ⋯ = ε. The geometric series has been quietly doing everybody's homework since forever.
Here's the proof the animation draws.
If $I_1, I_2, \ldots \in \mathcal{I}$ are pairwise disjoint and $P = \bigcup_{j=1}^{\infty} I_j \in \mathcal{I}$, then $\ell(P) = \sum_{j=1}^{\infty} \ell(I_j)$.
Proof. Empty $I_j$ contribute $0$ to the sum and nothing to the union, so drop them. If none are left, $P = \emptyset$ and both sides are $0$. Otherwise $P = [a, b)$ with $a < b$; write $I_j = [a_j, b_j)$ with $a_j < b_j$ for the ones left.
($\le$) Let $0 < \varepsilon < b - a$. Every point of $[a, b - \varepsilon]$ lies in $P$, so in some $I_j$, so in the larger open interval $(a_j - \varepsilon/2^j, \ b_j)$: $$[a, b - \varepsilon] \subseteq \bigcup_j \left(a_j - \tfrac{\varepsilon}{2^j}, \ b_j\right).$$ $[a, b - \varepsilon]$ is compact, so finitely many of these open intervals cover it, and the finite-cover exercise gives $$b - \varepsilon - a \le \sum_{\text{finitely many } j} \left(b_j - a_j + \tfrac{\varepsilon}{2^j}\right) \le \sum_j \ell(I_j) + \varepsilon.$$ So $\ell(P) = b - a \le \sum_j \ell(I_j) + 2\varepsilon$ for every small $\varepsilon > 0$, hence $\ell(P) \le \sum_j \ell(I_j)$.
($\ge$) Take finitely many of the $I_j$ and number them so that $a_1 < a_2 < \cdots < a_N$ (disjoint nonempty intervals have different left endpoints). They are disjoint, so $a_{j+1} \notin [a_j, b_j)$, which forces $b_j \le a_{j+1}$; they lie in $[a, b)$, so $a \le a_1$ and $b_N \le b$. Then $$\sum_{j=1}^{N} (b_j - a_j) \le (a_2 - a_1) + (a_3 - a_2) + \cdots + (a_N - a_{N-1}) + (b_N - a_N) = b_N - a_1 \le b - a.$$ Every finite partial sum is at most $\ell(P)$, so the whole sum is too. $\blacksquare$
Compactness was the whole game in the first half. It's where the completeness of $\mathbb{R}$ enters, and it's why this works for the real line and not for, say, the rationals alone. With the lemma, length spreads to the whole ring $\mathcal{R}(\mathcal{I})$ in exactly one way.
There is exactly one measure $\mu$ on $\mathcal{R}(\mathcal{I})$ with $\mu = \ell$ on $\mathcal{I}$. It is given by $\mu(E) = \sum_{i=1}^{n} \ell(P_i)$ whenever $E = P_1 \cup \cdots \cup P_n$ with $P_1, \ldots, P_n \in \mathcal{I}$ pairwise disjoint.
Proof. Well defined. By the second exercise, every $E \in \mathcal{R}(\mathcal{I})$ is such a disjoint union. If also $E = S_1 \cup \cdots \cup S_m$ disjointly, then each $P_i$ is the disjoint union of the intervals $P_i \cap S_l$ ($l = 1, \ldots, m$), so the lemma gives $\ell(P_i) = \sum_l \ell(P_i \cap S_l)$, and $$\sum_i \ell(P_i) = \sum_i \sum_l \ell(P_i \cap S_l) = \sum_l \sum_i \ell(P_i \cap S_l) = \sum_l \ell(S_l).$$
Unique. Any measure on $\mathcal{R}(\mathcal{I})$ that equals $\ell$ on $\mathcal{I}$ is finitely additive, so it must be given by the formula.
A measure. $\mu(\emptyset) = 0$. Let $E_1, E_2, \ldots \in \mathcal{R}(\mathcal{I})$ be pairwise disjoint with $E = \bigcup_j E_j \in \mathcal{R}(\mathcal{I})$. Write $E = I_1 \cup \cdots \cup I_m$ and $E_j = I_{j, 1} \cup \cdots \cup I_{j, n_j}$ as disjoint unions of intervals. For each $s$, the interval $I_s$ is the disjoint union of the countably many intervals $I_s \cap I_{j, l}$, so by the lemma $$\ell(I_s) = \sum_{j} \sum_{l} \ell(I_s \cap I_{j, l}).$$ Sum over $s$ and rearrange (all terms are $\ge 0$): $$\mu(E) = \sum_s \ell(I_s) = \sum_j \left( \sum_s \sum_l \ell(I_s \cap I_{j, l}) \right) = \sum_j \mu(E_j),$$ because for each $j$ the intervals $I_s \cap I_{j, l}$ are disjoint with union $E \cap E_j = E_j$. $\blacksquare$
So the formula is the one you'd guess: cut into disjoint intervals and add. The only way to get it wrong is to forget the "disjoint".
Find $\mu([0, 1) \cup [3, 5))$ and $\mu([0, 2) \cup [1, 3))$.
