# 7.3: More on Set Families

- Page ID
- 19206

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.

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 \(\mathcal{P}(S).\)

(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).

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.,

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

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.

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.

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.

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.\(\quad \square\)

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.

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.)

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.