Good evening everybody. We are starting the real measure theory and the very important object here is Sigma Field. The definition is on page 8 of lecture notes.
A family of Sigma fields of a universal set Omega is Sigma field. If three axioms are satisfied.
The whole set belongs to F.
If I said belongs to air for, then compliment bill belongs to Earth and if you have a sequence of subsets or then their total union belongs to Sigma field as well.
Already here you can.
Practice for better understanding.
For example.
If.
If two cells belong to Sigma Field.
First of all, of their union of course belongs to Sigma Field Becausw Finite Union.
Is.
Is a special case of infinite union.
Well, that is absolutely clear.
Moreover
the difference for example.
Again belongs to Sigma Field Becausw.
It has the form a intersection complement.
After that you can apply the Morgan law.
Which is will be essentially used in this module.
But his eltingen a complement union.
Be compliment.
And this belongs to.
F.
Becausw
a compliment belongs to F.
According to Axiom 2.
B belongs to Earth.
Because it is given.
The Union A&B belongs to Earth because of action three as explained 2 minutes ago.
And this bracket complement belongs to.
F becausw of Axiom 2 again.
In the first assignment, that is another exercise.
Show that.
Show that the.
Total intersection.
Bilong still F.
Of course, in case.
Load them.
Belong to Sigma Field.
Simple exercise again. You can apply the Des Morgan Follower.
Here are a series of examples we have fine. It said one can consider collection of all subsets. Of course it will be Sigma Field Becausw. All the three axioms are satisfied.
Obviously.
But this.
Collection empty and Omega is also Sigma Field.
Becausw
only gonna be a long story if.
Compliment of empty is Omega. Complement of Omega is empty, so axiom on two is satisfied.
And if you take a infinite series of subsets from F2, then the result will be definitely either empty set. If you take a union of empty sets, or it can be Omega if only again.
Hey peers in this union.
So another Sigma field empty A&BCD.
Again, compliment to a is busy dear compliment of the city is a if you take unions of any sales from this list the result will be again in this list.
For, for instance, Union A and empty is a union, ABCD is Omega, it is again in the list.
So this is the minimal Sigma Field which contains Singleton.
Well, the element BCD must be included.
Because it is complement. Remember axiom 2.
If Omega belongs to.
If it is necessary then complement of Omega is empty. It must be in the list.
So the empty set and the full set Omega are always included in the Sigma Field in everyone Sigma Field.
And if you have Singleton A, then the complement must be included.
So as a result, you can guess that if you have.
Find it said or then everyone Sigma field has.
The number of elements of which is power of two.
It can be.
16 elements here, all subsets can be 2 elements, can be 4 elements.
Can be.
It elements if you think a little.
So here is another example.
Want to?
345678 elements in this Sigma field check at home accurately that F5 is also Sigma Field.
So if you take Cartesian product of two sets, 1 two ABC.
Then of course, one can construct many Sigma fields for the Cartesian product.
And the Sigma Field presented in these lines.
Is quote generated by the 2nd component?
So.
It means that.
You construct Sigma Field of all subsets of Omega 2.
ABC all pairs and the triple. So here all subsets of Omega 2.
And everyone element a or B or she is replaced with the player.
So this Singleton A.
Is replaced with Pier 1828. This Singleton B is replaced with.
Want you to be?
So the point is that elements.
12 appear always together we have been.
And both want to.
We have A&B and again a 12 is present everywhere.
Just read attentively this example.
So this.
Example #3.
Is a bit more challenging, will take.
Infinite set of nature on numbers.
And of course one can construct Sigma Field of all subsets.
But here.
Here we have a special Sigma Field.
Which is in fact.
Connected to the Sigma field generated by even numbers, so we take.
All possible.
Find it or infinite collections of even numbers, for example 2.
4 eight 1214.
And so on. Huge collection of all subsets of even numbers.
And.
Except for that or we.
Consider everyone subset of even numbers.
Complemented with the full collection of odd numbers.
So for example.
If you take empty set, then.
Komplement with.
You odd numbers will be collection of old numbers. If you take only element 2.
Like this?
Then we have 1235 and all further odd numbers.
So one can take this set and.
Complemented with.
Full collection of odd numbers 1357.
The next day will be 89.
11 and so on.
So this is again Sigma Field.
You can easily check.
All there.
Definitions.
So the full collection.
Omega, which is an.
Is included becausw so we can take the.
Full collection of even numbers plus the full collection of odd numbers as a result of have Omega complement is empty.
Empty cities.
Subset of collection of even numbers, so all axioms are satisfied.
OK, next definition four is rather formal.
The minimal Sigma Field which contains open intervals on the straight line is called Borel Sigma Field.
So we will discuss this.
A bit later, and Borel Sigma Field to bill again, they appear very often in this module.
So next definition is a field.
Compare this definition of with Sigma Field or the differences.
That
In definition of a field.
If two sets belong to.
Field then the union belongs to the field. By induction, any finite union belongs to field.
But in the definition of Sigma Field.
Infinite.
Countable.
Union of subsets must also belong to the Sigma Field.
So of course every.
Every field.
Is a Sigma field becausw union belongs to Sigma field, but not every.
Field is a Sigma Field.
So here are exercises.
For you.
Independent reading and understanding.
And I will.
Consider the following example, which shows that the notion of field and Sigma Field are really different. Of course they coincide if they said Omega is fine it.
If Omega is fine it then every field is automatically Sigma field.
And of course every Sigma field is a field that is.
Absolutely general.
Statement.
OK example showing that.
Are there can exist fields of which are not Sigma fields?
So consider for instance collection of nature numbers.
And.
Introduce the following field.