The first two intervals are disjoint: $\mu = 1 + 2 = 3$.
The second two overlap. As a disjoint union, $[0, 2) \cup [1, 3) = [0, 3)$, so $\mu = 3$, not $2 + 2 = 4$: adding the lengths would count $[1, 2)$ twice.
Length is now a measure, but only on finite unions of intervals, a tiny collection. To measure any other set, do the natural thing: cover it by countably many sets you can already measure, and take the cheapest cover. First, the sets that can be covered.
Let $\mathcal{R}$ be a ring of subsets of $X$. The hereditary σ-ring $\mathcal{H}(\mathcal{R})$ is the collection of all $E \subseteq X$ for which there are $E_1, E_2, \ldots \in \mathcal{R}$ with $E \subseteq \bigcup_{j=1}^{\infty} E_j$.
"Hereditary" means closed under subsets: a subset of a coverable set is coverable by the same cover. Here's what it looks like in the two running examples.
Find $\mathcal{H}(\mathcal{R}(\mathcal{I}))$ for $\mathbb{R}$, and $\mathcal{H}(\mathcal{R})$ for the ring $\mathcal{R}$ of finite subsets of $\mathbb{R}$.
$\mathbb{R} = \bigcup_n [-n, n)$, so every subset of $\mathbb{R}$ is covered by intervals: $\mathcal{H}(\mathcal{R}(\mathcal{I})) = \mathcal{P}(\mathbb{R})$. That's the point: the outer measure will measure every subset of $\mathbb{R}$.
A countable union of finite sets is countable, and every countable set is a countable union of singletons. So $\mathcal{H}(\mathcal{R})$ is the collection of countable subsets of $\mathbb{R}$, a σ-ring but not a σ-algebra.
The name promises a σ-ring, and it delivers. Check it, since the Carathéodory theorem below leans on it.
Show that $\mathcal{H}(\mathcal{R})$ is a σ-ring containing $\mathcal{R}$, and that every subset of a set in $\mathcal{H}(\mathcal{R})$ is in $\mathcal{H}(\mathcal{R})$.
Proof. Each $E \in \mathcal{R}$ is covered by $E, \emptyset, \emptyset, \ldots$, so $\mathcal{R} \subseteq \mathcal{H}(\mathcal{R})$; in particular $\mathcal{H}(\mathcal{R})$ is nonempty. If $F \subseteq E$ and $E \subseteq \bigcup_j E_j$, then $F \subseteq \bigcup_j E_j$ too, so subsets stay in. In particular $E \setminus F \subseteq E$ is in $\mathcal{H}(\mathcal{R})$ whenever $E$ is.
For countable unions: if $E_k \subseteq \bigcup_j E_{k, j}$ with every $E_{k, j} \in \mathcal{R}$, then $\bigcup_k E_k \subseteq \bigcup_{k, j} E_{k, j}$, a countable union of sets of $\mathcal{R}$. This also covers finite unions. So $\mathcal{H}(\mathcal{R})$ is a σ-ring. $\blacksquare$
Now the set functions that will live on $\mathcal{H}(\mathcal{R})$. They keep the two properties of a measure that survive overcounting, and drop additivity.
An outer measure on $\mathcal{H}(\mathcal{R})$ is a function $\nu\colon \mathcal{H}(\mathcal{R}) \to [0, \infty]$ such that (i) $\nu(E) \le \nu(F)$ whenever $E \subseteq F$; (ii) $\nu\left(\bigcup_{j=1}^{\infty} E_j\right) \le \sum_{j=1}^{\infty} \nu(E_j)$ for all $E_1, E_2, \ldots \in \mathcal{H}(\mathcal{R})$; and (iii) $\nu(\emptyset) = 0$.
When $\mathcal{R}$ contains $X$ (say $\mathcal{R} = \mathcal{P}(X)$), $\mathcal{H}(\mathcal{R}) = \mathcal{P}(X)$, so on a small finite set it's easy to cook up examples.
Let $X = \{1, 2, 3, 4\}$. Which of these are outer measures on $\mathcal{P}(X)$? (a) $\nu(E) = \min\{|E|, 1\}$. (b) $\nu(E) = \lceil |E| / 2 \rceil$. (c) $\nu(E) = |E|^2$.
All three are $0$ on $\emptyset$ and grow with $E$, so only subadditivity is in question. (On a finite set, a countable union is a finite union, and it's enough to check two sets at a time.)
(a) Yes: $\min\{|E \cup F|, 1\} \le \min\{|E|, 1\} + \min\{|F|, 1\}$, since the left side is $1$ only when one of $E$, $F$ is nonempty.
(b) Yes: $|E \cup F| \le |E| + |F|$, and $\lceil (m + n)/2 \rceil \le \lceil m/2 \rceil + \lceil n/2 \rceil$.
(c) No: $\nu(\{1, 2\}) = 4$, but $\nu(\{1\}) + \nu(\{2\}) = 2$.
Those are toys. The outer measure that matters comes from a measure on the ring, by the cheapest-cover recipe.
Outer measure: the set's measure if you're allowed to wear a really big coat.
