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Smallest Sigma Algebra ... Axler, Example 2.28 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Sheldon Axler's book: Measure, Integration & Real Analysis ... and I am focused on Chapter 2: Measures ...

I need help in order to make a meaningful start on verifying the first part of Axler, Example 28 ...

The relevant text reads as follows:





Axler - Borel Subsets of R ... including Example 2.28 .png




Can someone please help me to make a meaningful start on verifying Example 2,28 ... that is, to show that the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is the set of all subsets $E$ of $X$ such that $E$ is countable or $X \setminus E$ is countable ... ...



Help will be much appreciated ...

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
Can someone please help me to make a meaningful start on verifying Example 2,28 ... that is, to show that the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is the set of all subsets $E$ of $X$ such that $E$ is countable or $X \setminus E$ is countable ... ...
Let $\mathcal{S}$ be be set of all subsets of $X$ that are either countable or co-countable (where "countable" is understood to include finite or empty, and "co-countable" means having a countable complement). Then $\mathcal{S}$ is a $\sigma$-algebra. To see that, notice that it certainly contains the empty set and is closed under complementation. Also every subset of a countable set is countable, and (by complementation) every superset of a co-countable set is co-countable. Now suppose that $E_1,E_2,\ldots$ is a sequence of sets in $\mathcal{S}$. If any one of those sets is co-countable then so is the union $\bigcup E_n$. Alternatively, if they are all countable then so is the union. That shows that $\mathcal{S}$ is closed under countable unions and is therefore a $\sigma$-algebra.

But it is easy to see that any $\sigma$-algebra $\mathcal{S}$ that contains $\mathcal{A}$ must contain all the countable subsets of $X$ and their complements. So it contains $\mathcal{S}$. Therefore $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Let $\mathcal{S}$ be be set of all subsets of $X$ that are either countable or co-countable (where "countable" is understood to include finite or empty, and "co-countable" means having a countable complement). Then $\mathcal{S}$ is a $\sigma$-algebra. To see that, notice that it certainly contains the empty set and is closed under complementation. Also every subset of a countable set is countable, and (by complementation) every superset of a co-countable set is co-countable. Now suppose that $E_1,E_2,\ldots$ is a sequence of sets in $\mathcal{S}$. If any one of those sets is co-countable then so is the union $\bigcup E_n$. Alternatively, if they are all countable then so is the union. That shows that $\mathcal{S}$ is closed under countable unions and is therefore a $\sigma$-algebra.

But it is easy to see that any $\sigma$-algebra $\mathcal{S}$ that contains $\mathcal{A}$ must contain all the countable subsets of $X$ and their complements. So it contains $\mathcal{S}$. Therefore $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$.

Thanks Opalg ...

Still reflecting on what you have written ...

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Let $\mathcal{S}$ be be set of all subsets of $X$ that are either countable or co-countable (where "countable" is understood to include finite or empty, and "co-countable" means having a countable complement). Then $\mathcal{S}$ is a $\sigma$-algebra. To see that, notice that it certainly contains the empty set and is closed under complementation. Also every subset of a countable set is countable, and (by complementation) every superset of a co-countable set is co-countable. Now suppose that $E_1,E_2,\ldots$ is a sequence of sets in $\mathcal{S}$. If any one of those sets is co-countable then so is the union $\bigcup E_n$. Alternatively, if they are all countable then so is the union. That shows that $\mathcal{S}$ is closed under countable unions and is therefore a $\sigma$-algebra.

But it is easy to see that any $\sigma$-algebra $\mathcal{S}$ that contains $\mathcal{A}$ must contain all the countable subsets of $X$ and their complements. So it contains $\mathcal{S}$. Therefore $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$.

Thanks again, Opalg ...

You write:

" ... ... But it is easy to see that any $\sigma$-algebra $\mathcal{S}$ that contains $\mathcal{A}$ must contain all the countable subsets of $X$ and their complements. So it contains $\mathcal{S}$. Therefore $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$. ... "

I am finding it difficult to see exactly why $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$. ...

Can you elaborate/explain further ... ?

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
The structure of the argument consists of these two parts:
1) $\mathcal{S}$ is a $\sigma$-algebra containing $\mathcal{A}$;
2) Every $\sigma$-algebra that contains $\mathcal{A}$ must contain $\mathcal{S}$.
Those two facts together say that $\mathcal{S}$ is the smallest $\sigma$-algebra containing $\mathcal{A}$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks Opalg ....

Peter