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

In summary: S}##, then ##E:=\bigcup_n E_n## is also in ##\mathcal{S}##.Let us show this. Supose that ##E_n## is countable for all ##n##, then ##E## is also countable since countable unions of countable sets remain countable. In that case we are done. In the other case, there exists ##m## such that ##E_m^c## is countable! But then ##E^c = \bigcap_n E_n^c \subseteq E_m^c## is countable so again ##E \in \mathcal{S}##. The other axioms of ##
  • #1
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I need help in order to make a meaningful start on verifying the first part of Axler, Example 28 ...
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
 
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  • #2
You should show an attempt.

Can you at least show that ##\mathcal{S}:=\{E \subseteq X: E \mathrm{\ countable \ or \ E^c \ countable}\}## is a ##\sigma##-algebra containing ##\mathcal{A}##? Why is it the smallest?
 
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  • #3
Math_QED said:
You should show an attempt.

Can you at least show that ##\mathcal{S}:=\{E \subseteq X: |E| < \infty \lor |E^c| < \infty\}## is a ##\sigma##-algebra containing ##\mathcal{A}##? Why is it the smallest?
Given that ##X## is a set and ##\mathcal{A} = \{ \{ x \} : x \in X \}## ... ...

Require to demonstrate that ##\mathcal{S}:=\{E \subseteq X: |E| < \infty \lor |E^c| < \infty\}## is a ##\sigma##-algebra containing ##\mathcal{A}##Proof:

First show that ##\mathcal{S}## contains ##\mathcal{A}## ...

Now ##E \subseteq \mathcal{A} \Longrightarrow E = \{ x \}## for some ##x \in X## ...

... ##\Longrightarrow \mid E \mid = 0 \lt \infty##

... ##\Longrightarrow E \subseteq \mathcal{S}##
Now ... show ##\mathcal{S}## is a ##\sigma##-algebra ...

##\bullet## Require ##\emptyset \in \mathcal{S}## ...

... suppose ##E = \emptyset## ...

... then ##\mid E \mid = 0 \lt \infty## ...

... so ##\emptyset \in \mathcal{S}## ...##\bullet## Require that if ##E \in \mathcal{S}## then ##E^c \in \mathcal{S}## ...

Suppose ##E \in \mathcal{S}## ...

Then ##E^c \in \mathcal{S}## ... since ##\mid E^c \mid = 0 \lt \infty## or ##\mid (E^c)^c \mid = \mid E \mid \lt \infty## ... one of which is true since ##E \in \mathcal{S}## ...

... so ##E \in \mathcal{S}## then ##E^c \in \mathcal{S}## ...
##\bullet## Require that if ##E_1, E_2##, ... ... is a sequence of elements of ##\mathcal{S}## ... then ##\bigcup_{ k = 1 }^{ \infty } \in \mathcal{S}## ...

Assume that ##E_1, E_2##, ... ... is a sequence of elements of ##\mathcal{S}## ...

Then ##\mid E_1 \mid = n_1 \lt \infty## ... or ... ##\mid E_1^c \mid = m_1 \lt \infty## ... ...

... and ##\mid E_2 \mid = n_2 \lt \infty## ... or ... ##\mid E_2^c \mid = m_2 \lt \infty## ... ...

.. and so on ...

... But how to proceed from here ...?

Can you help further ...?

Peter

===========================================================================

My thoughts on proceeding ...We could divide ##E_1, E_2##, ... ... into two subsequences ##E_{k_1}, E_{k_2}##, ... and ##E_{h_1}, E_{h_2}, ...## where the subsequence ##E_{k_1}, E_{k_2}##, ... is composed of all those elements of ##\mathcal{S}## ... where ##\mid E_k \mid \lt \infty## ... and where the subsequence ##E_{h_1}, E_{h_2}##, ... is composed of all those elements of ##\mathcal{S}## ... where ##\mid E_k^c \mid \lt \infty## ... and try to show using countable additivity that the unions of the elements of both subsequences are less than infinity and try to relate these results to the union of the elements of E_1, E_2, ... ...
 
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  • #4
I'm very sorry. I made a mistake in my definition of ##\mathcal{S}## in post #2 (now fixed), which is probably why your attempt doesn't work because my definition of ##\mathcal{S}## contained a mistake. I guess that happens when I give hints without writing down anything on paper ;)

But as you observed correctly, the crux is to show that if ##(E_n)_n## is a sequence in ##\mathcal{S}##, then ##E:=\bigcup_n E_n## is also in ##\mathcal{S}##.

Let us show this. Supose that ##E_n## is countable for all ##n##, then ##E## is also countable since countable unions of countable sets remain countable. In that case we are done. In the other case, there exists ##m## such that ##E_m^c## is countable! But then ##E^c = \bigcap_n E_n^c \subseteq E_m^c## is countable so again ##E \in \mathcal{S}##. The other axioms of ##\sigma##-algebra are trivially satisfied and I won't write them down anymore.

Hence, we have established that ##\mathcal{S}## is a ##\sigma##-algebra containing ##\mathcal{A}##. This implies that the ##\sigma##-algebra generated by ##\mathcal{A}##, write it as ##\mathcal{M}(\mathcal{A})##, is contained in ##\mathcal{S}##, that is ##\mathcal{M}(\mathcal{A}) \subseteq \mathcal{S}## (why?).

Now, you also have to show that the other inclusion ##\mathcal{S}\subseteq \mathcal{M}(\mathcal{A})## holds. Can you see why that is true?
 
