
7.3: More on Set Families


Lebesgue extended his theory far beyond $$\mathcal{C}_{\sigma}$$-sets. For a deeper insight, we shall consider set families in more detail, starting with set rings. First, we rephrase and supplement our former definition of that notion, given in §1.

Definition 1

A family $$\mathcal{M}$$ of subsets of a set $$S$$ is a ring or set ring (in $$S)$$ iff

(i) $$\emptyset \in \mathcal{M},$$ i.e., the empty set is a member; and

(ii) $$\mathcal{M}$$ is closed under finite unions and differences:

$(\forall X, Y \in \mathcal{M}) \quad X \cup Y \in \mathcal{M} \text { and } X-Y \in \mathcal{M}.$

(For intersections, see Theorem 1 below.)

If $$\mathcal{M}$$ is also closed under countable unions, we call it a $$\sigma$$-ring (in $$S).$$ Then

$\bigcup_{i=1}^{\infty} X_{i} \in \mathcal{M}$

whenever

$X_{i} \in \mathcal{M} \text { for } i=1,2, \ldots.$

If $$S$$ itself is a member of a ring ($$\sigma$$-ring) $$\mathcal{M},$$ we call $$\mathcal{M}$$ a set field ($$\sigma$$-field), or a set algebra ($$\sigma$$-algebra), in $$S$$.

Note that $$S$$ is only a member of $$\mathcal{M}, S \in \mathcal{M},$$ not to be confused with $$\mathcal{M}$$ itself.

The family of all subsets of $$S$$ (the so-called power set of $$S$$) is denoted by $$2^{S}$$ or $$\mathcal{P}(S).$$

Examples

(a) In any set $$S, 2^{S}$$ is a $$\sigma$$-field. (Why?)

(b) The family $$\{\emptyset\},$$ consisting of $$\emptyset$$ alone, is a $$\sigma$$-ring; $$\{\emptyset, S\}$$ is a $$\sigma$$-field in $$S.$$ (Why?)

(c) The family of all finite (countable) subsets of $$S$$ is a ring ($$\sigma$$-ring) in $$S$$.

(d) For any semiring $$\mathcal{C}, \mathcal{C}_{s}^{\prime}$$ is a ring (Theorem 2 in §1). Not so for $$\mathcal{C}_{\sigma}$$ (Problem 5 in §2).

Theorem $$\PageIndex{1}$$

Any set ring is closed under finite intersections.

A $$\sigma$$-ring is closed under countable intersections.

Proof

Let $$\mathcal{M}$$ be a $$\sigma$$-ring (the proof for rings is similar).

Given a sequence $$\left\{A_{n}\right\} \subseteq \mathcal{M},$$ we must show that $$\bigcap_{n} A_{n} \in \mathcal{M}$$.

Let

$U=\bigcup_{n} A_{n}.$

By Definition 1,

$U \in \mathcal{M} \text { and } U-A_{n} \in \mathcal{M},$

as $$\mathcal{M}$$ is closed under these operations. Hence

$\bigcup_{n}\left(U-A_{n}\right) \in \mathcal{M}$

and

$U-\bigcup_{n}\left(U-A_{n}\right) \in \mathcal{M},$

or, by duality,

$\bigcap_{n}\left[U-\left(U-A_{n}\right)\right] \in \mathcal{M},$

i.e.,

$\bigcap_{n} A_{n} \in \mathcal{M}. \quad \square$

Corollary $$\PageIndex{1}$$

Any set ring (field, $$\sigma$$-ring, $$\sigma$$-field) is also a semiring.

Indeed, by Theorem 1 and Definition 1, if $$\mathcal{M}$$ is a ring, then $$\emptyset \in \mathcal{M}$$ and

$(\forall A, B \in \mathcal{M}) \quad A \cap B \in \mathcal{M} \text { and } A-B \in \mathcal{M}.$

Here we may treat $$A-B$$ as $$(A-B) \cup \emptyset,$$ a union of two disjoint $$\mathcal{M}$$-sets. Thus $$\mathcal{M}$$ has all properties of a semiring.

Similarly for $$\sigma$$-rings, fields, etc.

In §1 we saw that any semiring $$\mathcal{C}$$ can be enlarged to become a ring, $$\mathcal{C}_{s}^{\prime}.$$ More generally, we obtain the following result.

Theorem $$\PageIndex{2}$$

For any set family $$\mathcal{M}$$ in a space $$S\left(\mathcal{M} \subseteq 2^{S}\right),$$ there is a unique "smallest" set ring $$\mathcal{R}$$ such that

$\mathcal{R} \supseteq \mathcal{M}$

("smallest" in the sense that

$\mathcal{R} \subseteq \mathcal{R}^{\prime}$

for any other ring $$\mathcal{R}^{\prime}$$ with $$\mathcal{R}^{\prime} \supseteq \mathcal{M}$$).

The $$\mathcal{R}$$ of Theorem 2 is called the ring generated by $$\mathcal{M}.$$ Similarly for $$\sigma$$-rings, fields, and $$\sigma$$-fields in $$S$$.

