A (concrete) Boolean algebra is a pair $(X, {\mathcal B})$, where X is a set, and ${\mathcal B}$ is a collection of subsets of X which contain the empty set $\emptyset$, and which is closed under unions $A, B \mapsto A \cup B$, intersections $A, B \mapsto A \cap B$, and complements $A \mapsto A^c := X \backslash A$. The subset relation $\subset$ also gives a relation on ${\mathcal B}$. Because the ${\mathcal B}$ is concretely represented as subsets of a space X, these relations automatically obey various axioms, in particular, for any $A,B,C \in {\mathcal B}$, we have:

1. $\subset$ is a partial ordering on ${\mathcal B}$, and A and B have join $A \cup B$ and meet $A \cap B$.
2. We have the distributive laws $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ and $A \cap (B \cup C) = A \cup (B \cap C)$.
3. $\emptyset$ is the minimal element of the partial ordering $\subset$, and $\emptyset^c$ is the maximal element.
4. $A \cap A^c = \emptyset$ and $A \cup A^c = \emptyset^c$.

(More succinctly: ${\mathcal B}$ is a lattice which is distributive, bounded, and complemented.)

We can then define an abstract Boolean algebra ${\mathcal B} = ({\mathcal B}, \emptyset, \cdot^c, \cup, \cap, \subset)$ to be an abstract set ${\mathcal B}$ with the specified objects, operations, and relations that obey the axioms 1-4. [Of course, some of these operations are redundant; for instance, intersection can be defined in terms of complement and union by de Morgan’s laws. In the literature, different authors select different initial operations and axioms when defining an abstract Boolean algebra, but they are all easily seen to be equivalent to each other. To emphasise the abstract nature of these algebras, the symbols $\emptyset, \cdot^c, \cup, \cap, \subset$ are often replaced with other symbols such as $0, \overline{\cdot}, \vee, \wedge, <$.]

Clearly, every concrete Boolean algebra is an abstract Boolean algebra. In the converse direction, we have Stone’s representation theorem (see below), which asserts (among other things) that every abstract Boolean algebra is isomorphic to a concrete one (and even constructs this concrete representation of the abstract Boolean algebra canonically). So, up to (abstract) isomorphism, there is really no difference between a concrete Boolean algebra and an abstract one.

Now let us turn from Boolean algebras to $\sigma$-algebras.

A concrete $\sigma$-algebra (also known as a measurable space) is a pair $(X,{\mathcal B})$, where X is a set, and ${\mathcal B}$ is a collection of subsets of X which contains $\emptyset$ and are closed under countable unions, countable intersections, and complements; thus every concrete $\sigma$-algebra is a concrete Boolean algebra, but not conversely. As before, concrete $\sigma$-algebras come equipped with the structures $\emptyset, \cdot^c, \cup, \cap, \subset$ which obey axioms 1-4, but they also come with the operations of countable union $(A_n)_{n=1}^\infty \mapsto \bigcup_{n=1}^\infty A_n$ and countable intersection $(A_n)_{n=1}^\infty \mapsto \bigcap_{n=1}^\infty A_n$, which obey an additional axiom:

5. Any countable family $A_1, A_2, \ldots$ of elements of ${\mathcal B}$ has supremum $\bigcup_{n=1}^\infty A_n$ and infimum $\bigcap_{n=1}^\infty A_n$.

As with Boolean algebras, one can now define an abstract $\sigma$-algebra to be a set ${\mathcal B} = ({\mathcal B}, \emptyset, \cdot^c, \cup, \cap, \subset, \bigcup_{n=1}^\infty, \bigcap_{n=1}^\infty )$ with the indicated objects, operations, and relations, which obeys axioms 1-5. Again, every concrete $\sigma$-algebra is an abstract one; but is it still true that every abstract $\sigma$-algebra is representable as a concrete one?

The answer turns out to be no, but the obstruction can be described precisely (namely, one needs to quotient out an ideal of “null sets” from the concrete $\sigma$-algebra), and there is a satisfactory representation theorem, namely the Loomis-Sikorski representation theorem (see below). As a corollary of this representation theorem, one can also represent abstract measure spaces $({\mathcal B},\mu)$ (also known as measure algebras) by concrete measure spaces, $(X, {\mathcal B}, \mu)$, after quotienting out by null sets.