Let $\mu$ be a measure on a ring $\mathcal{R}$. The outer measure induced by $\mu$ is the function $\mu^*$ on $\mathcal{H}(\mathcal{R})$ given by $$\mu^*(E) = \inf\left\{\sum_{j=1}^{\infty} \mu(E_j) : E_1, E_2, \ldots \in \mathcal{R}, \ E \subseteq \bigcup_{j=1}^{\infty} E_j\right\}.$$
The name has to be earned: $\mu^*$ really is an outer measure.
Let $\mu$ be a measure on a ring $\mathcal{R}$. Then $\mu^*$ is an outer measure on $\mathcal{H}(\mathcal{R})$.
Proof. (iii) Cover $\emptyset$ by $\emptyset, \emptyset, \ldots$: the sum is $0$, so $\mu^*(\emptyset) = 0$.
(i) Let $E \subseteq F$ in $\mathcal{H}(\mathcal{R})$. Every countable cover of $F$ by sets of $\mathcal{R}$ is also a cover of $E$, so the infimum defining $\mu^*(E)$ runs over at least as many sums as the one defining $\mu^*(F)$. Hence $\mu^*(E) \le \mu^*(F)$.
(ii) Let $E_1, E_2, \ldots \in \mathcal{H}(\mathcal{R})$ and $E = \bigcup_j E_j$, which is in $\mathcal{H}(\mathcal{R})$ by the exercise above. If some $\mu^*(E_j) = \infty$, the right side is $\infty$ and there is nothing to prove. Otherwise let $\varepsilon > 0$, and for each $j$ choose $E_{j, 1}, E_{j, 2}, \ldots \in \mathcal{R}$ covering $E_j$ with $$\sum_l \mu(E_{j, l}) < \mu^*(E_j) + \frac{\varepsilon}{2^j}.$$ All the $E_{j, l}$ together are a countable cover of $E$ by sets of $\mathcal{R}$, so $$\mu^*(E) \le \sum_{j, l} \mu(E_{j, l}) \le \sum_j \left(\mu^*(E_j) + \frac{\varepsilon}{2^j}\right) = \sum_j \mu^*(E_j) + \varepsilon.$$ Let $\varepsilon \to 0$. $\blacksquare$
An outer measure built from $\mu$ had better agree with $\mu$ on the sets $\mu$ already knew. It does, and this is the second place countable additivity on the ring is essential.
Let $\mu$ be a measure on a ring $\mathcal{R}$. Show that $\mu^*(E) = \mu(E)$ for every $E \in \mathcal{R}$.
Proof. ($\le$) $E, \emptyset, \emptyset, \ldots$ is a cover of $E$ with sum $\mu(E)$.
($\ge$) Let $E \subseteq \bigcup_j E_j$ with $E_j \in \mathcal{R}$. Put $G_j = (E \cap E_j) \setminus (E_1 \cup \cdots \cup E_{j-1})$. These are in $\mathcal{R}$, pairwise disjoint (a point of $E$ lands in $G_j$ for the first $j$ with $E_j$ containing it), and their union is $E$, which is in $\mathcal{R}$. Countable additivity on $\mathcal{R}$ and monotonicity ($G_j \subseteq E_j$) give $$\mu(E) = \sum_j \mu(G_j) \le \sum_j \mu(E_j).$$ Taking the infimum over all covers, $\mu(E) \le \mu^*(E)$. $\blacksquare$
That pieces trick, $G_j = E_j$ minus everything before it, keeps coming back. Here it is on its own.
Now specialize to length. Covers by sets of $\mathcal{R}(\mathcal{I})$ are the same as covers by intervals (split each finite union into disjoint intervals, and the sum doesn't change), so the induced outer measure has a very concrete formula.
The Lebesgue outer measure is the outer measure $\mu^*$ induced by the length measure $\mu$ on $\mathcal{R}(\mathcal{I})$. It is defined on every $E \subseteq \mathbb{R}$, by $$\mu^*(E) = \inf\left\{\sum_{j=1}^{\infty} (b_j - a_j) : E \subseteq \bigcup_{j=1}^{\infty} [a_j, b_j)\right\}.$$
First consequence: a single point costs nothing, since $\{p\} \subseteq [p, p + \varepsilon)$ for every $\varepsilon > 0$. Countably many points cost nothing either, but that takes the geometric series again.
The animation is the whole proof, so let's write it down once.
Show that every countable set $E = \{q_1, q_2, \ldots\} \subseteq \mathbb{R}$ has $\mu^*(E) = 0$. In particular $\mu^*(\mathbb{Q} \cap [0, 1]) = 0$, and $\mu^*(\mathbb{Q}) = 0$.
Let $\varepsilon > 0$ and cover $q_j$ by $[q_j, q_j + \varepsilon/2^j)$. The total length is $\sum_j \varepsilon/2^j = \varepsilon$, so $\mu^*(E) \le \varepsilon$ for every $\varepsilon > 0$. Hence $\mu^*(E) = 0$. ($\mathbb{Q}$ is dense, so every interval meets it. Dense is not the same as big.)