Proof

We give the proof for $$\sigma$$-fields; it is similar in the other cases.

There surely are $$\sigma$$-fields in $$S$$ that contain $$\mathcal{M};$$ e.g., take $$2^{S}.$$ Let $$\left\{\mathcal{R}_{i}\right\}$$ be the family of all possible $$\sigma$$-fields in $$S$$ such that $$\mathcal{R}_{i} \supseteq \mathcal{M}.$$ Let

$\mathcal{R}=\bigcap_{i} \mathcal{R}_{i}.$

We shall show that this $$\mathcal{R}$$ is the required "smallest" $$\sigma$$-field containing $$\mathcal{M}$$.

Indeed, by assumption,

$\mathcal{M} \subseteq \bigcap_{i} \mathcal{R}_{i}=\mathcal{R}.$

We now verify the $$\sigma$$-field properties for $$\mathcal{R}$$.

(1) We have that

$(\forall i) \quad \emptyset \in \mathcal{R}_{i} \text { and } S \in \mathcal{R}_{i}$

(for $$\mathcal{R}_{i}$$ is a $$\sigma$$-field, by assumption). Hence

$\emptyset \in \bigcap_{i} \mathcal{R}_{i}=\mathcal{R}.$

Similarly, $$S \in \mathcal{R}.$$ Thus

$\emptyset, S \in \mathcal{R}.$

(2) Suppose

$X, Y \in \mathcal{R}=\bigcap_{i} \mathcal{R}_{i}.$

Then $$X, Y$$ are in every $$\mathcal{R}_{i},$$ and so is $$X-Y.$$ Hence $$X-Y$$ is in

$\bigcap_{i} \mathcal{R}_{i}=\mathcal{R}.$

Thus $$\mathcal{R}$$ is closed under differences.

(3) Take any sequence

$\left\{A_{n}\right\} \subseteq \mathcal{R}=\bigcap_{i} \mathcal{R}_{i}.$

Then all $$A_{n}$$ are in each $$\mathcal{R}_{i}. \bigcup_{n} A_{n}$$ is in each $$\mathcal{R}_{i};$$ so

$\bigcup_{n} A_{n} \in \mathcal{R}.$

Thus $$\mathcal{R}$$ is closed under countable unions.

We see that $$\mathcal{R}$$ is indeed a $$\sigma$$-field in $$S,$$ with $$\mathcal{M} \subseteq \mathcal{R}.$$ As $$\mathcal{R}$$ is the intersection of all $$\mathcal{R}_{i}$$(i.e., all $$\sigma$$-fields $$\supseteq \mathcal{M}$$), we have

$(\forall i) \quad \mathcal{R} \subseteq \mathcal{R}_{i};$

so $$\mathcal{R}$$ is the smallest of such $$\sigma$$-fields.

It is unique; for if $$\mathcal{R}^{\prime}$$ is another such $$\sigma$$-field, then

$\mathcal{R} \subseteq \mathcal{R}^{\prime} \subseteq \mathcal{R}$

(as both $$\mathcal{R}$$ and $$\mathcal{R}^{\prime}$$ are "smallest"); so

$\mathcal{R}=\mathcal{R}^{\prime}. \quad \square$

Note 1. This proof also shows that the intersection of any family $$\left\{\mathcal{R}_{i}\right\}$$ of $$\sigma$$-fields is a $$\sigma$$-field. Similarly for $$\sigma$$-rings, fields, and rings.

Corollary $$\PageIndex{2}$$

The ring $$\mathcal{R}$$ generated by a semiring $$\mathcal{C}$$ coincides with

$\mathcal{C}_{s}=\{\text {all finite unions of } \mathcal{C}-\text {sets}\}$

and with

$\mathcal{C}_{s}^{\prime}=\{\text {disjoint finite unions of } \mathcal{C}-\text {sets}\}.$

Proof

By Theorem 2 in §1, $$\mathcal{C}_{s}^{\prime}$$ is a ring $$\supseteq \mathcal{C}$$; and

$\mathcal{C}_{s}^{\prime} \subseteq \mathcal{C}_{s} \subseteq \mathcal{R}$

(for $$\mathcal{R}$$ is closed under finite unions, being a ring $$\supseteq \mathcal{C}$$).

Moreover, as $$\mathcal{R}$$ is the smallest ring $$\supseteq \mathcal{C},$$ we have

$\mathcal{R} \subseteq \mathcal{C}_{s}^{\prime} \subseteq \mathcal{C}_{s} \subseteq \mathcal{R}.$

Hence

$\mathcal{R}=\mathcal{C}_{s}^{\prime}=\mathcal{C}_{s},$

as claimed.$$\quad \square$$

It is much harder to characterize the $$\sigma$$-ring generated by a semiring. The following characterization proves useful in theory and as an exercise.