In the rest of this post, I will state and prove these representation theorems. They are not actually used directly in the rest of the course (and they will also require some results that we haven’t proven yet, most notably Tychonoff’s theorem), and so these notes are optional reading; but these theorems do help explain why it is “safe” to focus attention primarily on concrete $\sigma$-algebras and measure spaces when doing measure theory, since the abstract analogues of these mathematical concepts are largely equivalent to their concrete counterparts. (The situation is quite different for non-commutative measure theories, such as quantum probability, in which there is basically no good representation theorem available to equate the abstract with the classically concrete, but I will not discuss these theories here.)

— Stone’s representation theorem —

We first give the class of Boolean algebras the structure of a category:

Definition 1. (Boolean algebra morphism) A morphism $\phi: {\mathcal A} \to {\mathcal B}$ from one abstract Boolean algebra to another is a map which preserves the empty set, complements, unions, intersections, and the subset relation (e.g. $\phi(A \cup B) = \phi(A) \cup \phi(B)$ for all $A, B \in {\mathcal A}$. An isomorphism is an morphism $\phi: {\mathcal A} \to {\mathcal B}$ which has an inverse morphism $\phi^{-1}: {\mathcal B} \to {\mathcal A}$. Two Boolean algebras are isomorphic if there is an isomorphism between them.

Note that if $(X,{\mathcal A}), (Y, {\mathcal B})$ are concrete Boolean algebras, and if $f: X \to Y$ is a map which is measurable in the sense that $f^{-1}(B) \in {\mathcal A}$ for all $B \in {\mathcal B}$, then the inverse of f is a Boolean algebra morphism ${}f^{-1}: {\mathcal B} \to {\mathcal A}$ which goes in the reverse (i.e. contravariant) direction to that of f. To state Stone’s representation theorem we need another definition.

Definition 2. (Stone space) A Stone space is a topological space $X = (X, {\mathcal F})$ which is compact, Hausdorff, and totally disconnected. Given a Stone space, define the clopen algebra ${}Cl(X)$ of X to be the concrete Boolean algebra on X consisting of the clopen sets (i.e. sets that are both closed and open).

It is easy to see that $Cl(X)$ is indeed a concrete Boolean algebra for any topological space X. The additional properties of being compact, Hausdorff, and totally disconnected are needed in order to recover the topology ${\mathcal F}$ of X uniquely from the clopen algebra. Indeed, we have

Lemma 1. If X is a Stone space, then the topology ${\mathcal F}$ of X is generated by the clopen algebra $Cl(X)$. Equivalently, the clopen algebra forms an open base for the topology.

Proof. Let $x \in X$ be a point, and let $K$ be the intersection of all the clopen sets containing x. Clearly, K is closed. We claim that K={x}. If this is not the case, then (since X is totally disconnected) K must be disconnected, thus K can be separated non-trivially into two closed sets $K = K_1 \cup K_2$. Since compact Hausdorff spaces are normal, we can write $K_1 = K \cap U_1$ and $K_2 = K \cap U_2$ for some disjoint open $U_1, U_2$. Since the intersection of all the clopen sets containing x with the closed set $(U_1 \cup U_2)^c$ is empty, we see from the finite intersection property that there must be a finite intersection K’ of clopen sets containing x that is contained inside $U_1 \cup U_2$. In particular, ${}K' \cap U_1$ and $K' \cap U_2$ are clopen and do not contain K. But this contradicts the definition of K (since x is contained in one of ${}K' \cap U_1$ and $K' \cap U_2$). Thus K = {x}.

Another application of the finite intersection property then reveals that every open neighbourhood of x contains at least one clopen set containing x, and so the clopen sets form a base as required. $\Box$

Exercise 1. Show that two Stone spaces have isomorphic clopen algebras if and only if they are homeomorphic. $\diamond$

Now we turn to the representation theorem.

Theorem 1. (Stone representation theorem) Every abstract Boolean algebra ${\mathcal B}$ is equivalent to the clopen algebra $Cl(X)$ of a Stone space X.