So far everything has been small. The real test is whether a big set gets its honest size: is $\mu^*([0, 1])$ actually $1$, or could some clever countable cover sneak in under it?
Show that $\mu^*([0, 1]) = 1$, and that $\mu^*([a, b]) = \mu^*((a, b)) = b - a$ for all $a < b$.
($\le$) $[0, 1] \subseteq [0, 1 + \varepsilon)$ for every $\varepsilon > 0$, so $\mu^*([0, 1]) \le 1$.
($\ge$) Let $[0, 1] \subseteq \bigcup_j [a_j, b_j)$, and let $\varepsilon > 0$. Then $[0, 1] \subseteq \bigcup_j (a_j - \varepsilon/2^j, \ b_j)$, an open cover of a compact set, so finitely many of these intervals cover $[0, 1]$. The finite-cover exercise gives $1 \le \sum_j (b_j - a_j) + \varepsilon$. That holds for every cover and every $\varepsilon$, so $\mu^*([0, 1]) \ge 1$.
The same argument works for any $[a, b]$. Since $(a, b)$, $[a, b)$, and $[a, b]$ differ by at most two points, which have outer measure $0$, monotonicity and subadditivity give them all the value $b - a$.
Compactness again: without it, nothing stops a cover of $[0, 1]$ from being cheaper than $1$. Now for an uncountable set that is still null.
Each row is a cover of the Cantor set, and each one is cheaper than the last. Here's the argument with numbers.
Let $C_0 = [0, 1]$, and let $C_{n+1}$ be $C_n$ with the open middle third of each of its intervals removed. Show that the Cantor set $C = \bigcap_n C_n$ has $\mu^*(C) = 0$.
$C_n$ is $2^n$ closed intervals of length $3^{-n}$. Covering each by a half-open interval of length $3^{-n} + \varepsilon 2^{-n}$ shows $\mu^*(C_n) \le (2/3)^n + \varepsilon$, so $\mu^*(C_n) \le (2/3)^n$. Since $C \subseteq C_n$ for every $n$, monotonicity gives $\mu^*(C) \le (2/3)^n \to 0$.
Yet $C$ is uncountable: it is the set of numbers in $[0, 1]$ with a base-3 expansion using only the digits $0$ and $2$, and those expansions are in bijection with infinite sequences of two symbols.
Covers are the whole definition, so build some. The cheapest cover wins.
Two rules came out of that: overlaps are paid for only once, and countable pieces are free. Compute a few.
Loading practice engine...
$\mu^*$ measures every subset of $\mathbb{R}$, but it is only subadditive. On some disjoint pairs it overcounts: Lebesgue Measure builds Vitali's nonmeasurable set, where $\mu^*$ fails to add. So we cut back to the sets where nothing goes wrong. Carathéodory's idea: a good set $E$ is one that splits every other set cleanly, with no overcount.
Let $\nu$ be an outer measure on $\mathcal{H}(\mathcal{R})$. A set $E \in \mathcal{H}(\mathcal{R})$ is $\nu$-measurable if $$\nu(A) = \nu(A \cap E) + \nu(A \cap E^c) \quad \text{for every } A \in \mathcal{H}(\mathcal{R}).$$ Write $\mathcal{M}_\nu$ for the collection of $\nu$-measurable sets.
Half of the test is free: $A \subseteq (A \cap E) \cup (A \cap E^c)$, so subadditivity always gives $\nu(A) \le \nu(A \cap E) + \nu(A \cap E^c)$. Only "$\ge$" needs checking, and only for $A$ with $\nu(A) < \infty$. (Complements are taken in $X$, but only through $A \cap E^c = A \setminus E$, which stays in $\mathcal{H}(\mathcal{R})$.) Here's a pass and a fail.
Carathéodory's test: slice every subset of the universe with E and check nothing leaks. Every one. I hope he had the afternoon free.
On a finite set you can run every test by hand.
Let $X = \{1, 2, 3, 4\}$. (a) For $\nu(E) = \lceil |E| / 2 \rceil$, find every $\nu$-measurable set. (b) Let $\mu$ be the measure on the ring generated by $\{1, 2\}$, $\{3\}$, $\{4\}$ with $\mu(\{1, 2\}) = 2$, $\mu(\{3\}) = 1$, $\mu(\{4\}) = 0$. Is $\{1\}$ $\mu^*$-measurable? Is $\{3, 4\}$?
(a) $\emptyset$ and $X$ always pass. A one-point set $\{p\}$ fails at $A = X$: $\nu(X) = 2$, but $\nu(\{p\}) + \nu(X \setminus \{p\}) = 1 + 2 = 3$. A three-point set fails the same way, through its complement. A two-point set, say $\{1, 2\}$, passes at $A = X$ ($2 = 1 + 1$) but fails at $A = \{1, 3\}$: $\nu(\{1, 3\}) = 1$, while $\nu(\{1\}) + \nu(\{3\}) = 2$. So only $\emptyset$ and $X$ are $\nu$-measurable, and checking $A = X$ alone would have been fooled.
