Hello everybody, know this lecture is about the monotone class theorem.
You must read the definition and understand what it means. The theorem says that the smallest matter in class containing a field.
Coincides with the Sigma field generated by a. Remember, the definitions first of all, not every field is a Sigma field. There was an example, so the Sigma Field is in some sense bigger.
A field must be extended.
To become a Sigma field. So in this stadium.
Gives you a tool how to do it.
We want us to construct.
Or the smallest mother don't class and it will coincide with the smallest Sigma Field which contains a generated by a means or the smallest Sigma Field which contains a.
The proof is equipped with asterisk means it is not necessary for this module, but it would be good if you understand it and they will go through the proof very quickly. The prophets in the election notices or other detailed.
So first of all.
Any Sigma Field is the monotone class.
Check on this at home.
JA is the smallest monotone class, so we have J subset of FA.
I will write down the key expressions.
In this weird file. So let us show that.
JA is a field.
The first axiom of a field.
Says that Omega must belong to J and that is absolutely obvious. Omega belongs to A.
Becausw A is a field.
And A is subset OJ so Omega belongs to JA. Now that is trivial.
So next, which is less trivial, we're going to show that.
All that.
If a set belongs to Jay, then complement belongs to Jay to do this.
We introduce.
The following set G.
Which contains all sets such that complement is in G.
So firstly.
Or the initial field is subset.
Becausw
calligraphic a is a field.
If a belongs to calligraphic a, then complement belongs to the field a.
So.
So.
Telegraphically initial field is subset of G.
Secondly,
G is a monotone class and the proof is a rather detailed presented in this.
Two paragraphs if we have increasing sequence of subsets from G.
Then compliments are decreasing.
And.
The intersection of complements belongs to J.
Becausw JA is a monotone class, so we see that.
The complement of the union.
Belongs to Jay.
And what is G?
G is collection of search such that complement belongs to GE. It means that.
The Union.
Belongs to G.
So we proved that if we have increasing sequence of subsets, then the union belongs to G.
Absolutely similarly.
If we have decreasing sequence of subsets then.
Compliments will be increasing.
And.
Well, they.
Union or the compliments belongs to JA means that the complement of the intersection belongs to J, meaning that the intersection belongs to G.
So G is a monotone class.
But
JA.
Is this smallest monotone class which contains a?
So we have.
We have
we have.
JA.
Again, the problem.
OK, one moment.
So since the A is the smallest monotone class containing a approved that.
JA.
Is subset of G.
What does it mean? It means that.
If.
Be.
Belongs to Jay.
Or then.
Then be.
Belongs to G.
And it follows that.
It follows all that.
Big compliment belongs to Jay.
According to the definition of G.
So it means.
Then the second axiom of a field.
Is.
So just right.
We need to.
Proof away.
Lust.
Axiom of.
A field.
So let us show that.
Let us show that.
For any.
Since he be.
From
from.
JA.
Their union.
Belongs to Jay.
So this proof follows the same pattern as above.
So we introduce the set.
Which satisfies what we want for the first case.
For their first case.
A.
Belongs to initial field, simple case.
So the field.
A.
Is.
Subset of G prime.
Well, I need to what is written if B belongs to the field then the union belongs to the field so remember.
A.
Belongs to the field.
First step.
And this set the prime is monotone class absolutely similarly to a water was presented.
2 minutes ago consider increasing sequence.
Consider the unions and so on.
We always use the property of the Monotone Class. GA is monotone class.
And again in decreasing sequence.
So G prime.
Is a monotone class.
And G Prime contains a like previously.
JA is the smallest monotone class which contains a, so GA is subset of G Prime.
And.
So far we have proved that if.
Set a billong Stew with a field.
Then
for an arbitrary be.
A union B.
Put in the battery. Be from GE.
The union belongs to GA.
So that is good, but not enough. We must consider arbitrary.
Arbitrary from GA.
Rich
remember is bigger than he.
Like previously
we introduce this set.
Collection of.
Subsets she.
Search event
a Union C belongs to J.
And very similar argument to the presented above shows of that or this said.
This said, which has known attention but it.
Includes.
The initial field.
And it is monotone class, so it is monotone class.
Which contains a.
Like previously GA.
Is the smallest monotone class which contains a, so it is subset of this.
And.
No. Well, they said capital A. It was arbitrary.
As a result of this observation, is the result of this inclusion, where we see that one arbitrary fixed.
Capital A.
And she.
Their union belongs to J, so JA is a field.
Well, this is.
Well, this is established.
We have proved that.
The complement belongs to the end.
Union of find it. Find it union.
Of two sets, for example, belongs to J, so that is a field, but that is not the end. We need to show that the A is a Sigma Field.
And that is rather simple, all radio.
So Omega belongs to Jay or was.
Show long ago ago.
If we have if we take a compliment then it belongs to Jay Becausw. Jay is a field and the third most important axiom of a Sigma field is about.
Infinite collection of subsets.
So now the subsets of this type.
Are increasing every next one is bigger.
And all of them are in JA Becausw Jay-Z field, so every.
Find it union of this or that and further.
Are in J because J is a field field is about finite unions.
But the total union of this increasing sequence of subsets belongs to the a.
Becaused J is monotone class.
So.
So if we have.
Increasing whoa, arbitrary sequence of subsets.
Well, then there from JA.
Well then there total union belongs to J. That is the last important axiom of a Sigma Field.
So.
Or what do we know now we know that.
We all know that.
Gee.
He
is.
Is Sigma field.
And the A of course.
Contains a.
Initial field is the subset.
So.
FA by definition.
Is the minimal.
Sigma field.
Containing
the initial field a.
What does it mean?
It means that.
Well, this minimal Sigma Field.
Cannot be bigger.
All that then GA.
So J is a Sigma Field.
Containing a FA is the minimal Sigma field, so halfway.
Is.
Subset of GE.
At the very beginning.
At the very beginning it was said that JA is subset of FA.
At the beginning.
So it means that.
JA.
Equals.
OK.
Well, that is the end of the proof.
So in the lecture notes, The Proof is detailed enough. Hopefully my explanations are helpful.
It would be good if you understand the proof, but as I said.
All the paragraphs equipped with asterisk are not necessary for this module.
Thank you for your attention and good luck.