Proof. We will need the binary abstract Boolean algebra $\{0,1\}$, with the usual Boolean logic operations. We define $X := \hbox{Hom}({\mathcal B}, \{0,1\})$ be the space of all morphisms from ${\mathcal B}$ to \{0,1\}. Observe that each point $x \in X$ can be viewed as a finitely additive measure $\mu_x: {\mathcal B} \to \{0,1\}$ that takes values in $\{0,1\}$. In particular, this makes X a closed subset of $\{0,1\}^{\mathcal B}$ (endowed with the product topology). The space $\{0,1\}^{\mathcal B}$ is Hausdorff, totally disconnected, and (by Tychonoff’s theorem) compact, and so X is also; in other words, X is a Stone space. Every $B \in {\mathcal B}$ induces a cylinder set $C_B \subset \{0,1\}^{\mathcal B}$, consisting of all maps $\mu: {\mathcal B} \to \{0,1\}$ that map B to 1. If we define $\phi(B) := C_B \cap X$, it is not hard to see that $\phi$ is a morphism from ${\mathcal B}$ to $Cl(X)$. Since the cylinder sets are clopen and generate the topology of $\{0,1\}^{\mathcal B}$, we see that $\phi({\mathcal B})$ of clopen sets generates the topology of X. Using compactness, we then conclude that every clopen set is the finite union of finite intersections of elements of $\phi({\mathcal B})$; since $\phi({\mathcal B})$ is an algebra, we thus see that $\phi$ is surjective.

The only remaining thing to check is that $\phi$ is injective. It is sufficient to show that $\phi(A)$ is non-empty whenever $A \in {\mathcal B}$ is not equal to $\emptyset$. But by Zorn’s lemma, we can place A inside a maximal proper filter (i.e. an ultrafilter) p. The indictator $1_p: {\mathcal B} \to \{0,1\}$ of p can then be verified to be an element of $\phi(A)$, and the claim follows. $\Box$

Remark 1. If ${\mathcal B} = 2^Y$ is the power set of some set Y, then the Stone space given by Theorem 1 is the Stone-Čech compactification of Y (which we give the discrete topology). $\diamond$

Remark 2. Lemma 1 and Theorem 1 can be interpreted as giving a duality between the category of Boolean algebras and the category of Stone spaces, with the duality maps being ${\mathcal B} \mapsto \hbox{Hom}({\mathcal B},\{0,1\})$ and $X \mapsto Cl(X)$. (The duality maps are (contravariant) functors which are inverses up to natural transformations.) It is the model example of the more general Stone duality between certain partially ordered sets and certain topological spaces. The idea of dualising a space X by considering the space of its morphisms to a fundamental space (in this case, $\{0,1\}$) is a common one in mathematics; for instance, Pontryagin duality in the context of Fourier analysis on locally compact abelian groups provides another example (with the fundamental space in this case being the unit circle ${\Bbb R}/{\Bbb Z}$). Other examples include the Gelfand representation in $C^*$ algebras (here the fundamental space is the complex numbers ${\Bbb C}$) and the ideal-variety correspondence that provides the duality between algebraic geometry and commutative algebra (here the fundamental space is the base field $k$). In fact there are various connections between all of the dualities mentioned above. $\diamond$

Exercise 2. Show that any finite Boolean algebra is isomorphic to the power set of a finite set. (This is a special case of Birkhoff’s representation theorem.) $\diamond$

— The Loomis-Sikorski representation theorem —

Now we turn to abstract $\sigma$-algebras. We can of course adapt Definition 1 to define the notion of a morphism or isomorphism between abstract $\sigma$-algebras, and to define when two abstract $\sigma$-algebras are isomorphic. Another important notion for us will be that of a quotient $\sigma$-algebra.

Definition 3. (Quotient $\sigma$-algebras) Let ${\mathcal B}$ be an abstract $\sigma$-algebra. A $\sigma$-ideal in ${\mathcal B}$ is a \subset ${\mathcal N}$ of ${\mathcal B}$ which contains $\emptyset$, is closed under countable unions, and is downwardly closed (thus if $N \in {\mathcal N}$ and $A \in {\mathcal B}$ is such that $A \subset N$, then $A \in {\mathcal N}$). If ${\mathcal N}$ is a $\sigma$-ideal, then we say that two elements of ${\mathcal B}$ are equivalent modulo ${\mathcal N}$ if their symmetric difference lies in ${\mathcal N}$. The quotient of ${\mathcal B}$ by this equivalence relation is denoted ${\mathcal B}/{\mathcal N}$, and can be given the structure of an abstract $\sigma$-algebra in a straightforward manner.