(b) $\mu^*(E)$ adds up the weights of the blocks $E$ touches, since the cheapest cover uses exactly those blocks. Test $\{1\}$ with $A = \{1, 2\}$: $\mu^*(\{1, 2\}) = 2$, but $\mu^*(\{1\}) + \mu^*(\{2\}) = 2 + 2 = 4$. It fails: it cuts a block. For $E = \{3, 4\}$, every $A$ splits into $A \cap \{3, 4\}$ and $A \cap \{1, 2\}$, which touch different blocks, so the weights add exactly. It passes.
The pattern in (b) is the general one: a set passes when it never cuts a block of positive weight. Try to fool the test.
Did you notice $\{4\}$ passes under $\nu_1$? It has outer measure $0$, and null sets always pass. That's completeness, and Carathéodory's construction gets it for free.
On Measures, completing a measure space was a separate construction: add every subset of a null set by hand. Here it's built in.
Now the theorem. It holds for any outer measure, and when the outer measure comes from a measure on a ring, it also catches the ring.
Let $\nu$ be an outer measure on $\mathcal{H}(\mathcal{R})$. Then $\mathcal{M}_\nu$ is a σ-ring, the restriction of $\nu$ to $\mathcal{M}_\nu$ is a measure, and every $E \in \mathcal{H}(\mathcal{R})$ with $\nu(E) = 0$ is in $\mathcal{M}_\nu$ (so the measure is complete). If $\nu = \mu^*$ for a measure $\mu$ on $\mathcal{R}$, then also $\mathcal{R} \subseteq \mathcal{M}_{\mu^*}$, hence $\mathcal{S}(\mathcal{R}) \subseteq \mathcal{M}_{\mu^*}$, and $\mu^* = \mu$ on $\mathcal{R}$.
Proof. Throughout, $A$ is any set in $\mathcal{H}(\mathcal{R})$, and every intersection $A \cap \cdots$ below is in $\mathcal{H}(\mathcal{R})$ because it is a subset of $A$. $\emptyset \in \mathcal{M}_\nu$, since $\nu(A) = \nu(\emptyset) + \nu(A)$.
The four-piece identity. Let $E, F \in \mathcal{M}_\nu$. Testing $E$ on $A$, and then $F$ on $A \cap E$ and on $A \cap E^c$, gives for every $A$ $$\nu(A) = \nu(A \cap E \cap F) + \nu(A \cap E \cap F^c) + \nu(A \cap E^c \cap F) + \nu(A \cap E^c \cap F^c). \tag{$*$}$$
Unions. Apply $(*)$ to the test set $A \cap (E \cup F)$ in place of $A$: its fourth piece is empty, so $\nu(A \cap (E \cup F))$ is the sum of the first three pieces of $(*)$. Substituting back, $$\nu(A) = \nu(A \cap (E \cup F)) + \nu(A \cap E^c \cap F^c) = \nu(A \cap (E \cup F)) + \nu(A \cap (E \cup F)^c).$$ So $E \cup F \in \mathcal{M}_\nu$.
Differences. Apply $(*)$ to the test set $A \cap (E \setminus F)^c$. Since $(E \setminus F)^c = E^c \cup F$, its pieces are $A \cap E \cap F$, $\emptyset$, $A \cap E^c \cap F$, and $A \cap E^c \cap F^c$. So $\nu(A \cap (E \setminus F)^c)$ is the sum of every piece of $(*)$ except $\nu(A \cap E \cap F^c) = \nu(A \cap (E \setminus F))$, and $$\nu(A \cap (E \setminus F)) + \nu(A \cap (E \setminus F)^c) = \nu(A).$$ So $E \setminus F \in \mathcal{M}_\nu$, and $\mathcal{M}_\nu$ is a ring.
Disjoint countable unions. Let $E_1, E_2, \ldots \in \mathcal{M}_\nu$ be pairwise disjoint, $E = \bigcup_j E_j$ (in $\mathcal{H}(\mathcal{R})$, a σ-ring), and $G_n = E_1 \cup \cdots \cup E_n \in \mathcal{M}_\nu$. Testing $E_n$ on $A \cap G_n$ gives $\nu(A \cap G_n) = \nu(A \cap E_n) + \nu(A \cap G_{n-1})$, so by induction $\nu(A \cap G_n) = \sum_{j \le n} \nu(A \cap E_j)$. Testing $G_n$ on $A$, and using $G_n^c \supseteq E^c$, $$\nu(A) = \nu(A \cap G_n) + \nu(A \cap G_n^c) \ge \sum_{j=1}^{n} \nu(A \cap E_j) + \nu(A \cap E^c).$$ Let $n \to \infty$, and use subadditivity on $A \cap E = \bigcup_j (A \cap E_j)$: $$\nu(A) \ge \sum_{j=1}^{\infty} \nu(A \cap E_j) + \nu(A \cap E^c) \ge \nu(A \cap E) + \nu(A \cap E^c). \tag{$\dagger$}$$ The reverse inequality is subadditivity, so $E \in \mathcal{M}_\nu$.