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  • #5
Math_QED said:
I'm very sorry. I made a mistake in my definition of ##\mathcal{S}## in post #2 (now fixed), which is probably why your attempt doesn't work because my definition of ##\mathcal{S}## contained a mistake. I guess that happens when I give hints without writing down anything on paper ;)

But as you observed correctly, the crux is to show that if ##(E_n)_n## is a sequence in ##\mathcal{S}##, then ##E:=\bigcup_n E_n## is also in ##\mathcal{S}##.

Let us show this. Supose that ##E_n## is countable for all ##n##, then ##E## is also countable since countable unions of countable sets remain countable. In that case we are done. In the other case, there exists ##m## such that ##E_m^c## is countable! But then ##E^c = \bigcap_n E_n^c \subseteq E_m^c## is countable so again ##E \in \mathcal{S}##. The other axioms of ##\sigma##-algebra are trivially satisfied and I won't write them down anymore.

Hence, we have established that ##\mathcal{S}## is a ##\sigma##-algebra containing ##\mathcal{A}##. This implies that the ##\sigma##-algebra generated by ##\mathcal{A}##, write it as ##\mathcal{M}(\mathcal{A})##, is contained in ##\mathcal{S}##, that is ##\mathcal{M}(\mathcal{A}) \subseteq \mathcal{S}## (why?).

Now, you also have to show that the other inclusion ##\mathcal{S}\subseteq \mathcal{M}(\mathcal{A})## holds. Can you see why that is true?
No problem ...

I have to say your help has been invaluable to me ...

Reflecting on what you have written ...

Thanks again ...

Peter
 
  • #6
Math_QED said:
I'm very sorry. I made a mistake in my definition of ##\mathcal{S}## in post #2 (now fixed), which is probably why your attempt doesn't work because my definition of ##\mathcal{S}## contained a mistake. I guess that happens when I give hints without writing down anything on paper ;)

But as you observed correctly, the crux is to show that if ##(E_n)_n## is a sequence in ##\mathcal{S}##, then ##E:=\bigcup_n E_n## is also in ##\mathcal{S}##.

Let us show this. Supose that ##E_n## is countable for all ##n##, then ##E## is also countable since countable unions of countable sets remain countable. In that case we are done. In the other case, there exists ##m## such that ##E_m^c## is countable! But then ##E^c = \bigcap_n E_n^c \subseteq E_m^c## is countable so again ##E \in \mathcal{S}##. The other axioms of ##\sigma##-algebra are trivially satisfied and I won't write them down anymore.

Hence, we have established that ##\mathcal{S}## is a ##\sigma##-algebra containing ##\mathcal{A}##. This implies that the ##\sigma##-algebra generated by ##\mathcal{A}##, write it as ##\mathcal{M}(\mathcal{A})##, is contained in ##\mathcal{S}##, that is ##\mathcal{M}(\mathcal{A}) \subseteq \mathcal{S}## (why?).

Now, you also have to show that the other inclusion ##\mathcal{S}\subseteq \mathcal{M}(\mathcal{A})## holds. Can you see why that is true?
We have that ##X## is a set and ##\mathcal{A}## is the set of subsets of ##X## that contain exactly one element of ##X##; that is

##\mathcal{A} = \{ \{ x \} : x \in X \}##

Now the ##\sigma##-algebra of ##X## generated by ##\mathcal{A}##, denoted ##\mathcal{M}(\mathcal{A})##, is the intersection of all ##\sigma##-algebras of ##X## that contain ##\mathcal{A}##.

Therefore, given ##\mathcal{S}## is a ##\sigma##-algebra of ##X## that contains ##\mathcal{A}##, we have that ##\mathcal{M}(\mathcal{A}) \subseteq \mathcal{S}## ...

But ... ##\mathcal{M}(\mathcal{A})## is a ##\sigma##-algebra of X that contains ##\mathcal{A}## ... and so contains all the countable subsets of ##X## and their complements ... so ##\mathcal{S}\subseteq \mathcal{M}(\mathcal{A})##...is that correct ... ?

Peter
 

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

1. What is the "smallest sigma algebra"?

The smallest sigma algebra, also known as the trivial sigma algebra, is the smallest possible collection of subsets of a given set that still satisfies the properties of a sigma algebra. It contains only the empty set and the entire set itself.

2. How is the smallest sigma algebra related to measure theory?

The smallest sigma algebra is an important concept in measure theory because it is the foundation for constructing measures on a given set. By defining a measure on the smallest sigma algebra, we can extend it to a larger sigma algebra and ultimately to the entire set.

3. What is the significance of Example 2.28 in Axler's "Measure, Integration & Real Analysis"?

Example 2.28 in Axler's book demonstrates the construction of a measure on the smallest sigma algebra for a given set. This example is important because it illustrates how measures can be defined on more complex sets, such as the real numbers, by starting with a simple sigma algebra.

4. How does the smallest sigma algebra differ from other sigma algebras?

The smallest sigma algebra differs from other sigma algebras in that it contains the fewest possible subsets of a given set while still satisfying the properties of a sigma algebra. Other sigma algebras may contain more subsets, which can lead to more complex measures.

5. Can the smallest sigma algebra be used to define a measure on any set?

Yes, the smallest sigma algebra can be used to define a measure on any set. However, this measure may not be very interesting or useful for more complex sets, as it only considers the empty set and the entire set itself. It is more commonly used as a starting point for constructing measures on larger sigma algebras.

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