Dear Sumant,

Its 19hrs and I am in my office. I did some stretches as a part of yoga and meditation for over 15 minutes. The point is I want to study Analysis and I think staying in office is better than going home at this point. Anyway the purpose of this blog will also be to put my thoughts. So without wasting any further time lets get down to business.

Q1. What is Algebra ?

An algebra *“a”* over a given set X is a non empty collection of subsets of X and has the following two properties.

- When set A and B belongs to this collection then so is their union.
- When set A belongs to this collection then so is A complement.

Note: here X is a set of real numbers

Q2. What is Sigma Algebra? Is it same as Borel Field ?

An algebra *“a”* is called Sigma Algebra if every union of countable collection of sets in *“a”* is again in *“a”*. Yes Sigma Algebra is same as Borel Field.

Remark : Note the way Sigma Algebra is defined. It says that every union of countable collection of sets in *“a”* is in *“a”*.

Q3. What is about smallest Sigma Algebra concept ?

First there is a non empty set X. Then we create all its subsets i.e Power Set P(X). Let E be a non-empty subset of P(X). Now consider all sigma algebras that contain E and we say that this set is non empty because there is always at least one algebra (E,E’,X,phi). Take the intersection over all sigma algebra, which contain set E. The claim is the resulting set is also a sigma algebra.

Q4. How will you show that this resulting Sigma Algebra is unique?

Let M(E) denote the intersection of all Sigma Algebras. Let us assume there is another Sigma Algebra N(E) smaller than M(E). Therefore N(E) has to be inside M(E) ?? It’s because of closure property of sigma algebra, since M(E) contains E and N(E) also claims to contain E and being smaller than M it has to be a subset of M(E). Thus N(E) is a subset of M(E). However M(E) comes from the intersection of all algebras containing E. Therefore M(E) is a subset of all algebras including N(E). Thus we proved both M(E) and N(E) being subsets of each other. Hence M(E)=N(E) and therefore M(E) is unique.

Q5. M(E) is called ?

M(E) is called Sigma Algebra generated over X by E.

Q6. What are the two sets always present in any Sigma Algebra ?

The Empty Set and Set X itself.

Q7. What do you call the algebra formed by Empty Set and Set X itself ?

Trivial Sigma Algebra.

Q8. What is Borel Sigma Algebra ?

The Borel Sigma Algebra is the Sigma Algebra generated by all open sets.

Q9. What are the four equivalent definitions of Borel Sigma Algebra ?

B: Collection of all open sets

B1: Collection of all closed sets

B2: Collection of all open intervals

B3: Collection of all closed intervals

Q10. How will you prove B1 from B ?

Let A belongs to B1 ie A is a closed set => A’ is open => A’ belongs to B (as A’ is open) => (A’)’ =A belongs to B (as B is sigma algebra, so it contains complements). Thus we showed that any element of B1 also belongs to B => B1 is a subset of B.

Conversely let U belongs to B (i.e U is open) => U’ is closed => U’ belongs to B1 (as U’ is closed) => (U’)’ = U belongs to B1 also as B1 is a sigma algebra. Thus again we started with a random element of B and showed it to be an element of B1 ie B is a subset of B1 and hence B=B1.

Q11. How will you prove B2 and B are equal ?

First B2 is a collection of all open intervals and since the union of open intervals is going to be an open set. B2 is a subset of B. Let us take any element U from B. Now U is an open set so it can be written as countable union of open intervals but then it becomes just a element of B2. Thus every element of B is an element of B2. Hence B2 = B.

Q12. That means we have now B = B1 and B = B2

ð B = B1 = B2.

Q13. What will you show B3 equal to?

We see that its easy to show B3 equal to B1. Now B1 is a collection of all close intervals. But we only can say about intersection of close intervals is close set. Take A and element of B3, thus A is closed => A’ is open. Now A’ = union of countable open sets. => A = intersection of complement of all these sets (by DeMorgan’s Law) and then each of these open intervals can be written as semi open union of close intervals. Let me illustrate this with a fig : if (a,b) interval is denoted by the following fig

----------------------(---------)-----------------

a b

Then (a,b)’ is denoted by

(-inf,a]U[b,+inf) as below

(-inf------------------] [------------------+inf)

a b

now the –inf and +inf can be replaced by

U(n=1 to inf ) [-n,a]U[b,n] and run n from 1 to infinity. (Note U here is Union). So we now have closed sets and intersection of any number of closed sets is closed. Thus we started from a closed set and showed it to be equal to countable number of closed intervals. Hence B1 is subset of B3.

While writing this I did go to Jimmy John’s to grab a Vegetarian Sandwich.

That’s for today.

Sumant Sumant