Every element of an algebra generated from a semi-algebra can be written as a finite disjoint union of elements

---

Concept

If a semi-algebra is used to generate an Algebra, we can exactly express which sets belong to the algebra:

We can proof that all sets which belong to the generated algebra can be written as finite unions of elements from the semi-algebra.

(Side-note: this is not possible for a Sigma-Algebra)

Definition

Let be a semi-algebra and a generated algebra.

() If A is an element of , A can be written as a finite union of disjoint sets from the semi-algebra () The union of any finite number of sets from the semi-algebra is also an element of the generated algebra .

Proof

Let there be a finite number of sets () from the semi-algebra .

Because the sets are elements of the semi-algebra, they are also elements of the generated algebra.

Since algebras are closed under the union operation and the ‘s are elements of the generated algebra, their union also is an element of the generated algebra

To proof this direction, we must show that every element from can be written as a finite union of elements from .

To do this, we must digress a little:

Based on the semi-algebra, we can construct a new collection of sets , which contains all sets that can be written as a finite union of sets (this is our initial goal for btw):

We will now show that is indeed an algebra on . To do that, we need to proof that

2. is an algebra

Proof that

The first proposition is easy, because every element can be written as a union of only itself - which indeed is a finite union, and therefore:

Proof that is an algebra

To proof this, we must check the three algebra conditions

1. This is true, because of the first proof we know that

2. Suppose . Then we would be able to write them as two finite intersections

   And their intersection can be written as:

   Now since the semi-algebra is closed under intersection (second property see here) and , their intersection also is an element of . Since it is an element of , it is also an element of . Hence, is closed under intersections

3. To clarify, we first notice that since , we can write as a finite union with . And its complement . By the third semi-algebra property we know that each complement can also be written as a finite union of sets . This allows us to reduce the complement of to an intersection of the subsets from each of the complements :

   Since the semi-algebra is closed under intersection we find an element: This reduces the complement of to a finite union of sets from :

   That is, is a finite union of sets from semi-algebra . Since is defined to contain all sets that can be written as a finite union of sets and is such a set:

This proves, that is an algebra .

Wrapping the proof

Since is an algebra and it contains , the generated algebra must be a subset of by definition.

1. All elements of can be written as finite unions of sets from .
2. All elements of are also elements of . Therefore, all elements of can be written as finite unions

Examples

---

Appendix

• Tags:
  • Measure Theory
• Reference:
  • ; Measure Theory Lecture 2
• Related:
  • Semi-Algebra, Borel Sigma-Algebra

---