Example 1. If $(X, {\mathcal B}, \mu)$ is a measure space, then the collection ${\mathcal N}$ of sets of measure zero is a $\sigma$-ideal, so that we can form the abstract $\sigma$-algebra ${\mathcal B}/{\mathcal N}$. This freedom to quotient out the null sets is only available in the abstract setting, not the concrete one, and is perhaps the primary motivation for introducing abstract $\sigma$-algebras into measure theory in the first place. $\diamond$

One might hope that there is an analogue of Stone’s representation theorem holds for $\sigma$-algebras. Unfortunately, this is not the case:

Proposition 1. Let ${\mathcal B}$ be the Borel $\sigma$-algebra on [0,1], and let ${\mathcal N}$ be the $\sigma$-ideal consisting of those sets with Lebesgue measure zero. Then the abstract $\sigma$-algebra ${\mathcal B}/{\mathcal N}$ is not isomorphic to a concrete $\sigma$-algebra.

Proof. Suppose for contradiction that we had an isomorphism $\phi: {\mathcal B}/{\mathcal N} \to {\mathcal A}$ to some concrete $\sigma$-algebra $(X,{\mathcal A})$; this induces a map $\phi: {\mathcal B} \to {\mathcal A}$ which sends null sets to the empty set. Let x be a point in X. (It is clear that X must be non-empty.) Observe that any Borel set E in [0,1] can be partitioned into two Borel subsets whose Lebesgue measure is exactly half that of E. As a consequence, we see that if there exists a Borel set B such that $\phi(B)$ contains x, then there exists another Borel set B’ of half the measure with $\phi(B')$ contains x. Iterating this (starting with [0,1]) we see that there exist Borel sets B of arbitrarily small measure with $\phi(B)$ containing x. Taking countable intersections, we conclude that there exists a null set N whose image $\phi(N)$ contains x; but $\phi(N)$ is empty, a contradiction. $\Box$

However, it turns out that quotienting out by ideals is the only obstruction to having a Stone-type representation theorem. Namely, we have

Theorem 2. (LoomisSikorski representation theorem) Let ${\mathcal B}$ be an abstract $\sigma$-algebra. Then there exists a concrete $\sigma$-algebra $(X,{\mathcal A})$ and a $\sigma$-ideal ${\mathcal N}$ of ${\mathcal A}$ such that ${\mathcal B}$ is isomorphic to ${\mathcal A}/{\mathcal N}$.

Proof. We use the argument of Loomis. Applying Stone’s representation theorem, we can find a Stone space X such that there is a Boolean algebra isomorphism $\phi: {\mathcal B} \to Cl(X)$ from ${\mathcal B}$ (viewed now only as a Boolean algebra rather than a $\sigma$-algebra to the clopen algebra of X. Let ${\mathcal A}$ be the Baire $\sigma$-algebra of X, i.e. the $\sigma$-algebra generated by $Cl(X)$. The map $\phi$ need not be a $\sigma$-algebra isomorphism, being merely a Boolean algebra isomorphism one instead; it preserves finite unions and intersections, but need not preserve countable ones. In particular, if $B_1, B_2, \ldots \in {\mathcal B}$ are such that $\bigcap_{n=1}^\infty B_n = \emptyset$, then ${}\bigcap_{n=1}^\infty \phi(B_n) \in {\mathcal A}$ need not be empty.

Let us call sets $\bigcap_{n=1}^\infty \phi(B_n)$ of this form basic null sets, and let ${\mathcal N}$ be the collection of sets in ${\mathcal A}$ which can be covered by at most countably many basic null sets.

It is not hard to see that ${\mathcal N}$ is a $\sigma$-ideal in ${\mathcal A}$. The map $\phi$ then descends to a map $\phi: {\mathcal B} \to {\mathcal A}/{\mathcal N}$. It is not hard to see that $\phi$ is a Boolean algebra morphism. Also, if $B_1,B_2,\ldots \in {\mathcal B}$ are such that $\bigcap_{n=1}^\infty B_n=\emptyset$, then from construction we have $\bigcap_{n=1}^\infty \phi(B_n) = \emptyset$. From these two facts one can easily show that $\phi$ is in fact a $\sigma$-algebra morphism. Since $\phi({\mathcal B}) = Cl(X)$ generates ${\mathcal A}$, $\phi({\mathcal B})$ must generate ${\mathcal A}/{\mathcal N}$, and so $\phi$ is surjective.