All countable unions. For any $E_1, E_2, \ldots \in \mathcal{M}_\nu$, the sets $F_j = E_j \setminus (E_1 \cup \cdots \cup E_{j-1})$ are in the ring $\mathcal{M}_\nu$, pairwise disjoint, with the same union. So $\mathcal{M}_\nu$ is a σ-ring.
A measure. With the $E_j$ disjoint and $A = E$, $(\dagger)$ reads $\nu(E) \ge \sum_j \nu(E_j) + \nu(\emptyset)$. Subadditivity gives "$\le$", so $\nu$ is countably additive on $\mathcal{M}_\nu$, and $\nu(\emptyset) = 0$.
Complete. If $\nu(F) = 0$, then $\nu(A \cap F) \le \nu(F) = 0$ and $\nu(A \cap F^c) \le \nu(A)$, so $\nu(A \cap F) + \nu(A \cap F^c) \le \nu(A)$, and $F \in \mathcal{M}_\nu$. Any subset of a null set is in $\mathcal{H}(\mathcal{R})$ and null, so it is in $\mathcal{M}_\nu$ too.
The ring is measurable. Now let $\nu = \mu^*$, and let $E \in \mathcal{R}$. Take $A$ with $\mu^*(A) < \infty$ (if $\mu^*(A) = \infty$, "$\ge$" is automatic). Given $\varepsilon > 0$, choose $E_1, E_2, \ldots \in \mathcal{R}$ covering $A$ with $\sum_j \mu(E_j) < \mu^*(A) + \varepsilon$. The sets $E_j \cap E$ and $E_j \setminus E$ are in $\mathcal{R}$, and they cover $A \cap E$ and $A \cap E^c$. So $$\mu^*(A \cap E) + \mu^*(A \cap E^c) \le \sum_j \mu(E_j \cap E) + \sum_j \mu(E_j \setminus E) = \sum_j \mu(E_j) < \mu^*(A) + \varepsilon,$$ using additivity of $\mu$ on $\mathcal{R}$. Let $\varepsilon \to 0$: $E \in \mathcal{M}_{\mu^*}$. Since $\mathcal{M}_{\mu^*}$ is a σ-ring containing $\mathcal{R}$, it contains $\mathcal{S}(\mathcal{R})$. Finally, $\mu^* = \mu$ on $\mathcal{R}$ is the exercise above. $\blacksquare$
That's existence: the restriction of $\mu^*$ to $\mathcal{S}(\mathcal{R})$ is a measure that extends $\mu$. For length, it extends $\ell$ to a measure on every set in $\mathcal{S}(\mathcal{R}(\mathcal{I}))$, and (spoiler) that's every Borel set. Before uniqueness, run the test yourself.
Loading practice engine...
Is this the only way to extend $\mu$? Not always. Uniqueness needs one more hypothesis, the one that rules out sets of infinite measure with nothing finite inside them.
A measure $\mu$ on a ring $\mathcal{R}$ is σ-finite if for every $E \in \mathcal{R}$ there are $E_1, E_2, \ldots \in \mathcal{R}$ with $E \subseteq \bigcup_{j=1}^{\infty} E_j$ and $\mu(E_j) < \infty$ for every $j$.
Most measures you'd write down pass. Here are two that do and one that doesn't.
Which are σ-finite? (a) The length measure $\mu$ on $\mathcal{R}(\mathcal{I})$. (b) Counting measure on the ring of finite subsets of $\mathbb{R}$. (c) On the ring of all subsets of $\{1, 2\}$, $\mu(\emptyset) = 0$, $\mu(\{1\}) = 1$, and $\mu(E) = \infty$ when $2 \in E$.
(a) Yes, for the dullest reason: every set in the ring already has finite length, so it covers itself.
(b) Yes, the same way: a finite set has finite counting measure.
(c) No. (It is a measure: the value is the sum of the point weights $1$ and $\infty$.) Any cover of $\{2\}$ has a set containing $2$, and that set has measure $\infty$. A point of infinite weight can't be cut down to finite pieces.
So σ-finite means every set can be cut into countably many pieces of finite measure. Now the payoff for the monotone class section.
Let $\mu$ be a σ-finite measure on a ring $\mathcal{R}$. If $\nu_1$ and $\nu_2$ are measures on $\mathcal{S}(\mathcal{R})$ with $\nu_1 = \nu_2 = \mu$ on $\mathcal{R}$, then $\nu_1 = \nu_2$. In particular, the restriction of $\mu^*$ to $\mathcal{S}(\mathcal{R})$ is the only measure on $\mathcal{S}(\mathcal{R})$ extending $\mu$.