Calligraphic a
so subsets of Naturile numbers belong to.
Calligraphic a
in two cases. If a is finite or empty.
And in case.
Complement is finite or empty.
So two types of subsets.
And here is.
For the simple proof.
Showing that this collection of subsets calligraphic a is a really a field.
If two sets.
Build onto this.
Collection or then the union.
Belongs to A and difference belong to a becausw. All these sets are finite union of finite sets.
And difference of finite sets.
Find it.
And those are first type.
Elements of calligraphic a.
If we have two.
Elements of the second type.
Then again, here is the proof that their union.
Belongs to A and difference belongs to A.
So again, the Morgan allows are in use. The Union is.
The complement of the intersection of compliments. So this is fine. It second is finite intersection is finite.
And.
Compliment to find it.
Belongs to a.
So you can check.
In dependently
the third case, if you have two subsets of different types, type one and Type 2, then again the union belongs to a.
Becausw compliment is fine it.
And that difference Becausw belongs to a becausw.
Becausw it is again fine it so the point is to pass to find it sets using the de Morgan laws.
So this.
On this.
Family A.
Is a field.
But not a Sigma Field.
Becausw
if for example of it take.
A1 being.
Who won?
8 two being 3.
A3 being.
5.
And so on and then.
Their total union.
Infinite Union.
What is it?
This collection of odd numbers.
It is not first type said not find it.
And its complement.
Even numbers.
Not fine it.
Means that.
The total union.
Does not belong to a.
So on this example.
In this example of this era that this.
Welldefined
collection
of subsets of natural numbers is a field, but not a Sigma Field.
So next theorem looks very general. It is very general when the proof is or rather simple.
Intersection.
Of any family of Sigma Fields.
Or an arbitrary universal set is a Sigma Field.
So the proof is.
Presented only 6 lines.
So suppose all of this F Alpha are Sigma fields and Alpha is from the.
Babytree index set.
So introduce total intersection of such.
Sigma fields.
Only God belongs to.
Every one Sigma field for all Alpha.
So Omega belongs to this intersection.
So Omega belongs to her. I am checking on the 1st Axiom of a Sigma Field. I want to show that.
This calligraphic F total intersection is a Sigma Field. The first axiom is satisfied.
If we have.
Is set from calligraphic F.
When a belongs to all.
Sigma fields for all Alpha.
So it means that for any Alpha.
Complement belongs to F Alpha.
Why?
Becausw if Alpha is Sigma Field.
If a set a belongs to F, Alpha and then complement a, she belongs to her father and this is valid for all Alpha.
So the complement belongs to F becausw it belongs to.
All sets of Alpha.
And similar reasoning.
You have a sequence of subsets which belongs to if then am belongs to to F Alpha for all N and for all Alpha.
So take.
But Ticular Alpha.
And we see that.
For particular Alpha or this union belongs to Earth Alpha and this is valid for arbitrary Alpha.
So this union belongs to F Alpha for arbitrary Alpha, so this union belongs to F because F is.
Intersection of all of Alpha.
So here is illustration to this theorem.
Illustration using example of finite set Omega.
After one.
Has how many?
812345678 elements.
And F2 has.
8 elements but different.
You see the Singleton she is absent in the first Sigma field that is present in the second signature.
A&BAB
is present in the 2nd and in the 1st.
A separately is present in the first, but not in this second.
So intersection of this Sigma fields has what?
Empty set which is in common as usual.
Set a B is present in the first set in the first Sigma field and in the second Sigma Field.
CD is present in the first Sigma field and in the second Sigma Field.
And the total set Omega is present everywhere as usual.
So intersection of these two Sigma Fields has four elements.
And that is of course Sigma Field.
Check the definitions. Please remember the definitions.
And.
Next definition.
So if we have.
Collection family of subsets.
When Alpha is indexes from arbitrary index set.
Then
and then one can construct many Sigma fields.
Which contain.
All.
Subsets.
8A.
So there are many Sigma Fields which contain all subsets a Alpha.
But if we take intersection of all those Sigma fields and then.
The result will be again Sigma Field.
Intersection of Sigma Fields will be Sigma Field.
And it will be cold generated by this.
Collection of subsets.
So look at the previous examples.
F1 is generated by two.
Subsets A&AB
Becausw.
If we have a beer.
Well then, are they?
Sigma field must include the complement, maybe right CD.
If we have element a or, then the complement BCD must be in the list.
If we have.
A&CD
no then this.
AAA CD
must be included, so this is the minimal Sigma Field which contains.
Set a one and set a 2.
Welcome construct many other Sigma Fields which contain this to cells.
But F1 is the minimal Sigma Field which contains.
A1 and a two capital E one capital A2 similarly.
Oh there Sigma Field F2?
Is generated by.
Subsets B1B2
check yourself.
If we have.
AB.
In the list.
And then they compliment.
She must be in the list.
You could have.
Subset B2.
In the list, let us see Singleton, then complement Abd must be in the list.
The Union.
Of a B&C?
Must be in the list.
So everyone.
Sigma field over a finite universal set contains power of two elements. So when you construct Sigma fields.
Generated minimal Sigma fields then please count.
Minimal Sigma Field which contains BO1B2.
In fact, hazz.
8.
Subsets 12345678.
When you deal with Sigma fields over finite sets, then every time calculate the number of subsets in that Sigma Field.
So next.
Theorem next lecture will be about monotone class theorem.
You're welcome to read definition and theory.
But as they said on the 1st lecture, are they?
Material equipped with asterisk is not necessary for this module.
But
it will be good if you understand so on the next lecture.
I synchronous lecture. I will discuss this proof.
That is all for today.
Thank you for their your questions.