Theorem $$\PageIndex{3}$$

The $$\sigma$$-ring $$\mathcal{R}$$ generated by a semiring $$\mathcal{C}$$ coincides with the smallest set family $$\mathcal{D}$$ such that

(i) $$\mathcal{D} \supseteq \mathcal{C}$$;

(ii) $$\mathcal{D}$$ is closed under countable disjoint unions;

(iii) $$J-X \in \mathcal{D}$$ whenever $$X \in \mathcal{D}, J \in \mathcal{C},$$ and $$X \subseteq J$$.

Proof

We give a proof outline, leaving the details to the reader.

(1) The existence of a smallest such $$\mathcal{D}$$ follows as in Theorem 2. Verify!

(2) Writing briefly $$A B$$ for $$A \cap B$$ and $$A^{\prime}$$ for $$-A,$$ prove that

$(A-B) C=A-\left(A C^{\prime} \cup B C\right).$

(3) For each $$I \in \mathcal{D},$$ set

$\mathcal{D}_{I}=\{A \in \mathcal{D} | A I \in \mathcal{D}, A-I \in \mathcal{D}\}.$

Then prove that if $$I \in \mathcal{C},$$ the set family $$\mathcal{D}_{I}$$ has the properties (i)-(iii) specified in the theorem. (Use the set identity (2) for property (iii).)

Hence by the minimality of $$\mathcal{D}, \mathcal{D} \subseteq \mathcal{D}_{I}.$$ Therefore,

$(\forall A \in \mathcal{D})(\forall I \in \mathcal{C}) \quad A I \in \mathcal{D} \text { and } A-I \in \mathcal{D}.$

(4) Using this, show that $$\mathcal{D}_{I}$$ satisfies (i)-(iii) for any$$I \in \mathcal{D}$$.

Deduce

$\mathcal{D} \subseteq \mathcal{D}_{I};$

so $$\mathcal{D}$$ is closed under finite intersections and differences.

Combining with property (ii), show that $$\mathcal{D}$$ is a $$\sigma$$-ring (see Problem 12 below).

By its minimality, $$\mathcal{D}$$ is the smallest $$\sigma$$-ring $$\supseteq \mathcal{C}$$ (for any other such $$\sigma$$-ring clearly satisfies (i)-(iii)).

Thus $$\mathcal{D}=\mathcal{R},$$ as claimed.$$\quad \square$$

Definition 2

Given a set family $$\mathcal{M},$$ we define (following Hausdorff)

(a) $$\mathcal{M}_{\sigma}=\{$$all countable unions of $$\mathcal{M}$$-sets$$\}$$ (cf. $$\mathcal{C}_{\sigma}$$ in §2);

(b) $$\mathcal{M}_{\delta}=\{$$all countable intersections of $$\mathcal{M}$$-sets$$\}$$.

We use $$\mathcal{M}_{s}$$ and $$\mathcal{M}_{d}$$ for similar notions, with "countable" replaced by "finite."

Clearly,

$\mathcal{M}_{\sigma} \supseteq \mathcal{M}_{s} \supseteq \mathcal{M}$

and

$\mathcal{M}_{\delta} \supseteq \mathcal{M}_{d} \supseteq \mathcal{M}.$

Why?

Note 2. Observe that $$\mathcal{M}$$ is closed under finite (countable) unions iff

$\mathcal{M}=\mathcal{M}_{s}\left(\mathcal{M}=\mathcal{M}_{\sigma}\right).$
Verify! Interpret $$\mathcal{M}=\mathcal{M}_{d}\left(\mathcal{M}=\mathcal{M}_{\sigma}\right)$$ similarly.

In conclusion, we generalize Theorem 1 in §1.

Definition 3

The product

$\mathcal{M} \dot{ \times} \mathcal{N}$

of two set families $$\mathcal{M}$$ and $$\mathcal{N}$$ is the family of all sets of the form

$A \times B,$

with $$A \in \mathcal{M}$$ and $$B \in \mathcal{N}$$.

(The dot in $$\dot{ \times}$$ is to stress that $$\mathcal{M} \dot{ \times} \mathcal{N}$$ is not really a Cartesian product.)

Theorem $$\PageIndex{4}$$

If $$\mathcal{M}$$ and $$\mathcal{N}$$ are semirings, so is $$\mathcal{M} \dot{ \times} \mathcal{N}$$.

The proof runs along the same lines as that of Theorem 1 in §1, via the set identities

$(X \times Y) \cap\left(X^{\prime} \times Y^{\prime}\right)=\left(X \cap X^{\prime}\right) \times\left(Y \cap Y^{\prime}\right)$

and

$(X \times Y)-\left(X^{\prime} \times Y^{\prime}\right)=\left[\left(X-X^{\prime}\right) \times Y\right] \cup\left[\left(X \cap X^{\prime}\right) \times\left(Y-Y^{\prime}\right)\right].$

Proof

The details are left to the reader.

Note 3. As every ring is a semiring (Corollary 1), the product of two rings (fields, $$\sigma$$-rings, $$\sigma$$-fields) is a semiring. However, see Problem 6 below.