Proof. Step 1: inside a set of finite measure. Fix $F \in \mathcal{R}$ with $\mu(F) < \infty$, and let $$\mathcal{K}_F = \{E \in \mathcal{S}(\mathcal{R}) : \nu_1(E \cap F) = \nu_2(E \cap F)\}.$$ It contains $\mathcal{R}$, because $E \cap F \in \mathcal{R}$ when $E \in \mathcal{R}$. It is a monotone class: if $E_n \uparrow E$ in $\mathcal{K}_F$, then $E_n \cap F \uparrow E \cap F$, and continuity from below gives $\nu_1(E \cap F) = \lim \nu_1(E_n \cap F) = \lim \nu_2(E_n \cap F) = \nu_2(E \cap F)$. If $E_n \downarrow E$, then $E_n \cap F \downarrow E \cap F$, and continuity from above applies because $\nu_i(E_1 \cap F) \le \nu_i(F) = \mu(F) < \infty$. By the monotone class theorem, $\mathcal{K}_F \supseteq \mathcal{K}(\mathcal{R}) = \mathcal{S}(\mathcal{R})$.
Step 2: every set has a countable cover of finite measure. Let $\mathcal{C}$ be the collection of $E \in \mathcal{S}(\mathcal{R})$ that lie in a countable union of sets of $\mathcal{R}$ of finite measure. It contains $\mathcal{R}$ (that's σ-finiteness), it contains $E \setminus F$ whenever it contains $E$, and a countable union of countable covers is a countable cover. So $\mathcal{C}$ is a σ-ring containing $\mathcal{R}$, and $\mathcal{C} = \mathcal{S}(\mathcal{R})$.
Step 3. Let $E \in \mathcal{S}(\mathcal{R})$ and $E \subseteq \bigcup_j F_j$ with $F_j \in \mathcal{R}$, $\mu(F_j) < \infty$. The sets $G_j = F_j \setminus (F_1 \cup \cdots \cup F_{j-1})$ are in $\mathcal{R}$, have finite measure, are pairwise disjoint, and $E = \bigcup_j (E \cap G_j)$. By Step 1, $$\nu_1(E) = \sum_j \nu_1(E \cap G_j) = \sum_j \nu_2(E \cap G_j) = \nu_2(E).$$ The last sentence follows because the restriction of $\mu^*$ to $\mathcal{S}(\mathcal{R})$ is one such extension, by Carathéodory's theorem. $\blacksquare$
The σ-finite hypothesis isn't decoration. Without it, two different measures can agree on the whole ring.
On $X = \mathbb{Q}$, let $\mathcal{R}$ be the ring of finite unions of sets $[a, b) \cap \mathbb{Q}$, and let $\mu(E) = \infty$ for every nonempty $E \in \mathcal{R}$ (and $\mu(\emptyset) = 0$). Find two different measures on $\mathcal{S}(\mathcal{R})$ that extend $\mu$.
$\mu$ is a measure on $\mathcal{R}$: if a disjoint union of ring sets is a nonempty ring set, one of the pieces is nonempty, and both sides are $\infty$. It is not σ-finite, since the only set of finite measure is $\emptyset$.
$\mathcal{S}(\mathcal{R})$ contains each $\{q\} = \bigcap_n \left([q, q + \tfrac{1}{n}) \cap \mathbb{Q}\right)$, and every subset of $\mathbb{Q}$ is a countable union of these, so $\mathcal{S}(\mathcal{R}) = \mathcal{P}(\mathbb{Q})$. Counting measure $c$ extends $\mu$, because every nonempty set in $\mathcal{R}$ is infinite. So does $\nu(E) = \infty$ for every nonempty $E$. They disagree on $\{0\}$: $c(\{0\}) = 1$, while $\nu(\{0\}) = \infty$. (Carathéodory picks $\nu$: here $\mu^*(\{0\}) = \infty$.)
Back to the real line, where everything is σ-finite and the story ends cleanly.
$\mathcal{S}(\mathcal{R}(\mathcal{I})) = \mathcal{B}(\mathbb{R})$. The restriction of the Lebesgue outer measure to $\mathcal{B}(\mathbb{R})$ is a measure with $\mu^*([a, b)) = b - a$ for all $a \le b$, and it is the only measure on $\mathcal{B}(\mathbb{R})$ with that property.
Proof. Any σ-ring containing $\mathcal{I}$ contains its finite unions, so $\mathcal{S}(\mathcal{R}(\mathcal{I})) = \mathcal{S}(\mathcal{I})$. It contains $\mathbb{R} = \bigcup_n [-n, n)$, so it is a σ-algebra, and then it is the σ-algebra $\sigma(\mathcal{I})$ generated by $\mathcal{I}$ (each is a σ-ring containing $\mathcal{I}$, and each is the smallest of its kind).
Each $[a, b) = \bigcap_n (a - \tfrac{1}{n}, b)$ is a countable intersection of open sets, so it is Borel, and $\sigma(\mathcal{I}) \subseteq \mathcal{B}(\mathbb{R})$. Conversely, let $U$ be open. Each $x \in U$ has some $(x - r, x + r) \subseteq U$, and then rationals $p \le x < q$ with $[p, q) \subseteq (x - r, x + r)$. So $U$ is the union of the intervals $[p, q) \subseteq U$ with rational endpoints, a countable union. Every open set is in $\sigma(\mathcal{I})$, so $\mathcal{B}(\mathbb{R}) \subseteq \sigma(\mathcal{I})$.