The only remaining task is to show that $\phi$ is injective. As before, it suffices to show that $\phi(A) \neq \emptyset$ when $A \neq \emptyset$. Suppose for contradiction that $A \neq \emptyset$ and $\phi(A) = \emptyset$; then $\phi(A)$ can be covered by a countable family $\bigcap_{n=1}^\infty \phi(A_n^{(i)})$ of basic null sets, where $\bigcap_{n=1}^\infty A_n^{(i)} = \emptyset$ for each i. Since $A \neq \emptyset$ and $\bigcap_{n=1}^\infty A_n^{(1)} = \emptyset$, we can find $n_1$ such that $A \backslash A_{n_1}^{(1)} \neq \emptyset$ (where of course $A \backslash B := A \cap B^c$). Iterating this, we can find $n_2, n_3, n_4,\ldots$ such that $A \backslash (A_{n_1}^{(1)} \cup \ldots \cup A_{n_k}^{(k)}) \neq \emptyset$ for all k. Since $\phi$ is a Boolean space isomorphism, we conclude that $\phi(A)$ is not covered by any finite subcollection of the $\phi( A_{n_1}^{(1)} ), \phi( A_{n_2}^{(2)} ), \ldots$. But all of these sets are clopen, so by compactness, $\phi(A)$ is not covered by the entire collection $\phi( A_{n_1}^{(1)} ), \phi( A_{n_2}^{(2)} ), \ldots$. But this contradicts the fact that $\phi(A)$ is covered by the $\bigcap_{n=1}^\infty \phi(A_n^{(i)})$. $\Box$

Remark 3. The Stone representation theorem relies in an essential way on the axiom of choice (or at least the boolean prime ideal theorem, which is slightly weaker than this axiom). However, it is possible to prove the Loomis-Sikorski representation theorem without choice; see for instance this paper. $\diamond$

Remark 4. The construction of $X, {\mathcal A}, {\mathcal N}$ in the above proof was canonical, but it is not unique (in contrast to the situation with the Stone representation theorem, where Lemma 1 provides uniqueness up to homeomorphisms). In particular it does not seem to provide a duality similar to Stone duality. [Removed, Jan 13, in view of comments: the functoriality of the construction can be used to provide a Stone-type duality.] $\diamond$

Remark 5. The proof above actually gives a little bit more structure on $X, {\mathcal A}$, namely it gives X the structure of a Stone space, with ${\mathcal A}$ being its Baire $\sigma$-algebra. [Added, Jan 13: Furthermore, the ideal ${\mathcal N}$ constructed in the proof is in fact the ideal of meager Baire sets; see comments.] $\diamond$

A (concrete) measure space $(X,{\mathcal B},\mu)$ is a concrete $\sigma$-algebra $(X,{\mathcal B})$ together with a countably additive measure $\mu: {\mathcal B} \to [0,+\infty]$. One can similarly define an abstract measure space $({\mathcal B},\mu)$ (or measure algebra) to be an abstract $\sigma$-algebra ${\mathcal B}$ with a countably additive measure $mu: {\mathcal B} \to [0,+\infty]$. (Note that one does not need the concrete space $X$ in order to define the notion of a countably additive measure.)

One can obtain an abstract measure space from a concrete one by deleting X and then quotienting out by some $\sigma$-ideal of null sets – sets of measure zero with respect to $mu$. (For instance, one could quotient out the space of all null sets, which is automatically a $\sigma$-ideal.) Thanks to the Loomis-Sikorski representation theorem, we have a converse:

Exercise 3. Show that every abstract measure space is isomorphic to a concrete measure space after quotieting out by a $\sigma$-ideal of null sets (where the notion of morphism, isomorphism, etc. on abstract measure spaces is defined in the obvious manner.) $\diamond$

[Update, Jan 13: some corrections, and formatting fixes.]