**Reference: Introductory Real Analysis, Kolmogorov and Fomin. Dover Publications. Translated and edited by Richard A. Silverman:**

**Problem 1:**

Let X be an uncountable set, and let be the ring consisting of all finite subsets of X and their complements. Is a -ring?

**Solution 1:**

Recall the definition: A ring of sets is called -ring if it contains the union whenever it contains the sets .

There are countably many finite subsets of X (given uncountable set) and their complements (also especially) so that the union of all these is the whole set X. Hence, X is a -ring as it is closed under countably many set unions. QED.

**Problem 2:**

Are open intervals Borel sets?

**Solution 2:**

Recall the following:

(1) Borel sets or B-sets are the subsets of the real line belonging to the minimal B-algebra generated by the set of all closed intervals .

Problem 3:But a B-algebra is also a ring and is closed under differences/complements. Open intervals are complements of closed intervals. And, they belong hence to B-algebra. Note we still have to answer the question whether open sets are Borel sets and so we have to check if they are contained in a minimal B-algebra. As per the following theorem: Given any non-empty system of sets , there is a unique irreducible (minimal) B-algebra generated by the system or the Borel closure of . So, yes indeed open sets are Borel sets. QED.

**Problem 3A:**

Let be a function defined on a set M and taking values in a set N. Let be a system of subsets of M, and let denote the system of all images of sets . Moreover, let be a system of subsets of N, and let denote the system of all preimages of sets . Prove that:

i) If is a ring, so is .

ii) If is an algebra, so is

iii) If is a B-algebra, so is

iv) where stands for a ring.

v) where is irreducible or minimum Borel algebra generated by a non-empty system of sets .

**Problem 3B:**

Prove the following:

i) If is a ring, so is .

ii) If is an algebra, so is .

iii) If is Borel-algebra, so is .

iv)

v) where is irreducible minimum Borel algebra generated by a non-empty system of sets .

**Solution 3A:**

Part 1: Given mapping , that is, , Let be subsets of N and let be subsets of M. Then, as it is given that is a ring then :

The above implies:

, that is, , that is, .

Also, so that , that is, , that is, we conclude .

Also, as is a ring; so that so that for some , which is

So, we have shown that the system of sets is closed under set union, set intersection, set difference, and hence, also set symmetric difference. Hence, it is is a ring. QED.

Part ii:

We have already shown that is a ring, now is also an algebra meaning a ring of sets with a unit element. Let it contain a unit E such that if is any set belonging to , then . so that , hence, , which in turn implies that is a unit of . Hence, is a ring of sets with a unit element, or in other words, an algebra of sets.

Part iii:

To prove that: if is a Borel algebra, so is . From parts i and ii above, is a ring with a unit or in other words, is an algebra. Moreover, if it contains whenever it contains the sets all of which are members of .

But for an arbitrary (finite or infinite) collection of sets:

Now,

hence

Consequently, , which in turn implies that is a Borel algebra also. QED.

Part iv and part v:

Using parts i and iii above, these can be very easily proved.

**Solutions 3b:**

i) Prove: If is a ring, so is . Note that , . Proof (i): As is a system of subsets of M. and it is a ring, so if , and , it implies that ; ; . We know that and . And, quite obviously, so that it is true that if is a ring, then is also a ring.

ii) Check if it is true or false: if is an algebra, so is . Argument: As is an algebra, it is already a ring. But additionally, contains an element E such that (whenever but but it need not be true that is a unit element of . So, we can say that in case is an algebra, then may or may not be an algebra. QED.

iii) Check if it is true or false: If is a Borel algebra, then so also is also a Borel algebra. Argument: As in previous case, may not have a unit element. So, need not be a Borel algebra. QED.

iv) . Check if this is true or false. Argument: Let , . By definition of a ring, , , , hence we get the following: ; and . This in turn means that , and . Hence, . Similarly, we can show that . QED.

v) . Check if this is true or false. (We will discuss this later; meanwhile, if you wish, please try as homework and let me know).

Cheers,

Nalin Pithwa.

Comments are most welcome.