By Carathéodory's theorem, $\mathcal{B}(\mathbb{R}) = \mathcal{S}(\mathcal{R}(\mathcal{I})) \subseteq \mathcal{M}_{\mu^*}$, so $\mu^*$ restricted to $\mathcal{B}(\mathbb{R})$ is a measure, and $\mu^*([a, b)) = \mu([a, b)) = b - a$. If $\lambda$ is another measure on $\mathcal{B}(\mathbb{R})$ with $\lambda([a, b)) = b - a$, then $\lambda$ agrees with $\mu$ on $\mathcal{R}(\mathcal{I})$ by finite additivity, $\mu$ is σ-finite, and the uniqueness theorem gives $\lambda = \mu^*$ on $\mathcal{B}(\mathbb{R})$. $\blacksquare$
So "length on intervals" pins down a measure on every Borel set, with no choices left. Carathéodory also hands you more than the Borel sets: $\mathcal{M}_{\mu^*}$ is complete, so it contains every subset of a null Borel set, like every subset of the Cantor set. Those are the Lebesgue measurable sets.
Lebesgue Measure builds a measure $m$ from the Riesz theorem with $m([a, b)) = b - a$. By the corollary, $m$ and the restriction of $\mu^*$ are the same measure on $\mathcal{B}(\mathbb{R})$: two roads, one destination.
Last, the proofs on this page in pieces: each question below shows part of one with a step missing.
Loading practice engine...
Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: a claim about anything on this page, some of them the kind that sounds right and isn't. Decide, then read why.
Loading practice engine...
Next, the mixed set. Every question type on this page except true or false goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; the types whose choices are sets or proof lines need their answer choices, and they'll say so.
Loading practice engine...
Last, the Spicy Bonus. Carathéodory's σ-ring can be much bigger than $\mathcal{S}(\mathcal{R})$, but not in any interesting way: at least for sets of finite outer measure, every $\mu^*$-measurable set is a set from $\mathcal{S}(\mathcal{R})$ with a null set taken away.
Let $\mu$ be a measure on a ring $\mathcal{R}$, and let $E \in \mathcal{H}(\mathcal{R})$ with $\mu^*(E) < \infty$. (a) Show that there is $F \in \mathcal{S}(\mathcal{R})$ with $E \subseteq F$ and $\mu^*(F) = \mu^*(E)$. (b) Prove that $E$ is $\mu^*$-measurable if and only if $E = F \setminus N$ for some $F \in \mathcal{S}(\mathcal{R})$ and some $N \subseteq F$ with $\mu^*(N) = 0$.
Proof. (a) For each $n$, choose $E_{n, 1}, E_{n, 2}, \ldots \in \mathcal{R}$ covering $E$ with $\sum_j \mu(E_{n, j}) < \mu^*(E) + \tfrac{1}{n}$, and let $F_n = \bigcup_j E_{n, j} \in \mathcal{S}(\mathcal{R})$. Then $E \subseteq F_n$, and the $E_{n, j}$ are a cover of $F_n$, so $\mu^*(F_n) \le \sum_j \mu(E_{n, j}) < \mu^*(E) + \tfrac{1}{n}$. Let $F = \bigcap_n F_n$, which is in $\mathcal{S}(\mathcal{R})$ because a σ-ring is closed under countable intersections. Then $E \subseteq F \subseteq F_n$, so $$\mu^*(E) \le \mu^*(F) \le \mu^*(F_n) < \mu^*(E) + \tfrac{1}{n}$$ for every $n$, and $\mu^*(F) = \mu^*(E)$.
(b) ($\implies$) Suppose $E \in \mathcal{M}_{\mu^*}$, and take $F$ from (a). By Carathéodory's theorem $F \in \mathcal{S}(\mathcal{R}) \subseteq \mathcal{M}_{\mu^*}$, so $N = F \setminus E \in \mathcal{M}_{\mu^*}$, and $\mu^*$ is additive on $\mathcal{M}_{\mu^*}$: $$\mu^*(F) = \mu^*(E) + \mu^*(N).$$ Since $\mu^*(F) = \mu^*(E) < \infty$, we can subtract: $\mu^*(N) = 0$. And $E = F \setminus N$, because $E \subseteq F$.
($\impliedby$) Suppose $E = F \setminus N$ with $F \in \mathcal{S}(\mathcal{R})$, $N \subseteq F$, and $\mu^*(N) = 0$. Then $N \in \mathcal{H}(\mathcal{R})$ (it is a subset of $F$, and $\mathcal{S}(\mathcal{R}) \subseteq \mathcal{H}(\mathcal{R})$ since $\mathcal{H}(\mathcal{R})$ is a σ-ring containing $\mathcal{R}$), so $N \in \mathcal{M}_{\mu^*}$ by completeness. Also $F \in \mathcal{M}_{\mu^*}$. A ring is closed under differences, so $E = F \setminus N \in \mathcal{M}_{\mu^*}$.
So, for sets of finite outer measure, $\mathcal{M}_{\mu^*}$ is exactly the completion of $\mathcal{S}(\mathcal{R})$ under $\mu^*$: nothing new beyond sets from $\mathcal{S}(\mathcal{R})$ and null sets. $\blacksquare$